Structural Steel & Timber Design3

STRUCTURAL STEEL & TIMBER DESIGN III

STUDY GUIDE

(SSD301C)

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STUDY GUIDE

COPYRIGHT FIRST EDITION, 2001 REVISED EDITION, 2010 FLORIDA, SOUTH AFRICA

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STUDY GUIDE

STRUCTURAL STEEL & TIMBER DESIGN III STUDY GUIDE (SSD301C)

COMPILED BY: G. PARROTT MODERATED BY: (ORIGINAL MATERIAL: C.A. HOLLENBACH) REVISED BY: BD IKOTUN

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COPYRIGHT DECLARATION In accordance with the Copyright Act, 98 of 1978 no part of this material may be reproduced, published, redistributed, transmitted, screened or used in any form without prior written permission from Unisa. When materials have been used from other sources permission must be obtained directly from the original source. FIRST EDITION, 2001 REVISED EDITION, 2010 FLORIDA, SOUTH AFRICA

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STRUCTURAL STEEL & TIMBER DESIGN III (SSD301C) OBJECTIVES The objectives of this section are to introduce the student to the course material, explain what is expected of the student, and how the course will be evaluated. The study guide is aimed to direct the learner through the contents of the syllabus by defining the learning outcomes and leading the learner through each section. There are a number of typical examination questions covering each section which is intended for self evaluation. COURSE COMPOSITION: The course material is divided into two parts: MODULE 1: STRUCTURAL STEEL DESIGN This module consists of a study guide, a compulsory prescribed textbook and relevant SANS codes. The study guide for this module contains brief highlight on each topic, specifying the objectives and specific outcomes. Students are directed to the prescribed book for the details of each topic. MODULE 2: STRUCTURAL TIMBER DESIGN This module consists of study notes and compulsory prescribed relevant SANS codes. Using the course Material To successfully complete the course, the following procedure is recommended. Work methodically through each section in turn and make your own rough notes as you progress. Contact the mentor if you are unable to clarify any important points. Once you are comfortable with the section, attempt the self evaluation questions. Do not move onto the next section until you have mastered the present one. Do not attempt and submit the assignment or project until you fully understand the course materials required. 5

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MODULE 1 STRUCTURAL STEEL DESIGN

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MODULE 1 STRUCTURAL STEEL DESIGN Introduction The primary objective of this part of the course is to introduce you to the basics of designing Structural steel elements such as trusses, beams and columns. This part also covers the loading on structures including the effects o wind. You will be lead through the theory behind each section before getting into the actual design of the elements.

Course Format The prescribed text is ideally suited to the course and should be followed closely together with the code of practice when working through each section. The textbook indicates references to the code on the right hand side of the page thus: [4.5.11] to enable the student to quickly refer to the relevant parts of the code. There are self evaluation questions provided for each section which are typical examination questions for which the final answers are given (but not the worked solution) verify your solution.

Prescribed text book: •

Structural steel design to SANS10162:1-2005-1, Parrot, G.K, Shades Technical Publications. SA 2006 ISBN1-919858-13-X.

Relevant SANS codes and construction handbook. •

South African Steel Construction Handbook (Limit states design). The South African Institute of Steel Construction.

•

SANS 10162-1:2005: The structural use of steel - Part 1: Limit-state design of hotrolled steelwork. ISBN 0-626-16165-7.

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•

SANS 10160: The general procedures and loadings to be adopted in the design of buildings.

Further reading: There is a wide variety of books available for structural steel design. The students must however ensure that any book used for additional reading is based on the relevant codes of practice.

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SECTION 1.1: LOADING & LIMIT STATE DESIGN The correct determination of loads acting on a structure is obviously a critical component of the overall design process as any error in load calculations will lead to errors in the effects of load on the structure and make element strength calculation meaningless.

1.1.1 Objectives The objective of this section is to ensure that the learner is able to determine the intensity of loading on structural elements and also provide an understanding of the “limit-states” approach to structural design.

1.1.2 Specific Outcomes At the end of this section the learner will be able to make a reasonable assessment of the intensity of basic loads acting on structural elements. The learner will also have an understanding of “Limit State Design” and be able to convert the loads calculated into ultimate and serviceability limit state values.

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SECTION 1.2: ANALYSIS This is basically a revision section since most of the material covered has previously been covered in the prerequisite subjects “Applied Mechanics I” and “Theory of Structures II”. The additional work looks at the understanding of the analysis of frames specifically the second order effects resulting from sway of the structure which is a design requirement of the code of practice used.

1.2.1 Objectives The primary of this section is to ensure that the learner is able to carry out the basic analysis of structures (Statically determinate).

1.2.2 Specific Outcomes At the end of this section the learner will be able to calculate the load effects on beams of support reactions, shear forces and bending moments, and the load effects on pin-jointed trusses of axial forces (tension and compression). Given the first order analysis calculated by computer, the learner will also be able to determine the second order effect on framed structures.

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SECTION 1.3: CONNECTIONS The type of connection used can affect the analysis and strength of a structural element which makes it important to consider connection design before designing the elements themselves.

1.3.1 Objectives The objective of this section is to show the learner the different types of connections used in structural steel and to provide the necessary equations to determine their strength.

1.3.2 Specific Outcomes At the end of this section the learner will know of the different ways of connecting structural steel and be able to determine the strength of bolted and welded connections.

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SECTION 1.4: TIES & STRUTS These structural elements generally make up trusses or girders and are used extensively in practice for roofs and bridges. It is important to remember here that the strength of these members is also largely dependent on the type and strength of the connected ends.

1.4.1 Objectives The objective of this section is to derive and provide the learner with the equations necessary to determine the strength of members both in tension and compression.

1.4.2 Specific Outcomes At the end of this section the learner will be able to determine the different modes of failure and calculate the factored resistance of members subjected to a pure axial load either in tension or compression.

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SECTION 1.5: BEAMS Beams are used extensively in structural steel building frames where they provide the support structure to the floors. It is very important to clearly understand that the compression flange of a beam wants to buckle laterally, and that the type of restraint offered has a large influence on the strength of a beam.

1.5.1 Objectives The objectives of this section are to give the learner an understanding of the behaviour of beams and to provide him/her with the necessary information required to determine the strength (resistance) of hot-rolled steel beams and plate girders subjected to bending and shear effects of loading.

1.5.2 Specific Outcomes At the end of this section the learner will understand and be able to take into account the various types of restraint offered to a beam. The learner will be able to determine the factored bending resistance of hot-rolled and welded plate girder beams as well as the factored shear resistance including the provision and design of web stiffeners where required. The learner will also be able to check the serviceability limit state of deflection.

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SECTION 1.6: COLUMNS These elements are encountered often and are generally referred to as “beam-columns”. The learner will already know how to determine the factored resistance for bending and axial force independently, and will now look at the interaction of these load effects.

1.6.1 Objectives The objective of this section is to demonstrate the effects of combined stresses acting on a member and to enable the learner to assess the suitability of a given section subjected to this combined stress.

1.6.2 Specific Outcomes At the end of this section the learner will be able to assess the strength and suitability of structural steel elements that are subjected to a combination of bending moment about one or both axes together with either axial compression or tension forces.

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SELF EVALUATION QUESTIONS The questions that follow are intended to assist the learner with self-evaluation of the course material. They cover the full structural steel component of the subject and are typical questions that may be expected in a final examination.

Answers to each question are provided at the end of each section for the learner to check his/her accuracy.

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SECTION 1: LOADING & LIMIT STATE DESIGN QUESTION 1. 1 The figure below shows the evaluation of part of a building frame. The frames are spaced at 6. 5 m centres over the length of the building. All columns are 254 x 107 H-section and all beams are 356 x 171 x 67 I-sections. The structure is braced and all connections should be considered as “pinned” (i.e. beams are simply supported). The floor consists of a 150 mm thick reinforced concrete slab with applied finishes that amount to 1 kN/m2 and is to be utilised for offices containing data processing equipment. The reactions at this floor level from columns A2 and B2 are given below: Dn

Ln

Column A2

25 kN

120 kN

Column B2

25kN

85kN

Determine the ultimate reaction of the base of column: “C1”.

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QUESTION 1. 2 The figure below shows the elevation of a sign board that is to be erected on Durban’s beachfront. The sign is required to withstand a wind force with a mean return period of 25 years.

(a) Calculate the wind velocity pressure. (b) Ignoring the self-weight of the sign and posts, calculate the maximum ultimate moment that each post would be subjected to as a result of the wind pressure.

QUESTION 1. 3 The figure below shows the line drawing representing a typical frame of a rectangular clad steel building with a mono-pitched roof. The structure is 25 m long with the frames at 5 m centres. The building is to be located at Durban international airport (site altitude 75 m) and will be used as a repair workshop. The interior of the building will be open plan and all four walls are equally permeable.

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(a) Calculate the free stream velocity pressure qz. (b) Draw a neat sketch of the frame to clearly indicate each of the in-plane internal and external pressure coefficients considering the wind only in the direction as indicated. (local external pressure coefficients may be ignored). (c) Draw the loading diagrams (units to be in kN/m) on the frame for each possible load case of this load.

QUESTION 1. 4 The figure below shows a typical frame for a rectangular clad industrial building which is to be constructed at a new airport at Beaufort west. The building is 60 m long with frames at 6 m c/c. The two gable ends are equally permeable and the other walls are impermeable. The site altitude is approximately 750 m above mean sea level. The roofing contractor has requested you to provide the required number of fixings per square metre of roof area if each fixing is capable of resisting a factored tensile force of 0, 2 kN. The roof sheeting has a mass of 9 kg/m2. Hint: Only the forces on the roof are required.

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QUESTION 1. 5 The figure below shows the section through a grandstand that is to be constructed at the proposed new sports centre in Beaufort west. The overall length of the grandstand will be 75 m and the frames are at 8 m centres. The sports centres will be situated in a developed suburb where the altitude is approximately 800 m above mean sea level. a. Determine the free-stream velocity pressure. b. Provide a neat sketch to indicate the ultimate wind loads (in kN/m) on the roof of a typical frame of the grandstand considering both downward and uplift conditions. c. List the combinations of loads (dead, live and wind) together with the partial load factors that may need to be considered for the analysis of the roof at the ultimate limit state.

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ANSWERS: 1.1

440, 2 kN

1.2

(a)

0, 95 kN/m2

(b)

83 kNm

(a)

0, 876 kN/m2

(b)

external: low wall 0, 7, high wall -0, 25, high roof -0, 4, low roof -1, 0

1.3

internal: -0, 3 or 0, 0 (c)

Case 1: low wall 4, 4, high 0, 22, high roof -0, 44, low roof -3, 1 Case 2: low wall 3, 1, high wall -1, 3, high roof -1, 8, low roof -4, 4

1.4

8 No.

1.5

(a)

0, 84 kN/m2

(b)

downward: 13, 95 kN/m and 16, 57 kN/m uplift: 1, 74 kN/m and 6, 10 kN/m

(c)

1, 5 Dn 1, 2 Dn + 1, 6 Ln 1, 2 Dn + 1, 3 Wn(135) 0, 9 Dn + 1, 3 Wn(0)

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SECTION 2: ANALYSIS Questions on analysis are covered adequately in the first part of each question in the following sections.

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SECTION 3 & 4: CONNECTIONS, TIES & STRUTS QUESTION 3. 1 The figure below shows the bolted connection to the end of a 160 x 65 PFC tie. M16 grade bolts are to be used and all holes will be drilled. The bolts will be of such a length that the thread will not be in the shear plane. (a) Check the spacing of the bolts for the given configuration. (b) Calculate the maximum factored tension resistance considering all possible modes of failure.

QUESTION 3. 2 The figure below shows a 90 x 65 x 6 unequal angle member in tension, spliced in the centre using a 12 mm plate and 6 No. grade 4, 8 M20 bolts. It may be assumed that the bolt thread will not be intercepted by the shear plane. The angles will be gas-cut to length and are connected through the short leg. All bolt holes will be punched. The ends are connected to thick rigid plates with 6 mm fillet welds over a length of 50 mm each using an E70XX electrode.

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Calculate the factored tension resistance of the member showing the calculations for all possible modes of failure.

QUESTION 3. 3 The figure below shows the elevation of a pin-jointed truss subjected to the loads shown which are at the ultimate limit state. (a) Calculate the support reaction and the nature and magnitude of the forces in member A-B, B-E, and E-F. (b) Check the suitability of a 120 x 120 x 10 equal leg angle to be used as the bottom chord if all of the nodes along the bottom chord are laterally restrained.

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(c) If a 80 x 80 x 6 equal leg angle is to be used as the top chord and connected to support “A” as detailed overleaf, calculate the factored tension resistance of this member. All bolts will be M20 grade 8. 8 and the holes in the leg of the angle will be punched. The gusset plate will be connected to the backing plate with a 6 mm fillets weld using an E70XX electrode.

QUESTION 3. 4 The figure below shows the elevation and isometric view of a grade 300W steel tie comprising two 70 x 70 x 6 equal leg angles connected back-to-back with a 10 mm plate that is 70 mm wide. Grade 4, 8 M20 bolts are used in punched holes with the thread not intercepting any shear plane, and the plate is connected at the end with a 6 mm continuous fillet weld using an E70XX electrode. Calculate the factored tensile resistance of the tie considering ALL possible modes of failure. 24

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QUESTION 3. 5 The figure below shows the elevation of a pin-jointed steel truss manufactured from grade 300W steel. The loads indicated are at the ultimate limit state. Lateral restraint will be provided at G, I and K. (a) Calculate the reactions and determine the magnitude and nature of the forces in members AB, HI, DI and AD. (b) If a 120 x 120 x 10 equal leg angle is to be used for the top member G-H-I-J-K, calculate the factored compressive resistance of this member and comment on your answer.

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(c) Determine the factored tensile resistance of member AB if a 25 mm solid square bar is to be used and connected at the ends using a 10 mm thick plate as detailed below using 2 No. M20 grade 8, 8 bolts in punched holes.

QUESTION 3. 6 The figure below shows the elevation of a pin-jointed truss that is manufactured from grade 300W steel. The loads shown represent the critical design combination of loads at the ultimate limit state.

(a) Calculate the reactions, then find the magnitude and nature of the forces in members 2-3, 8-16 and 11-12. (b) Comment on the suitability of a 90 x 90 x 6 equal leg to be used as the bottom chord of the truss (i.e. from node 10 to node 16) if lateral restraint is provided at the supports and at the node 13. Consider only the compressive force in member 11-12 to verify your answer. 26

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(c) If a 180 x 70 parallel flange channel tie is to be spliced as detailed below using a 12 mm thick plate and M24 grade 4, 8 bolts in punched holes, determine the factored tensile resistance of the member considering all possible modes of failure.

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ANSWERS 3.1

(a)

ok

(b)

Tr = 135, 7 kN (based on bolts shear)

3.2

Tr = 91, 0 kN (based on weld strength)

3.3

(a)

AB = 168, 0 kN (T), BE = 59, 4 kN (C), EF = 126, 0 kN (C)

(b)

Cr = 131, 4 kN

(c)

Tr = 190 kN (based on rupture)

3.4

Tr = 145, 6 kN (based on weld strength)

3.5

(a)

AB = 145 kN (T), HI = 150 kN (C), DI = 7, 07 kN (T), AD = 7, 07 kN (T)

(b)

Cr = 140, 8 kN

(c)

Tr = 145, 6 kN (based on weld strength)

(a)

2-3 = 26, 25 kN (T), 8-16 = 7, 0 kN (T), 11-12 = 54, 25 kN (C)

(b)

Cr = 52, 88 kN

(c)

Tr = 229, 2 kN (based on bolts shear)

3.6

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SECTION 5: BEAMS QUESTION 5. 1 A grade 300W rolled steel beam is required to support the loads as shown in the figure below. Lateral support to the compression flange will be provided at the position of the point loads, and the supports may be considered to be restrained against torsion. Initial calculations have shown that a 457 x 191 x 75 I-section should suffice.

a) Draw fully dimensioned design shear force and bending moment diagrams. b) Check the above beam for shear and flexure. c) Determine the magnitude of the maximum ultimate point load that could be placed on top of the beam directly above support D if the length of stiff bearing at the support is 100 mm. (consider yielding and crippling).

QUESTION 5. 2 The figure below shows the elevation of a 1420 x 400 (10W, 25F) welded plate girder manufactured form grade 300W steel.

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The nominal dead and live loads making up each of the point loads at B, C and D, that the girder is to support addition to its self weight, are tabled below: LOAD

Dn

Ln

P1

100 kN

100kN

P2

125 kN

150 kN

P3

150 kN

200 kN

The beams producing these point loads are connected to the top flanges of the girder and are capable of providing effective lateral restraint at these points. The supports offer full torsional restraint and simple rotational restraint. a) Draw fully dimensioned shear force and bending moment diagrams of the girder. b) If stiffeners are positioned at 2,5 centres over the entire length of the girder, check that this spacing meets the requirements of the code and check the girder for combined bending and shear at “D”. (the value of Mr may be taken from b above).

QUESTION 5. 3 A beam with a cantilever end is subjected to loads at the ultimate limit state (inclusive of selfweight) as shown below. Lateral restraint is provided at B and torsional restraint is provided at the supports. All loading may be considered “normal”

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a) Draw the shear force and bending moment diagram showing all critical values. b) The beam is to be welded plate girder manufactured form grade 300W steel using 150 x 16 flange plates and a 400 x 10 web. Given that J = 520 x 103 mm4 and cw = 389 x 109 mm6, calculate all other section properties required for checking flexure. (shape facto Zpl/Ze = 1,15) c) Check the beam for flexure considering the segment between A and B. d) Check the beam for shear.

QUESTION 5. 4 1420 x 300 (10W, 20F) welded plate girder manufactured from grade 300W steel is simply supported over a span of 12 m. Two point loads are applied at third-points, each having dead and live load components of 150 kN and 200 kN respectively. In addition to these point loads the beam is also subjected to a uniformly distributed dead load of 20 kN/m, which excluded the self-weight of the beam. Torsional restraint is provided at the supports, but not rotational restraint in plan. The applied loads are free to move laterally and should be considered as “destabilising”. Lateral restraint is provided at mid-span as well as the supports.

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a) Sketch the shear force and bending moment diagrams at the ultimate limit state. b) Check the given beam for flexure (bending) and deflection. c) Check the given beam for shear on the end panel. d) If the width of bearing at the supports is 90 mm, comment on the suitability of the 100 x 10 load bearing stiffeners which are placed in pairs at the supports and at thirdpoints.

QUESTION 5. 5 The figure below the elevation of a 1840 x 400 (12W, 35F) welded plate girder manufactured from grade 300W steel. The beam is subjected to two dead load point loads of 380 kN each as well as the load from the crane that will run on the bottom flange. The crane has a mass of 150 kg and working live load of 500 kN. There is no UDL applied to the beam, but allowance must be made for its self-weight. The top flange of the girder is laterally restrained at the supports as well as at “B” and “C”. The rotational restraint at these points is simple (pinned) and torsional restraint is provided at the ends. (a) Calculate the maximum bending moment (position the crane accordingly) and check the girder for bending. (b) If the crane is positioned 2 m from “A”, calculate the maximum shear force and check the girder for shear on the 4m panel only. 32

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ANSWERS: 5.1

5.2

5.3

5.4

5.5

(a)

V = 114 kN, M = 241,5 kNm

(b)

Mr = 251,88 kNm, Vr = 741,4 kN

(c)

210,2 kN

(a)

V = 671,2 kN, M = 4056 kNm

(b)

Mr = 4443,1 kNm, Vr = 1340,4 kN, interaction = 0,766

(a)

Vmax = 115 kN, Mmax = 188,2 kNm

(b)

Iyy = 9,033 x 106 mm4, Zplx = 1,39 x 106 mm3

(c)

Mr = 202,18 kNm

(d)

Vr = 712,8 kN

(a)

Vmax = 658,4 kN, Mmax = 2475,2 Nm

(b)

Mr = 2694,0 kNm

(c)

Vr = 686,9 kNm

(d)

Cr = 961,0 kN, Br = 929,0 kN

(a)

Mmax = 6973 kNm, Mr = 7144 kNm

(b)

Vmax = 1233 kN, Vr = 1789 kN

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SECTION 6: COLUMNS QUESTION 6.1 The figure below shows the elevation of a 254 x 254 x 73 H-section column subjected to bending about the major axis only. The column forms part of a multi-storey framed structure and the moments given are at the ultimate limit state and include allowance for second order effects. The ultimate axial load is 1200 kN and includes the self-weight of the member. The major axis is unrestrained over the full height of the column and the minor axis is restrained against lateral movement at third-points.

(a) Check the stability of the given section and comment on your answer.

QUESTION 6. 2 The figure below shows the elevation of a rigid building frame. All columns will bend about the strong axis in the plane shown. Weak axis bending in the column may be ignored since the out-of-plane beams are “pinned” to the web of the columns and shear walls will provide adequate bracing to the weak axis of the columns. A first order liner analysis of the structure produced an axial load of 1200 kN together with bending moment diagrams about the x-x axis of one of the columns between levels 2 and 3 (shown bold in the figure) as shown:

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(a) If the relative first-order lateral displacement between the second and third floors as found to be 52 mm, draw the final design bending moment diagram for the highlighted column allowing for second order effects. (b) Check the adequacy of a 254 x 254 x 107 H-section if it is to be used for this column which is subjected to combined bending and compression.

QUESTION 6. 3 The figure below shows the elevation of a rigid-framed multi-storey steel structure subjected to a critical load combination at the ultimate limit state. All beams are

305 x 165 x 41 I-

sections and all columns are 254 x 254 x 89 H-sections. A first order linear analysis of the structure produced the ultimate gravity and translational bending moments about the major axis of the centre column between the first and second floor as shown adjacent.

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(a) Draw the final bending moment diagram for the major axis of this column considering the amplification factor to allow for second order effects. Consider the level at the top of this column where the deflection relative to the bottom of the column was found to be 18 mm. (b) The minor axis is laterally restrained at mid-height and bent in double curvature with equal end moments of 25 kNm (including second-order effects). Comment on the suitability of the given column section subjected to combined bi-axial bending and axial load. The ultimate axial load on this column was found to be 1890 kN.

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ANSWERS: 6.1

(a)

Interaction a = 0,778, b = 0,994, c = 0,865

6.2

(a)

Mtop = 250 kNm, Mbot = 125 kNm

(b)

Interaction a = 0,856, b = 0,581, c = 0,954, d = 0,626

(a)

Mtop = 108, 4 kNm, Mbot = 63,4 kNm

(b)

Interaction a = 0,988, b = 0,846, c = 0,999, d = 0,488

6.3

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MODULE 2 STRUCTURAL TIMBER DESIGN

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MODULE 2 STRUCTURAL TIMBER DESIGN Relevant SANS codes •

SANS 10163-2:2001: The structural use of timber. Part 2, Allowable stress design / South African Bureau of Standards.

•

SANS 10160: The general procedures and loadings to be adopted in the design of buildings.

Further reading There are numerous reference works published, inter alia, by the South African Bureau of Standards, the National Building Research institute, the National Timber Research Institute, the South African Lumber Millers Association (SALMA) and the Institute for Timber Construction (ITC). Contact these organisations for more information.

Introduction The primary objective of this part of the course is to show the student that timber can serve as a useful alternative structural material. Timber need not be restricted to roof trusses which remain hidden from sight; it can also perform a variety of structural functions from buildings to bridges.

Course format A set of notes will be made available by UNISA which should be followed closely together with the code of practice when working through each section. Question for self-evaluation are provided at the end of each chapter in the set of notes provided.

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MODULE 2 STUDY NOTES CHAPTER 1 TIMBER AS A STRUCTURAL MATERIAL CONTENTS

PAGE

1.1

OBJECTIVES

43

1.2

TIMBER AS A STRUCTURAL MATERIAL

43

1.2.1

1.3

Introduction

43

1.2.2

Species

43

1.2.3

Manufactured timber products

44

1.2.4

Properties

44

1.2.5

Grading

46

DEFINITIONS

47

1.3.1

Material

47

1.3.2

Defects

53

1.3.3

Design

55

1.4

STANDARDISED CODES AND SPECIFICATIONS

57

1.5

INTRODUCTION TO TIMBER DESIGN

57

1.6

SPECIFIC OUTCOMES

60

1.7

REFERENCES

61

1.8

QUESTIONS FOR SELF-EVALUATION

62

1.9

SOLUTIONS TO THE SELF-EVALUATION QUESTIONS

63

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1.10

TUTORIAL

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1.1

OBJECTIVES:

The objective of this section is to introduce the learner to timber and define common terms used in timber and timber construction.

1.2

TIMBER AS A STRUCTURAL MATERIAL

1.2.1

Introduction

Timber is defined as being either a softwood or hardwood. Hardwood comes from deciduous (broad-leaved) trees and it is seldom used for common structures, whereas softwood comes from evergreen trees (conifers in particular) and is extensively used for timber structures. Although this course will mainly deal with softwoods it is to be noted that hardwoods are often more durable and stronger than softwoods as demonstrated by Rhodesian Teak and Jarrah woods used for railway sleepers, etc. Structural grade plywoods are marketed locally in both softwood and hardwood varieties. 1.2.2

Species

Owing to different climatic conditions in the various parts of South Africa, a regional usage factor has to be acknowledged. Different species is pine trees are grown and marketed in the different regions:

Table 1.1: SA pine species distribution and usage Botanical Name

Pinus

Pinus

Pinus

Pinus

Patula

Elliottii

Radiata

Pinaster

Common Name

Mexican

Slash

Monterey

Cape

Pine

Pine

Pine

pine

Gauteng

80%

20%

Nil

Nil

K. Natal

20%

80%

Nil

Nil

W. Cape

Nil

Nil

90%

10%

Regional Usage

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The differences need not generally be considered in designing structures with timber owing to the fact that Stress-graded timbers are used commercially and these are tested by the suppliers to comply with SANS 1783: Sawn softwood timber. Eucalyptus is widely grown and used in South Africa, but not generally for structures. Gum poles tend to be used as pit-props in mines, fencing, telephone and electricity poles. In addition they are commonly used in low-load structures such as thatched roofs on cottages, etc. The exception is Eucalyptus Saligna (Sydney blue gum or more commonly Saligna) which is used in glulam products.

1.2.3

Manufactured timber products

Laminated timber or Glulam is commonly used to resolve architectural requirements, the relevant code is SANS1460: Laminated timber (glulam). These products are marketed under different names according to their grades and composition, lengths of up to 15 m are commercially available. Although suppliers have a set of standard sizes up to 1200 mm deep, special orders can be made up on request. Plywood complying with SANS 929: Plywood and composite board may also be used in structural applications.

During the Second World War the British built the De Haviland Mosquito aircraft using a plywood and balsa sandwich for the fuselage. This Wooden Wonder was a very fast and versatile aircraft and served for the duration of the war in a number of roles from fighter to bomber.

1.2.4 Properties Only commercially available structural grades of South African softwood will be considered in these lectures, however the learner must be aware of the possibility of designing timber structures using other types and grades of wood. The learners should visit lumber yards to 44

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find out firsthand what different grades and sizes of structural timber are available in this area. Wood is not a homogenous material like steel; it is a natural product at the mercy of the elements both during growth and after reduction to boards. In order to ensure the end-user and designer of a uniform product, the lumber suppliers grade timber in accordance with a set of limitations specified by the SANS. Trees want to stand up straight and to that end if for example the prevailing winds generally come from one direction the tree will strengthen itself against tensile forces on the windward side and against compressive forces on the leeward side. The end result is plain to see when the tree is felled, the rings do not form perfect concentric circles and ring spacing is closer on the one side than on the other, similarly the density of the wood varies from one side to the other. These variations mean that the grader will find a range of grades from one tree’s lumber. The outer wood is stronger than the inner core. The distortion of the grain around knots adversely affects the strength of timber especially of tensile and bending members. Knots are therefore regarded as defects whether they are sound knots around living branches or the loose knots of dead branches. The strength and modulus of elasticity are greatest parallel to the grain, strengths and other properties vary considerably relative to the direction of the grain and are much weaker when measured perpendicular or tangential to the grain. Timber seems to loose strength with time and this loss is accelerated by high stresses. Time to failure decreases with increase in stress. Therefore the designer should concern himself more with long duration high stress loading than with long term low stress loadings. The design codes do in fact take cognisance of this fact. Timber with a high modulus of elasticity is more likely to be stronger than timber with a low modulus of elasticity. Grading of timber in terms of stiffness and thus modulus of elasticity is therefore possible. Timber creeps under long term load condition and may give the impression of being unsafe owing to excessive deflections. The design codes have addressed this issue.

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The advantages of timber as a structural material are: i.

Timber is relatively light, easy to handle and workable with normal tools and machinery.

ii.

It has a higher strength to weight ratio than for example concrete.

iii.

Corrosive environments generally have less effect on timber than steel.

iv.

Thermal expansion/contraction is low.

v.

It has a high fire resistance.

vi.

It has a high resistance to impact loadings.

vii.

Aesthetically pleasing it is suitable for many architectural features.

Disadvantages of timber as a structural material are: i.

Insect and fungus damage can be disastrous if the timber is not treated.

ii.

Deterioration owing to moisture changes may occur.

iii.

Distortions as a result of creep under long-term loads do occur.

iv.

High moisture content leads to a reduction in stiffness and strength.

v.

Load defects are found because of the existence of knots.

vi.

It is anisotropic – material properties differ in different directions.

vii.

It is hygroscopic – it tends to absorb moisture.

viii.

It is susceptible to wind, rain and ultra-violet radiation damage.

1.2.5

Grading

The SANS caries out spot-checks from time to time to ensure that products bearing their SANS mark do infact comply with the specifications. These markings are in signal red on 46

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each piece of timber and are marked according to the test method and stress grade. Visual stress graded timber is marked with a “V” and a number, e.g. V4 is a visually graded piece with a stress grade of 4 MPa. The minimum stress grade of structural timber is 4 MPa. Mechanical stress grades are marked with “M” and Proof-graded timber is marked with “P” e.g. M4 and P4 respectively. The relevant codes are: Visual grading – SANS 1783-2: Stress-graded structural timber and timber for frame wall construction. Mechanical grading - SANS 10149: The mechanical stress grading of softwood timber (flexural method). Proof-grading – SANS 1783-4: Brandering and battens.

1.3

DEFINITIONS

Although the following definitions comprise the most commonly used terminology of the timber industry in Southern Africa, the list is not necessarily complete and the learner should familiarise himself with all the definitions in the standardised specifications and those used by the industry. The following definitions have been extracted from SANS codes 10005, 10163, 10243 and 1783 as well as The Encyclopaedia of Wood and Mondi trade literature.

1.3.1

Material

Batten A piece of timber used to support roof coverings of tiles or slates, usually having crosssectional dimensions of 38 x 38 mm or 38 x 50 mm or 50 x 50 mm. Brandering Timber fixed to the underside of truss chords onto which ceiling boards are attached. Usually 38 x 38 mm in cross-section. 47

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Cambium The thin layer of tissue between the wood and the bark. It subdivides repeatedly to form new wood and bark cells. Cellulose The principle constituent of wood comprising 50% of the mass of the wood. It is a carbohydrate that forms the framework of wood cells. Charge The quantity of timber treated in one and the same treating cycle. Density group Density group D, is assigned to timber that exceeds 480 kg/m3 and timber that has a density between 400 kg/m3 and 480 kg/m3 is classified as density group D2. Flat-grained timber The grain patterns made by the annual growth rings make an angle of less than 450 with the surface of the piece. General slope of grain The slope of grain (as observed over a distance of at least 600 mm) on the face that is furthest away from the pith and tangential to a growth ring. Glued laminated member (Glulam) A member which has been manufactured by gluing a number of strips of timber together which have their grains essentially parallel to each other. Grade stress See stress grade

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Gum Non-volatile viscous plant exudates which, either swell or dissolve in water. The gum exuded by pines is in fact oleoresin. Heartwood Wood between the pith to the sapwood. It is no longer active and is identified easily by its darker colour owing to the presence of phenolic compounds, gums, resins and other materials which make it more resistant to decay than sapwood. Inner bark The physiologically active layer of tissues between the cambium and the phellogen, plus any cells of the phloem that remains alive. Lignin The second most abundant constituent of wood, located principally in the secondary wall and the middle lamella, which is the thin cementing layer between wood cells. It is an irregular polymer of substituted propylphenol groups. In softwoods it is 23 – 33% of the mass of the wood and 16 – 25% of the mass of hardwoods. Lot Not less than 50 and more than 10 000 pieces of structural timber of the same cross-section and stress-graded by the same method, from one manufacturer, submitted at any one time for inspection and testing. Lumber This is a general term used to describe timber which has been cut into boards or planks. It is more commonly used in USA than Southern Africa. Lumen The cell cavity of wood.

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Moisture content The mass of water in a sample of timber expressed as grams of moisture per kilogram (or percentage) of the oven-dried mass of the sample timber. Oleoresin In softwoods, especially pine, this is a solution of pine resin (rosin) in an essential oil (turpentine). Outer bark The layer of dead tissues (generally of a dry corky nature) outside the phellogen. Phloem The tissues if the inner bark, characterised by sieve tubes which act as conduits for foodbearing sap. Pith The soft core near the centre of any section of trunk or branch. Planned all round (PAR) See wrought timber. Plywood Plywood comprises and odd number of layers of veneers of wood glued together with the grains of each veneer perpendicular to the preceding one. Both softwood and hardwood plywood are available. Regularising The process whereby a piece of timber of rectangular cross-section is machined throughout its length to a uniform thickness or width or both.

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Resin Inflammable, water-soluble plant exudates common to both hardwoods and softwoods (e.g. gum trees and pines). Rough-sawn Timber sawn on a breakdown saw or similar machine and edged but not otherwise machined. Sapwood The pale coloured word near the bark, it is the part of the timber most vulnerable to decay. Seasoning The drying out of timber to improve its serviceability. It may be air-dried without artificial heat or kiln-dried using artificial heat. Sizing See thicknessing. Softwood structural timber (structural timber) Timber derived from coniferous trees grown in Southern Africa and having a nominal width of at least 75 mm, a nominal thickness of at least 38 mm, and a working stress in bending of at least 4 MPa. Stress grade (of a piece of structural timber) The numerical value of the working stress in bending in MPa that can be n safely sustained by the piece under long-term loading conditions. Sterilisation of timber The destruction of all live stages of insects or fungi or both.

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Tantalised timber Timber impregnated with copper-chrome-arsenic (CCA) or copper-gallium-arsenic as a preservative. The copper gives the treated timber a distinctive green colour. (“Tanalith” is a trade name). Thicknessing Is the process of converting rough-sawn timber to regularised timber by skimming a thin layer off the surface. This process does not necessarily produce smooth surfaces. (See Wrought timber). Timber In Southern Africa is the term used to describe wood that has been cut into boards and planks but in the USA it refers to trees which are still standing. Veneer A thin layer of wood, usually about 1 mm thick, cut from a log either by rotary-cutting in a lathe against a knife or by sawing or by slicing. Volatile solvent A solvent for wood preservatives that evaporates from the treated open-stacked timber to the extent of at least 90% of its volume within a period after treatment not exceeding two weeks. Wood preservative A pesticidal agent which, when impregnated into or otherwise applied to timber, renders it less susceptible to destruction by fungi, insects or marine borers. Wood protectant An agent that is fungicidal or insecticidal or both, applied to timber to protect it temporarily from attack during one or more stages of processing. Wrought timber A board planed on all sides to produce smooth surfaces. 52

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Xylem The portion between the cambium and the pith.

1.3.2

Defects

Bleeding – The exudation of creosote on the surface of treated timber. Blooming – The exudation of dry preservative compound on the surface of treated timber. E.g. Crystal formation in the case of timber treated with PCP (pentachlorophenol). Bow – Lengthways curvature in one plane along the edge of a piece of lumber. Check – A lengthways separation of the wood fibres along the grain forming a crack or fissure. Compression break – Minute ridges formed by the buckling of cell walls as a result of excessive compression along the grain. Cup – Curvature occurring in the transverse section of a board. Defective – A piece of structural timber that fails in one or more respects to comply with the requirements of the relevant specification. Discolouration – A change in the colour of a piece of lumber that affects only its appearance and is caused by fungal stain. Edge knot – A knot of which a cross-section occurs on an edge and which extends at least two thirds across or through the board. Hair check – A very fine check of width not exceeding 0, 5 mm. Inbark – Bark of thickness exceeding 1 mm (usually associated with a knot) that has been enclosed by the tree and subsequently exposed by reduction of the tree to lumber. Knot hole – A hole or cavity formed in a piece of lumber as a result of the absence of a knot or part thereof and the effect extends at least two-thirds across or through the board.

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Knot whorl – Four (or more) closely associated knots that occur on at least two faces of a piece of lumber and that originate in unexposed pith. Machine damage – Limited undersizing of timber by mechanical means during regularising. (Usually at the ends of boards due to movements of the timber during machining). Machine skip – An area of the surface of a regularised piece of timber, that was not removed during machining. Resin infiltrated area – An area on the surface where the wood is saturated with resin to an extent that causes a darkening of the wood to a colour that is deeper than that of sound knot in that board. Resin pocket – A cavity that contains or has contained resin. Shake – A separation along the grain, the greater part of which occurs between the annual growth rings. Sound knot – A knot that is free of decay, insect damage and inbark and is firmly attached along at least two thirds of its periphery to the surrounding wood on one face of the board. It may be shrunk away completely from the surrounding wood on the opposite face. Split – A separation of the wood fibres along the grain forming a crack or fissure that extends through the board and is visible on both faces. Spring – Lengthways curvature in its own plane on the face-side of a board. Superficial face splay knot – A splay knot that has been so cut that the knot does not penetrate the lumber to a depth exceeding one-eighth of the thickness of the board. Through face knot – A knot that penetrates the lumber from one face to the other but is not exposed on an arris (side). Twist – A form of warp that appears as lengthways spiral distortion. Wane – The original surface of a tree, with or without bark, visible on a piece of square sawn lumber.

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Warp – Any departure (may take the form of bow, spring or twist or any combination of these) from a true or plane surface of a board.

1.3.3

Design

Allowable stress or allowable joint force – The grade stress or basic joint force multiplied by the modification factors that are appropriate to the specific conditions under which a member, structure or joint will operate. Basic joint force – The force assigned to a joint or member of a joint to quantify the strength of the joint or member. Bottom chord – The lower edge of a truss. It usually carries combined bending and tension stresses. Built-up beam – A beam made up of two or more timber members. Calculated deflection – The deflection predicted for a structure based on elastic theory analysis. Camber – A specified upward displacement in the centre of a beam or bottom chord to compensate for deflection due to loads and/or self weight. Characteristic joint strength – The force assigned to joint or member of a joint to quantify its strength. Characteristic timber strength – The strength assigned to a timber member or product and below which not more than 5% of the tests fail. Connector – Any device, capable of transmitting specified loads, used for joining one or more pieces of timber together. Design deflection – The calculated deflection adjusted to account for creep, abnormal moisture content or abnormal fluctuation in moisture content. Factored resistance φ R – The product of the resistance R and the appropriate resistance factor. 55

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Gravity load – The mass of the supported object multiplied by acceleration due to gravity. Imposed load – The load due to intended usage, including loads due to movable partitions, cranes, winds, rain, ice, snow, earth and hydrostatic pressures. Limit-states – Those conditions of a structure at which the structure ceases to fulfil the function for which it was designed. The states concerning safety are called Ultimate limitstates and include exceeding of load- carrying capacity, overturning, uplift, sliding, fracture and fatique failure. Serviceability limit-states restrict intended use and occupancy of a structure and include excessive deflection and vibration and permanent deformation. Load effect – A force or moment induced in a member, due to ultimate loading, ultimate displacement or ultimate distortion. Member – Any structural component of either solid timber or built-up from pieces glued together to form particular sections (e.g. “I” beam with solid flanges and plywood web). Modification factor – A factor that is applied to the member resistance, the connection resistance or the calculated deflection, to allow for the specific condition/s under which a member or structure will operate and which will influence its structural behaviour. Nominal load – Those loads specified in SANS 10160. Resistance R – The resistance of a member, connection or structure, as calculated in accordance with SANS 10163-1, based on the specified material properties and nominal dimensions. Resistance factor φ - A factor, referred to in SANS 10163-1, that is applied to a specified material property or to the resistance of a member, connection or structure that, for the limit state under consideration, takes into account the variability of the material properties, dimensions, workmanship, type of failure and uncertainty in prediction of member resistance. Self-weight load – The load comprising the weight of all structural members and any permanent finishes attached to the structure. Serviceability load – The design load or action effect that pertains to the serviceability limit state 56

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Structure – Any assembly of timber members, including the detail parts, fasteners and other items required for the manufacture or erection of the assembly. Ultimate load – The design load or action effect that pertains to the ultimate limit state.

1.4

STANDARDISED CODES AND SPECIFICATIONS

The SANS codes and specifications which have already been mentioned form the basis of design in South Africa. The learner is not expected to learn these codes by rote but he is expected to be familiar with the content and to be able to apply them to this design. The SANS publications have lists of references and bibliographies which the learners should endeavour to read. In addition to these codes the Eurocodes and the British Standards institution codes are also to be studied by the learners. There is a growing international tendency towards standardisation and we may soon see the application of a single set of codes and specifications.

1.5

INTRODUCTION TO TIMBER DESIGN

Timber is probably the most versatile material, it has been used for millennia in vast quantities for every sort of application. Structural applications range from simple frames for dwellings to the war-machines of the ancient Greeks and Romans. Modern structural timber design has become a sophisticated extension of what previously was empirical design, but there are still a number of uncertainties for which the design methods have to make assumptions or adjustments. The designer therefore needs to know the material, the application and the situation in which the structure will be used. The demand for diversity from architects and the public in general has resulted in a range of timber structures, some of which are not at all economic in design. For example a traditional post-and-beam house uses far more timber than is necessary, but it achieves an effect which some people want. Some of the structural applications of timber are described in the following paragraphs (After Stalnaker & Harris). 57

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Pole structures Pine or gum poles may be used for falsework in concrete construction, frames of cheap housing and sheds, roof structures, fencing, sign-board supports, bridges, piers, jetties, etc. Timber-framed (or light-framed) structures Timber-framed buildings in the modern sense comprise a designed frame which may be clad with planking on the outside and dry-walling on the inside. Other exterior cladding includes corrugated iron or asbestos-cement or glass-fibre sheeting. Another variation uses a clay brick veneer on the exterior. Post-and-beam construction (Tudor style) This very heavy construction has stood the test of time, many fine examples are over 400 years old. Posts 150 – 250 mm square are placed at 2, 4 – 3, 6 m centres and horizontal members are spaced 1, 2 – 2, 4 m vertically. Often diagonal “knee braces” are used to provide rigidity.

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Log buildings The cost generally discourages this type of building but there are a few examples of log cabins in South Africa. They are seldom actually designed. Rough-hewn logs are stacked on top of each other and often rely on gravity and friction to hold them together rather then any conventional jointing or connecting systems. Prefabricated panel systems A variety of proprietary systems are available but these seem to be more popular overseas than in South Africa. One example is the stressed-skin panel which is a timber frame with plywood on both sides forming an integral part of the structural strength. Laminated structural components This is a relatively new branch of engineered timber structures where considerable cost saving can be effected by using timber beyond its traditional capacity. By using high quality adhesives, timber is glued together to form sections which are much stronger than equivalent sawn timber sections. This is however an aspect of design which still requires more scientific analysis and at present laminated members are basically designed according to the shape and grade of the component timber. A major attraction is the variety of length and shapes that can be achieved. Complex curved portal frames being the most common of these. Overseas laminated timber roof structures commonly span in excess of 30 m. Glued components are another “new” development. Plywood web l-beams with solid timber flanges also offer the user cost savings in both material and erection. Heavy timber construction Where the smallest dimensions of columns are greater than 200 mm and beams exceed 150 x 300 mm, the structure is regarded as heavy construction. This sort of work is rare owing to the cost of timber. Piers, jetties and bridges would usually fall under this category.

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Bridges Although not that common there are a few wooden bridges in South Africa which carry vehicular traffic on a daily basis. Forestry road bridges are also in use but these tend to be crude structures made from logs. In overseas there are timber bridges which still serve the public after 500 years of uses. Formwork and falsework Although largely replaced by steel and glass-fibre or plastic moulds, timber still plays an important role in formwork for concrete construction.

1.6

SPECIFIC OUTCOMES

At the end of this section the learner will be able to identify and name components of timber structure.

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1.7

REFERENCES 1. BURDZIK, Prof. WMG: personal communications. Pretoria: 1996. 2. FOREST PRODUCTS LABORATORY. Forest Service, US Dept. Of Agriculture. 1980. The Encyclopaedia of wood. New York: Sterling Publishing Co. Inc. 3. MOINDI TIMBERT PRODUCTS: Trade literature. Mondi Ltd, Stellenbosch. 4. SANS 10005: The preservative treatment of timber. 5. SANS 1783: Sawn softwood timber 6. SANS 10163-1: The structural use of timber. Part 1: Limit-states design. 7. SANS 10163-2: The structural use of timber. Part 2: Allowable stress design. 8. Stalnaker JJ & Harris EC: Structural design in wood. Van Nostrand Reinhold, New York, 1989. (Imperial units)

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1.8

QUETIONS FOR SELF-EVALUATION

These questions are intended to provide the learner with a means of determining whether he has understood the contents of the Lecture and generally attained sufficient knowledge in the subject to be able to pass the exams. Solutions to the questions are given on the next page. It should be noted that at this level of tertiary education the Lectures serves as a study guide and the learner must carry out his own research in order to acquire the expected levels of knowledge about the subject.

1.8.1

Draw a Fink W truss. Show and name all structural components.

1.8.2

Define the following terms:

a) Engineered truss design b) Jack rafter c) Lateral brace d) Node or node-point e) Vierendeel truss or girder

1.8.3

What types of bolts are acceptable and what are not for use in timber trusses?

1.8.4

Briefly discuss the effects of wind on a structure.

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1.9

SOLUTIONS TO THE SELF-EVALUATION QUESTIONS

1.9.1

Fink W truss from SANS 10243

19.2

Definitions from SANS 10243

a. “Engineered truss design: A truss design in which loading requirements, timber sizes, grades and plate requirements are detailed.” b. “Jack rafter: A special rafter used to form a hip.” c. “Lateral brace: A member placed and connected to a member of truss to prevent horizontal movement.” d. “Node; node-point: The point of intersection of the various members that make up the panels of the truss.” e. “Vierendeel truss/girder: A special form of truss that is not triangular but that has specially designed joints at member intersections and that are rigid enough to prevent the truss from deforming.” 1.9.3

\

Bolt types in trusses from SANS 10243 63

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“Bolts should preferably be hexagonal-head bolts. Cup-head or carriage bolts are not recommended.”

1.9.4 Wind can exert a horizontal force that is not necessarily perpendicular to the building but is assumed to be so for most design purposes. Wind can also exert an upward force acting on the plan area of a roof. Within the building both positive and negative pressures may result from wind. (Refer to SANS 10160 for more details on wind and other forms of structural loadings).

Having completed these questions the learner should now realise that a comprehensive working knowledge of the prescribed codes, and eventually the books, is essential for success in this subject.

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1.10

TUTORIAL

1.10.1 Describe in detail the section of a tree trunk, including all the components parts. Illustrate your answer with a sketch. (10) 1.10.2 Explain limit-states design and the various states with respect to structural timber design. (10) 1.10.3 Discuss the chemical composition of wood. (10) 1.10.4 Discuss the advantages and disadvantages of using timer as a structural material (10)

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MODULE 2 STUDY NOTES CHAPTER 2 ALLOWABLE STRESS DESIGN METHODS CONTENTS

PAGE

2.1

OBJECTIVES

68

2.2

DESIGN METHODS

68

2.2.1 Introduction

68

2.2.2 Working stress design

68

2.2.3 Limit-states design

69

2.2.4 Plastic design

70

REVISION OF DIPLOMA LEVEL TIMBER DESIGN

70

2.3.1 Allowable stress design

70

2.3.2 Duration of load factor

71

2.3

2.3.2.1 Example of duration of load factor 2.3.3 Load sharing factor

72

2.3.4 Type of structure factor

72

2.3.5 Quality of fabrication factor

73

2.3.6 Moisture content factor

73

2.3.6.1 Example of moisture content factor 2.3.7 Creep deflection factors 2.3.7.1 Example of deflection factor calculation 2.3.8 Design of tension members 2.3.8.1 Example of tension member design 2.3.9 Compression member design 2.3.9.1 Example of compression member design 2.4

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SPECIFIC OUTCOMES

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2.5

REFERENCES

82

2.6

QUESTIONS FOR SELF-EVALUATION

83

2.7

SOLUTIONS TO THE SELF-EVALUATION QUESTION

84

2.8

TUTORIAL

88

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2.1

OBJECTIVES

The objective of this section is to explain the basis of allowable stress design and the use of the factors associated with it. Lecture 2 introduces the design methods for structural timber design and reviews the National Diploma syllabus with respect to timber design.

2.2

DESIGN METHODS

2.2.1

Introduction

Essentially there are two main categories of design methodology for designers to use, working stress design and limit-states design. Plastic design is a variation of limit-states design. The following has been abstracted from Prof. Burdzik’s notes.

2.2.2 Working stress design Working stress, allowable stress, elastic and deterministic design are all terms which essentially mean the same thing. The material is assumed to be linearly elastic and a factor of safety has been applied to the yield or failure strength. The characteristic failure strength is that strength, for timber, that has a 95% probability of being exceeded. A nominal load is that load is assumed to have a 5% probability of being exceeded. The characteristic strength divided by some factor that includes a safety factor against overstress, give us the permissible stress. It is assumed that as long as the stresses due to applied nominal loads are less than the permissible stress, the structure will be safe. Characteristic strength value are determined by testing a large sample of the material in such a way to ensure that the probability of any specimen of the material having a strength of less than the nominal value will be less than 5%. SANS 10163-2: The structural use of timber. Part 2: Allowable stress design is the applicable design code for working stress design. Working stress codes demand that the stress due to applied loads must be less than the permissible stress. Stresses due to applied loads are calculated by analysis of the structure. Permissible stresses are quantified in the design code. 68

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Loads should be adjusted in accordance with the requirements and recommendations of SANS 10160: The general procedure and loadings to be adopted in the design of buildings. The code introduces factors for the duration of load, etc. as required by Clauses 4.1.2 and 5 of SANS 10163-2. Despite the international trend towards the use of limit-states design, working stress design is still endorsed by most countries as an acceptable method, albeit more expensive, for the design of small timber structures. This is probably because it is a tried and tested method which is far simpler to apply than limit-states.

2.2.3

Limit-states design

Limit-states (or load factor or unified code) design attempts to ensure uniform safety against failure. Assumptions include that both material and likely loads have some form of distribution of strength and dimensions. Loads are multiplied by factors (in the same way as allowable stress design using SANS 10160 procedures), but additional factors are introduced. These factors are the ultimate limit states and the serviceability limit states which so modify load combinations applied to a structure that a uniform safety factor against failure will result (SANS 10160). The probable distribution of dimensions and strength are taken up in a material strength factor. This method assures the designer of a more realistic appraisal of the true safety factor against structural failure. Limit-states, in addition to strength values, include deflections, crack size, serviceability, etc. The applicable design code in Southern Africa is SANS 10163-1: The structural use of timber. Part 1: Limit-states design. Limit-states design is a method which may be applied to any structural timber design however it is more specifically intended for substantial structures where a higher confidence level is expected. For example a 3-storey building. It would be unusual to apply this method to a small wooden bungalow design. Limit-states design is more simple and rational than elastic design and easier to apply to more sophisticated structures with the results being more predictable of actual in-service behaviour of the structure. Although opinion is divided, it is generally accepted that the use of limit-states design results in a more economical structure than one designed using allowable stress. 69

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2.2.4

Plastic design

The material is assumed to behave plastically and that when yield stresses are reached in flexural members the member will have enough rotational capacity to shed some of the load to sections that are not as heavily loaded. Failure will occur when sufficient plastic rotational hinges have formed for the structure to become a mechanism. As with limit-states design safety factors are applied to the loading. We will not be covering Plastic Design in this course.

2.3

REVISION OF DIPLOMA LEVEL TIMBER DESIGN

2.3.1 Allowable stress design The grade stresses referred to in SANS 10163-2, other than those for compression of slender members, are the 5th percentile strength value of a sample, divided by a factor of 2,22. The 5th percentile value is that value which has a 95% probability of being exceeded and the 2,22 factor includes a factor for the duration of load combined with a safety factor. The grade stresses in the code are therefore for long duration loading. Stresses for slender compression members are divided from a Perry-Robertson type formula which is based on Euler’s buckling formula that allows for initial lack of straightness, Permissible stresses are obtained either by multiplying SANS 10163-2 grade stresses by stress modification factors or the loads can be multiplied by the relevant coefficient or divided by the stress modification factors. The factors used in the design of members are:

−

the factor for the duration of load .

k1

=

k2

−

the factor of load sharing .

k3

−

the factor for the type of structure.

k4

−

the factor for quality of fabrication

k5

−

the factor for the moisure content.

k1

C fD

WD + W1 + WW ⋅ WD + C fI ⋅ WI + C fW ⋅ WW

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f b = p ⋅ k1 ⋅ k 2 ⋅ k 3 ⋅ k 4 (bending stress ) f t = p ⋅ k1 ⋅ k 2 ⋅ k 3 ⋅ k 4 (tensile stress parallel to the grain ) f v = p ⋅ k1 ⋅ k 2 ⋅ k 3 ⋅ k 4 (shear stress )

Allowable stress for compression: f c = p ⋅ k1 ⋅ k 2 ⋅ k 3 ⋅ k 4 ⋅ k 5 Where p = Grade stress

There are other modification factors in SANS 10163-2 but these will be dealt with later. Note that the symbols used throughout this course will be those in the relevant SANS codes.

2.3.2

Duration of load factor

Long duration loads appear to cause strength loss in timber. Permanent, medium and short duration loads are all considered when determining the value of k1 (SANS 10163-2: CI.6.3.3). Permanent loads are those on the structure for 3 months or longer and include the self weight of the structure. Medium duration loads (1-3 months) include imposed floor loads, etc., short duration loads of less than a day include wind and very short duration loads of less than one hour are the 0,9 kN point load. (SANS 10163-2: Table 7). Where wind loading is critical and works to counter the effect of other loads, use only the wind load to calculate k1. Where loads work in the same direction and have the same form, the actual loads can be inserted into the formula for the duration of load factor. It is therefore possible to modify the loading and keep the permissible stress equal to the grade stress.

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2.3.2.1 Example of duration of load factor

Calculate the k1 factor, for bending stresses, when a uniform loading is applied. Self-weight is 1.5 kN/m and the imposed load is 2.1 kN/m.

Bending Moment =

wL2 for both load cases, it therefore does not need to be calculated to 8

determine the factor. The factor will be the same for all lengths under this loading.

k1 =

1, 5 + 2,1 = 1, 247 1, 0 ⋅ 1, 5 + 0, 66 ⋅ 2,1

In this case wind loading was not considered so WW = 0 in the equation from SANS 10163-2: CI.6.3.3.

2.3.3

Load sharing factor

Care must be taken in applying this factor to ensure that the structure has a uniformly distributed load acting on four or more members spaced not more than 600 mm apart in such a way that they are restrained to the same deflection. If the structure complies with these parameters and it is not a truss or a group of laminated members then the modification factor k2 may be taken as 1, 15. In all other cases k2 = 1

2.3.4

Type of structure factor

Usually this factor k3 = 1 and it is applied to structural elements where failure would have little effect. However the factor’s values range from 0,85 to 1,15 for trusses and girders of different spans.

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2.3.5

Quality of fabrication factor

This modification factor k4 is usually taken as 1 unless the structure under consideration is a fabricated set of components complying with a SANS standard then k4 = 1, 05.

2.3.6

Moisture content factor

Care must be taken in applying this factor to timber treated with water-based preservatives such as CCA (“Tanalised”) owing to the strong possibility of moisture content being in excess of 200 g/kg (20%). If the moisture content is certain not to exceed 170 g/kg (17%) take k5 = 1 but if the moisture content can occasionally exceed 170 g/kg (17%) in a joint or compression member then take k5 = 0, 75. Normal moisture content of structural timber is in the order of 12% and fibre saturation point is around 25%.

2.3.6.1

Example of moisture content factor

Calculate the permissible axial stress for a column carrying a permanent load of 8 kN and a short duration load of 15 kN. The grade stress is 4 MPa and the moisture content is 25%.

k1 =

8 + 15 = 1, 285 1, 0 ⋅ 8 + 0, 66 ⋅ 15

k2 = 1, 0 k3 = 1, 0 k4 = 1, 0 k5 = 0, 75 (Moisture content > 17%)

Allowable stress

= fc

= p . k1 . k2 . k3 . k4 . k5

= 4, 0 . 1, 285 . 1, 0 . 1, 0 . 1, 0 .0, 75 73

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= 3, 855 MPa

2.3.7

Creep deflection factors

The biological variability of timber contributes to the fact that under prolonged loading deformation will increase with time. Some deformation will remain after the load is removed, this permanent deformation is called creep. Creep is a function of the stress level and duration to which the fibres have been subjected. It has been assumed that design stresses are equal to the permissible stresses. Elastic deflections have to be adjusted in relation to the duration and intensity of the load. The moisture content of 17% is a critical point above which creep also increases.

Note that omission of creep modification factors will lead to actual deflection being greater than those calculated.

∆ design deflection = ∆ calculated deflection . d1 . d2 ≤ ∆ allowable deflection where

∆ = deflection d1 = deflection modification for creep d2 = deflection modification for moisture content (See SANS 10163-2: Table 9). Refer to clause 6.4 SANS 10163-2.

2.3.7.1 Example of deflection factor calculation The effect of self weight and imposed load is the same in this case because they are both uniformly distributed loads.

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d1 =

2+3 = 1,19 0,6.2 + 1,0.3

2.3.8

Design of tension members

Tension members are those that are subject to axial tensile forces parallel to the grain of the timber member. Forces tend to stretch the member. Failure is not influenced by slenderness of the member.

σt =

P = ft A

where

σ t is the calculated tensile stress in MPa P is the total tensile force or applied load in N A is the net cross-sectional area in mm2 ft is the allowable tensile stress in MPa ft = p . k1 . k2 . k3 . k4 The maximum Slenderness Ratio of a tension member may not exceed 250, l e / b 〈 72

75

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2.3.8.1 Example of tension member design Calculate the maximum allowable force that may be applied to 38 x 152 SAP Grade V4 member with a 20 mm diameter hole drilled through its centre. The proportion of the permanent load to the imposed short duration load is 2. Effective area = A = (152 -20) . 38 = 5016 mm2 Grade stress = p = 2,2 MPa

(SANS 10163-2: Table 3)

Duration of load factor = k1 = (2 + 1) / (1, 0 . 2 + 0, 66 . 1) = 1, 128 Factors k2, k3 & k4 all equal 1, 0 Allowable tensile stress = ft = p . k1 . k2 . k3 . k4 = 1, 128 . 1 . 1 . 1 . 2, 2 = 2,48 MPa Total tensile force = P < = A . ft = 5016 . 2, 48 = 12 439 N = 12,439 kN

Note that had the hole been drilled through the 38 mm face i.e. a 152 mm long hole there would have been substantially less wood left in the section and the equation would have been: Effective area = A = 152 . (38 – 20) = 2 736 mm2 with obvious reductions in order results.

2.3.9

Compression member design

Members which are axially loaded parallel to the grain with purely compressive forces are compression members, these could be struts or columns. Forces tend to shorten the member. The failure force is a function of the length and cross-sectional dimensions. The strength of a compression member is usually written as a function of a slenderness ratio. This slenderness ratio for steel is defined as the ratio between effective length and radius of gyration whereas

timber column slenderness is described as the ratio between the effective length and the cross-sectional dimensions.

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A loaded timber compression member can fail in three ways depending on support conditions, length and dimensions: i.

A short thick member will have a low slenderness and when it fails the wood fibres buckle or fold together so that a ‘compression crease’ runs across the material. These creases may run diagonally or at 900 to the direction of the compressive force.

ii.

An intermediate slenderness column, loaded to failure will have local breaking and bending of wood fibres as well as sideways movement of central portion of the column.

iii.

A slender timber column will buckle sideways before any deformation of fibres can occur.

Elastic design principles are still valid for compression members:

σc =

P ≤ fc A

Where P = applied compression force A = cross-sectional area of the member

σ c = the compressive stress due to the design load fc = allowable compressive stress = p . k1 . k2 . k3 . k4 The Grade Compressive Stress is a function of the Effective Length, which in turn is a function of the End Support Conditions, and the Cross-sectional Dimensions. The end support conditions play an important role in the way a column buckles. The greater the fixity at the ends of a column the shorter the effective length becomes and thus the greater the failure or permissible stress becomes. The buckling mode or buckled shape gives us a good idea of the Euler effective length of a column. The Euler column formula assumes straight, homogeneous columns that have linearelastic properties with concentric loading.

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Timber does not satisfy most of these conditions and the ends can seldom be fully fixed, in addition effective lengths determined by Euler’s buckling formula were found to be too conservative it was therefore modified to the Perry-Robertson type formula now found in SANS 10163-2 for concentrically loaded pin-ended columns:

pc =

p cy + (η + 1) p e 2

 p cy + (η + 1) p e    − p cy ⋅ p e 2   2

−

Where Pc = the grade compressive stress. Pcy = the grade compressive stress parallel to the grain in a short strut, which is calculated at a theoretical value of slenderness ratio λ approaching zero in MPa

η = the eccentricity coefficient = 0, 002 λ Where: slenderness ratio = λ = l e / r Where: l e = effective length in mm r = radius of gyration in mm Pe = the Euler stress =

π 2 ⋅ Ee λ2

Where: Ee = effective modulus of elasticity = E / 2,22 Where: E is the average modulus of elasticity. In the case of rectangular sections,

r=

b

12

Where: r = radius gyration in mm. b = least dimension of the rectangular section in mm. In the case of round sections. 78

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r=

D 4

Where: D is the least diameter of the pole in mm. When determining the slenderness ratio l e / b or l e / D, use the least dimension of the section i.e. b is the smallest lateral dimension of a rectangular section and D is the least diameter of a pole. See Tables 4-6 in SANS 10163-2. According to Burdzik column or compression member design procedure is: i.

Determine the maximum axial compressive force on the member

ii.

Establish the support conditions, draw the buckled mode of the column and determine the effective length about the two major axes.

iii.

Assume that the most economical member can be found if the slenderness ratios about the two major axes are the same. Find the ratio between b, the width, and d, the depth.

iv.

Assume a value for the permissible stress, divide the force by the permissible stress and thereby determine a required cross-sectional area. Use the ratio between b and d and find the dimensions.

v.

Use the dimensions found in iv. to ascertain the grade stress. Determine the permissible stress by multiplying the grade stress by the k factors.

vi.

If the new permissible stress is more than 20% greater or less than the estimated permissible stress of iii. find the average between the estimated and the new permissible stress. Repeat the later part of step

vii.

And the whole of step

viii.

Until the permissible stress is within tolerances.

ix.

Standardise one of the dimensions and calculate and standardise the other. Check that the stresses due to the load are less than the permissible stress.

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Figure 2.1 shows some of the end conditions for a column or compression member. On the left of each sketch the actual member is shown indicating the fixity and on the right of the sketch the buckled shape and the Euler’s effective length are shown. These values are based on Euler’s assumptions that: i.

Yield stress is ignored and infinite elasticity is assumed.

ii.

The column is perfectly straight and loaded absolutely concentrically.

iii.

The member is stable and in equilibrium.

See Table 15 in SANS 10163-2 for the design effective lengths which differ from Euler’s effective lengths as discussed above.

2.3.9.1 Example of compression member design Determine the dimensions of a 3 m long column using SAP V4. The column is fixed at one end and hinged at the other. It is unlikely that the moisture content will exceed 120 g/kg. A permanent load of 7 kN and a short duration load of 15 kN is applied.

i.

Axial load 80

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ii.

Effective length

= 0, 85 . 3 000 (Table 15 SANS 10163-2) = 2 550 mm

iii.

Buckling length about both axes is the same, therefore b = d

iv.

Assume permissible stress = 3, 0 MPa Required sectional area

= 22 000 / 3, 0 = 7 333 mm2

with b = d we can take the root of the area to find the value of b

b = 85, 6 mm v.

Slenderness ratio

vi.

Grade stress = 2, 23 (Table 4, SANS 10163-2)

= 2 550 / 85, 6 = 29, 8 (say 30)

k1 = (7 + 15) / (1,0 . 7 + 0,66 . 15) = 1, 302

k2 to k5 all equal 1,0 fc = 2, 23 . 1, 302 . 1 . 1 . 1 . 1 = 2,90 MPa vii.

The difference between assumed and calculated stresses is < 20% so from Appendix B in SANS 10163-2 the thickness b is taken as 76 mm.

viii.

Slenderness ratio Grade stress

= 2 550 / 76 = 34 = 1, 77

Permissible stress Width Standardise

d

= 1, 302 . 1, 77 = 2, 30 MPa = 22 000 / (2,30 . 76) = 125, 9 mm

d = 152 mm

Final dimensions = 76 x 152 mm

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2.4

SPECIFIC OUTCOMES

The learner will be able to determine the factors for duration of load, load sharing, type of structure, quality of fabrication, moisture content and creep deflection after working though this section.

2.5

REFERENCES 1. BURDZIK, Prof. WMG: Personal communications. Pretoria 1996. 2. SANS 1783: Sawn softwood timber 3. SANS 10163-2 : The structural use of timber. Part 2: Allowable stress design.

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2.6

QUESTIONS FOR SELF-EVALUATION

Try these questions before turning the page to look at the solutions.

2.6.1

Calculate the value of the k1 factor for bending stresses for a uniform beam, 6 m long,

with a self weight of 1,5 kN/m and an imposed point load of 6 kN in the centre of the beam. Ignore wind loads.

2.6.2

Calculate the value of the creep factor for bending stresses for the 8 m centrally

loaded long beam shown in the sketch.

2.6.3

Calculate the size of a tension member for a truss with a self weight of 10 kN and an

imposed load of 8 kN. The trusses are at 600 mm centres and are made of M6 SAP. It is assumed that joints will deform plastically producing a load-deflection curve plateau which ensures that load sharing can occur despite the fact that the trusses will have similar stifnesses.

2.6.4

Give five factors which affect lateral buckling of a timber beam.

2.6.5

Determine a suitable cross-section for a 4 m long column of Stocklam which is hinged

at both ends but laterally supported 2 m above the bottom end along the minor axis and along the major axis it is hinged at one end and fixed

at the other. The long duration load is 30

kN and the short duration is 40 kN. Moisture content will remain below 10%.

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2.7

SOLUTIONS TO THE SELF-EVALUATION QUESTIONS

Having completed these questions the learner should now realise that a comprehensive working knowledge of the prescribed design codes is essential even if it is an open book exam.

2.7.1

wL2 1,5 ⋅ 6 2 = = = 6,75 kN .m 2 8

Moment due to self weight

wL 6,6 = = 9 kN .m 4 4

Moment due to imposed load =

k1

2.7.2

=

6,75 + 9 = 1,241 1,0 ⋅ 6,75 + 0,66 ⋅ 9

In this case the effects of the loads are not equal. Deflection due to self weight

=

5 ⋅ w ⋅ L4 106,7 = 384 ⋅ E ⋅ I E⋅I

Deflection due to imposed load

=

1 ⋅ w ⋅ L3 53,3 = 48 ⋅ E ⋅ I E⋅I

(E.I cancels out and may be omitted from the question for d) d1 =

2.7.3

106,7 + 53,33 = 1,36 0,6 ⋅ 106,7 + 1,0 ⋅ 53,33

Grade stress = 3, 6 MPa

(SANS 10163-2 : Table 3)

k1 = (10 + 8) / (1, 0 . 10 + 0, 66 . 8) = 1,178 k2 = 1, 15 k3 = 1, 0 and k4 = 1, 0 ft = 1, 178 . 1, 15 . 1, 0 . 1, 0 . 3, 6 = 4, 87 MPa 84

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Area required = p / ft = (18 . 103) / 4, 87 = 3 696 mm2 Using a standard width of 38 mm: Depth = 3 696 / 38 = 97 mm Therefore use the standard depth of 114 mm giving a member of 38 x 114 mm

2.7.4

The solution was found in the commentary of Clause 7.2.3.2 of SANS 10163-2. (This

illustrates the need for the learner to thoroughly acquaint himself with all the content of the relevant codes) i.

The depth-to width ratio.

ii.

The geometrical and physical properties of the beam section.

iii.

The nature of the applied load.

iv.

The position of the applied loading with respect to the neutral axis of the section.

v.

The degree of restraint provided at the vertical supports and at points along the span.

2.7.5 i.

Axial load = 30 + 40 = 70 kN

ii.

Effective lengths: Minor axis = 1, 0 . 2 000 = 2 000 mm Major axis = 0, 85 . 4 000 = 3 400 mm

iii.

Set slenderness ratio on major axis = slenderness ratio on minor axis 2 000 / b = 3 400 / d d = 1, 7 b

iv.

Assume permissible stress = 4 MPa 85

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v.

Area = Force / permissible stress = 70 000 / 4 = 17 500 mm2 b . 1, 7 . b = 17 500 therefore b = 101, 5 mm Therefore d = 1, 7 . 101, 5 = 172, 6 mm New slenderness ratios: Minor axis = 2 000 / 101, 5 = 19, 7 (say 20) Major axis = 3 400 / 172, 6 = 19, 7 (say 20)

vi.

Determine the permissible stresses: Grade stress = 4,32 MPa (Table 4 SANS 10163-2) k1 = (30 + 40) / (1,0 . 30 + 0,66 . 40) = 1,24 (for both axes) k2, k3, k5 = 1,0 k4 = 1,05 (Stocklam complies with an SANS code) Permissible stress = 1,24 . 1 .1 . 1,05 . 1 . 4,32 = 5,62 MPa This is > 20% more than the estimated stress.

vii.

New estimated permissible stress = (5,62 + 4) / 2 = 4,81 MPa

viii.

Area required = 70 000 / 4,81 = 14 553 mm2 b . 1,7 . b = 14 553 therefore b = 92,5 mm and d = 157, 3 mm New slenderness ratios: Minor axis = 2 000 / 92,5 = 21,6 (say 22) Major axis = 3 400 / 157,3 = 21.6 (say 22) 86

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ix.

Determine permissible stresses Grade stress = 3,78 MPa Permissible stress = 1,24 . 1 . 1 . 1,05 . 3,78 = 4,92 MPa Difference between new estimated and calculated stresses < 20% New dimensions: 1,7 . b2 = 70 000 / 4,92 therefore b = 91,48 mm Standardise from Appendix B : b = 100 mm New slenderness ratio about minor axis = 2 000 / 100 = 20

x Determine permissible stresses Grade stress = 4,32 MPa Permissible stress = 1, 24 . 1 . 1 . 1,0 5 . 4, 32 = 5, 62 MPa Depth of member = 70 000 / (5,62 . 100) = 124,6 mm Standardise 100 + 33,3 = 133,3 mm

Final dimensions = 100 x 133, 3 mm

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2.8

TUTORIAL

2.8.1 With reference to SANS 0163-2, discuss the effects of moisture content and the duration of load on structural timber members.

(5)

2.8.2 Briefly compare allowable stress and Limit-states design methods. 2.8.3 Calculate the k1 factor for bending stresses for a 7 m long uniform section timber beam with a self weight of 3 kN/m. The beam has two points’ loads of 9 kN acting at the third points.

(10)

2.8.4 Calculate the dimensions for a concentrically loaded, 3,6 m long column with a permanent load of 18 kN, medium term load of 10 kN and a short duration load of 16 kN. M8 SAP timber, with a moisture content of 165 g/kg is being used in structure where both ends are hinged.

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MODULE 2 STUDY NOTES CHAPTER 3 BASIC STRUCTURAL ELEMENTS & JOINTS

CONTENTS

PAGE

3.1

OBJECTIVES

91

3.2

BEAM DESIGN

91

3.2.1 Permissible stresses

91

3.2.2 Laterally supported beam design example

92

3.2.3 Laterally unsupported beams

94

3.2.4 Example of laterally unsupported beam design

98

3.3

JOINTS

100

3.3.1 Allowable joint force and design loads

100

3.3.2 Butt-joints in compression

102

3.3.3 Density groups

102

3.3.4 Fasteners

102

3.3.5 Loads at various angles to the grain

103

3.3.6 Nails

103

3.3.6.1 Example of joint design with nails 3.3.7 Bolts

105 106

3.3.7.1 Example of bolted joint design

108

3.4

TIMBER SECTION PROPERTIES

110

3.5

SPECIFIC OUTCOMES

114

3.6

REFERENCES

115

3.7

QUESTIONS FOR SELF-EVALUATION

116

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3.8

SOLUTIONS TO THE SELF-EVALUATION QUESTIONS

117

3.9

TUTORIAL

120

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3.1

OBJECTIVES

The objective of this section is: 1. To show the student how to apply the allowable stress method of design to design basic ties, struts and beams. 2. To explain the use of different types of connections in timber and to enable the student to determine the allowable joint forces and design loads.

3.2

BEAM DESIGN

The design of timber beams does not only needs to consider the usual bending, shear stresses, deflection and camber, but also the bearing area at the supports in relation to the allowable compressive stresses of the timber. The basic principles of timber beam design hold true for all timber products. This means that, as with other materials such as concrete and steel, timber beams can be designed to take a number of different forms and compositions. For example, the latest trend in South Africa is to use “I” beams which are composite (built-up) beams with solid timber flanges and plywood webs. Laminated timber beams have been in service for some time in this country.

3.2.1 Permissible stresses Calculate bending stress = σ b =

6⋅M ≤ fb b ⋅ d2

Where: M = applied moment in MPa b = width of member in mm d = depth of member in mm

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fb = Pb . k1 . k2 . k3 . k4 in MPa Calculate shear stress = σ v =

1, 5 ⋅ V ≤ fv b⋅d

Where M = applied shear force in N b = width of member in mm d = depth of member in mm fv = Pv . k1 . k2 . k3 . k4 in MPa Calculate bearing stress = σ cp =

R ≤ f cp b⋅w

Where R = reaction force in N b = width of member in mm w = width of support in mm fcp = Pcp . k1 . k2 . k3 . k4 in MPa (Allowable compressive stress perpendicular to the grain)

3.2.2

Laterally supported beam design example

Design a beam to a span 5 m with a permanent load of 1,5 kN/m, it is laterally supported on its compression edge along its full length. A 70 mm wide Stocklam is required. Calculate the required depth of the beam, final long term deflection and test for shear. (W Burdzik)

Referring to Table 3 SANS 10163-2: Grade stress in bending = pb = 5,2 MPa

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Allowable bending stress = fb = 5,2 . 1 . 1 . 1 . 1 = 5,2 MPa Allowable shear stress = fv = 0,7 . 1 . 1 . 1 . 1 = 0,7 MPa Allowable bearing stress = fcp = 2,1 . 1 . 1 . 1 . 1 =2,1 MPa

Design for bending. Test for shear and bearing: Bending stress = σ b =

6.M ≤ fb b ⋅ d2

Moment M = 1,5 . 52 / 8 = 4,69 kN.m Therefore d2 =

6 ⋅ 4, 69 ⋅ 10 6 so d = 278 mm 5, 2 ⋅ 70

Standardise the depth (Clause b.2.2 SANS 10163-2) 9 . 33,3 = 299 mm

Required support width: Design stress

≤ Allowable stress

R ≤ f cp b⋅w R = 1,5 . 5 / 2 = 3,75kN R / w . b = 2,1 therefore w = 3,75 . 103 / 2,1 . 70 = 25,5 mm

Test for shear stress: Design stress

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1,5 . V ≤ fv b⋅d 1,5 ⋅ 3, 75 ⋅ 10 3 ≤ 0, 7 70 ⋅ 299 0, 269 ≤ 0, 7 The shear is therefore satisfactory.

Calculate final deflection:

Design deflection = Elastic deflection . d1 . d2 ≤ ∆

∆ ALLOWABLE

ALLOWABLE

= L / 200

= 5000 / 200 = 25 mm (CI .4.1.2 SANS 1 0163 − 2 )

/ = b . h 3 / 12 = 70 . 299 3 / 12 = 155, 93 . 10 3 mm 4 5 ⋅ w ⋅ L4 5 ⋅ 1, 5 ⋅ 5000 4 = 10,04mm = 384 ⋅ E ⋅ I 384 ⋅ 7800 ⋅ 155, 93 ⋅ 10 WD + W1 + WW 1,5 + 0 + 0 = = 1.667 d1 = C dD .WD + C dl .W1 + C dW .WW 0,6.1,5 + 1,0.0 + 1,0.0

∆ ELASTIC =

d2 = 1,0 (Table 9 SANS 10163-2) Assume the Equilibrium Moisture Content applies, this is less than 17% anywhere in Southern Africa.

∆ = 10, 04 . 1, 667 . 1, 0 = 16, 74 mm ≤ ∆ ALLOWABLE Member size is 299 x 70 Grade 5 stocklam

3.2.3

Laterally unsupported beams

The lateral stability of timber beams is a function of the ratio between depth and width and the effective length of the beam. It is an aspect of design that should always be examined. Table 10 in SANS 10163-2 shows h / d ratios. In addition to a beam’s h / d ratios there are other factors which can introduce torsional effects which may results in buckling:

 Geometrical and physical properties of the beam. 94

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 The nature of the applied loadings.  Position of the applied loads with respect to the neutral axis.  The degree of lateral support at supports and along length of beam. A Slenderness Factor and a critical Slenderness Factor are used in timber beam design: le ⋅ d and C k = b2

Cs =

3⋅E 5 ⋅ Pb

Where Cs = Slenderness Factor CK = Critical Slender Factor l e = Effective length as multiple of Lu obtained from Table 3.1

Lu = Laterally unsupported length of beam d = Depth of beam b = Width of beam E = Average modulus of elasticity Pb = Grade Bending Stress Table 3.1: Effective l e for rectangular beams Type of beam span & nature of load

Effective length l e

Single span beam, load concentrated at centre

1,61 Lu

Single span beam, uniformly distributed load

1,92 Lu

Single span beam, equal end moments

1,84 Lu

Cantilever beam, load concentrated at unsupported end

1,69 Lu

Cantilever beam, uniformly distributed load

1,06 Lu

Single span or Cantilever beam, any load

1,92 Lu

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The designer has three options, firstly Figure 4 in SANS 10163-2 gives a modified bending stress, Pbm which may be read directly or secondly to use Table 3.2 of this study guide, which also gives Pbm, both of these were derived from formulae in Clause 7.2.3 of SANS 10163-2 and both simplify the calculation process to the form: Allowable bending stress = fb = Pbm . k1 . k2 . k3 . k4 . k5 The critical Slenderness factor has already been calculated in Table 3.2 of this study guide so all that remains is to determine the values of cs and l e , then read off Pbm from Table 3.2 where values above the thick line are intermediate beams (where 10 < Cs ≤ Ck) and those below the thick line are long beams (Ck < Cs < 52) Short beams are those where Cs < 10 and Pbm so no modification is needed.

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Table 3.2: Modified Grade Bending Stress Pbm for Slender Beams

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Deflection, shear and bearing all are designed for in the same way as laterally supported beams.

3.2.4

Example of laterally unsupported beam design

Calculate the size of a Grade 8 member spanning 6 m with a tiled-roof load, tile mass is 50 kg/m2. The members are at an angle of 250 and are spaced at 600 mm centres. There is no lateral bracing. Estimate self weight of beams and battens at 0,3 kN/m2. The short duration roof load for inaccessible roofs is 0,5 kN/m2 (SANS 10160).

Factoring loads as required by SANS 10160:

Tiles = 0,6 . 0,5 / cos 250 = 0,33 kN/m Beam = 0,6 . 0,3 = 0,18 kN/m Self weight = 0,33 + 0,18 = 0,51 kN/m Imposed load = 0,6 . 0,5 = 0,3 kN/m Total load = 0,51 + 0,3 = 0,81 kN/m Bending moment = M = 0,81 . 62 / 8 = 3,645 kN/m Dimensions are unknown, so assume a value for modified bending stress:

Pbm = 7,0 MPa K1 = (0,51 + 0,3) / (1,0 . 0,51 + 0,66 . 0,3) = 1,14 Fb = Pbm . k1 . = 7,0 . 1,14 = 7,98 MPa Set induced stress ≤ permissible stress

6⋅M ≤ fb b ⋅ h2 98

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Set h = 3 . b (From Table 10 : SANS 10163-2)

6 ⋅ 3,645 ⋅ 10 6 ≤ 7,98 9 ⋅ b3 67 ≤ b mm with a span of 6 m it is almost a foregone conclusion that glulam is used so we can set b = 70 mm then h = 3 . 70 = 210 mm, standardise to 7 . 33,3 = 233 mm From Table 3.1 (or SANS 10163-2 Table 11): l e = 1,92 . LU = 1,92 . 6 000 = 11520 mm

Calculate the slenderness factor:

le ⋅ d 11520 ⋅ 233 = = 23, 4 2 b 70 2 pro rata from Table 3.1 Pbm = 6,708 MPa

CS =

f b = Pbm ⋅ k1 = 6, 708 ⋅ 1,14 = 7, 647 MPa

σb =

6 ⋅ 3, 645 ⋅ 10 6 = 5,755 MPa 〈 f b therefore size is acceptable. 70 ⋅ 233 2

Check shear:

1, 5 ⋅ V ≤ fv b⋅d f v = p v ⋅ k1 = 1,0 ⋅ 1,14 MPa 1, 5 ⋅ 2, 43 ⋅ 10 3 Induced horizontal shear stress = σ v = = 0, 223 MPa 70 ⋅ 233

σ v 〈 f v which is satisfactory Check bearing stresses at support:

Design stress

≤ Allowable stress

R ≤ f cp 3,4 ⋅ 1,14 = 3, 87 MPa b⋅w R / w ⋅ 70 = 3, 87 therefore w = 2, 36 ⋅ 10 3 / 3, 87 ⋅ 70 = 8,7 mm, 99

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This is minimal so a nominal 38 x 75 mm wallplate will be more than adequate. Check deflection:

From SANS 10163 Table 1: ∆ ALLOWABLE = L / 250 = 24 mm Design deflection = elastic deflection ⋅ d1 ⋅ d 2 ≤ ∆ ALLOWABLE / = b ⋅ h 3 / 12 = 70 . 2333 / 12 = 73, 788 . 10 6 mm 4

∆ ELASTIC = d1 =

C dD

5 ⋅ w ⋅ L4 5 ⋅ 0, 81 ⋅ 6000 4 = 17, 64 mm = 384 ⋅ E ⋅ I 384 ⋅ 10500 ⋅ 73, 788 ⋅ 10 6 WD + WI + WW 0, 51 + 0, 3 = = 1, 34 ⋅ WD + WI + C dW ⋅ WW 0, 6 ⋅ 0, 51 + 1, 0 ⋅ 0, 3

d2 = 1,0 (Table 9 SANS 10163-2) Assume that EMC applies. ∆ = 17, 64 . 1, 34 . 1, 0 = 23, 64 mm ≤ ∆ ALLOWABLE

Member is 233 x 70 Grade 8 Stocklam on a 38 x 114 mm wallplate.

3.3

JOINTS

The term fastener, as used in SANS 10163-2, generally means screw, bolt or connector other than nails or nail-plates. Designers must be aware of the possible causes of joint failure so that these problems can be accommodated. Some of the more common problems include: joint slip or joint deformation, timber shrinkage, fungal decay and fastener corrosion. These potential problems are largely attributable to moisture. If adequate precautions are taken to ensure that the timber remains below 17% moisture content, preservatives (and or sealants) are applied and corrosionresistant (galvanised) fasteners are used, then most hazards are avoided.

3.3.1

Allowable joint forces and design loads

The basic joints according to SANS 10163-2 (Appendix C) is taken as the lesser of the following two values: 100

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a) The average ultimate force divided by 3 or the 5% lower exclusion limit divided by 2,22; or b) The average force at joint slip (timber to timber) of 0,76 mm divided by 1,6.

Tables 22, 23, 25, 27, 28, 31, 32 and 33 in SANS 10163-2 show basic joint forces for various fasteners, nails and density groups. The allowable joint force = basic force multiplied by the relevant modification factors (k1 to k5). The load acting on an individual fastener is: F =

(Fdx

+ Fmx ) + (Fdy + Fmy ) 2

2

Where Fdx = Fx / n Fdy = Fy / n Fmx = M . Ym / /p Fmy = M . Xm / /p Fx = force in the X direction Fy = force in the Y direction n = Number of fasteners M = Moment acting as a point Xm, Ym = x and y co-ordinate distances to the nail furthest from the joint centroid. /p = /x + /y = Polar moment of area of the nail group. /x = Second moment of area of the nail group about the X axis. /y = Second moment of area of the nail group about the Y axis.

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3.3.2

Butt-joints in compression

Butt-joints acting in one plane are fairly common in trusses. The maximum allowable gap is 1 mm between members. It may be assumed that 50% of the compressive forces acting in the member are transferred to the joint fasteners. The allowable compressive stresses should not be exceeded.

3.3.3

Density groups

The tables mentioned in 3.3.1 above give basic joint forces for the two density groups. Stress grades 4 and 5 generally fall under density group D2 and stress grades 6 to 16 fall under density group D1.

3.3.4

Fasteners

When groups of screws, bolts or connectors are used in a joint refer to SANS 10163-2: CI.8, 1.5. See the following example:

Six 12 mm bolts form two rows of three parallel to the grain in a V4 SAP 38 x 114 mm tension member subject to a permanent load of 8 kN, a short duration load of 5 kN. Will the bolts be strong enough for the load?

P6 = 6 . 2,7 . 1,0 = 16,2 kN (From table 21 and 27 : SANS 10163-2) K1 = (8 + 5) / (1,0 . 8 + 0,66 . 5) = 1,15 Ft = 2,2 . 1,15 . 1,0 . 1,0 . 1,0 = 2,53 MPa A = 38 . 114 = 4 332 mm2 P = 4 332 . 2,53 / 1000 = 10,96 kN

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The allowable joint force 16,2 kN > 10,96 kN load therefore this joint will be satisfactory.

3.3.5

Loads at various angles to the grain

The angle of load to the grain affects the strength of fasteners. Coach screws, bolts, toothed split-ring and shear-plate connectors which are subject to forces which are not parallel to the grain need to be modified by calculating the allowable force N1 for any angle against the grain. This is necessary because the compressive strength of wood differs with respect to the direction of the grain. The bearing pressure of a fastener meets relatively high resistance in the direction of the grain and a much lower resistance perpendicular to the grain. Nails, staples and wood screws are exempt from this because in their case, joint slip is a more important factor for failure than ultimate strength. The allowable force is determined by Hankinson’s formula:

N1 =

P⋅Q P ⋅ sin θ + Q ⋅ cos 2 θ 2

Where: P = allowable force for the fastener parallel to the grain in N. Q = allowable force for the fastener perpendicular to the grain in N

θ = the acute angle of load direction and the grain longitudinal direction.

SANS 10163-2 Clause 8.1.6 gives a fairly descriptive commentary and set of rules to be applied which will not be repeated here.

3.3.6

Nails

A variety of nails are available but we will only concern ourselves with plain wire nails complying with SANS 820: Mild steel nails. 103

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The minimum spacing of nails should be 10.D along the grain and 5.D across the grain and the edge distance should be 5.D for members subjected to axial forces and 10.D for members subjected to bending. (See CI.8.2 SANS 10163-2). By staggering the nails (or any fasteners), more can be placed in a given area and still comply with the requirements of the Code. See Figure 3.1.

Allowable shear load that each nail can carry is: Pv = Pa / (50 . D2) OR Where holes are pre-drilled (0,5 to 0,8.D (maximum) holes) Pv = Pa / (12,5 . D2) Where D = nominal diameter of nails in mm. Pa = basic force in single shear in kN. (Table 22 SANS 10163-2) The allowable strength calculated in this manner may be increased by 25% if metal sideplates are used. Read the commentary in SANS 10163-2 Clause 8.2.1 with regard to withdrawal forces. Baird and Ozelton state no load in withdrawal should be carried by a nail driven into the end grain of timber. They also give a formula for the ultimate withdrawal load in side grain: Fu = 47,6 . p2.5.

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Where Fu = Ultimate nail in N/mm of penetration. p = The specific gravity of the wood. {see below} D = The diameter of the nail in mm. Specific gravity is now called Relative density and it is the density of the substances in question divided by the density of water. The average density, given in Table 1 of SANS 10163-2, divided by 1 000 will give the required value. Note: The basic load should be taken as Fu / 6 and may be applied to both green and

seasoned timber. Whether this formula for ultimate withdrawal load is used or not is a matter of designer’s choice, it is not mentioned in the SANS codes for Allowable Stress or Limit-states design. These codes of practice are not comprehensive design documents, they are guidelines for minimum standards of practice and as can be seen from reading them there are many instances where the respective codes refer the reader to other literature.

3.3.6.1 Example of joint design with nails

Design a tension splice in a 38 x 114 mm V4 SAP member which has a permanent load of 4 kN and a short duration load of 3kN.

Assume that splice plates are 25 mm thick V4 SAP on each face of the member to be joined then: Nail length = 25 . 2 + 38 = 88 mm Use a 4,0 mm diameter nail 90 mm long Basic force = 0,23 kN/nail for D2 timber (Table 22 SANS 10163-2) K1 = 4 + 4 + 3 / (1 . 4 + 0,66 . 3) = 1,17 Allowable force per nail = 1,17 . 0,23 = 0,269 kN in single shear 105

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therefore in double shear force = 0,538 kN Number of nails required = Load / Allowable force = (4 + 3) / 0,538 = 13 say 14 nails. Spacing across the grain = 10 . D = 40 mm Spacing parallel to the grain = 5 . D = 20 mm Using staggered spacing the splice needs to be 480 mm long as illustrated in Figure 3.2.

This is because there are 6 x 40 mm spaces on each side in order to achieve the necessary spacing.

3.3.7

Bolts

Bolts should have washers under both head and nut to facilitate fixing and to spread the load. Clause 8.4.4 of SANS 10163-2 makes provision for modification of the basic forces when large washers are used. The minimum sizes of washers given as 3.D diameter and 0,3.D thickness by SANS for large washers are actually recommended by Baird & Ozelton as the minimum for any washers. (See Appendix A of SANS 10163-2) SANS is more conservative than the overseas standards with regards to spacing of bolts. 4.D is the general spacing, with 2.D allowed as edge distance on axially loaded members (1,5.D overseas), 4.D edge distance for members subjected to bending and 7.D from ends of members.

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Bolt holes should be the same diameter as the bolt with an allowable tolerance of + 1,0 mm. See clause 8.4.6 SANS 10163-2, with regard to joint slip, hole diameter and creep deflection. The stress modification factors k1 to k5 and the deflection factors d1 and d2 must be applied to bolts. The allowable load per unit area for bolts = Pv = Pa / (16. D2) Where D = Diameter of the bolt in mm. Pa = basic force in kN. (Tables 27 & 28 SANS 10163-2)

Note: It can be seen from these tables that both timber thickness and density group have a

direct bearing on the allowable bolt load. As a rule the minimum required bolt length will be the thickness of the members to be joined plus the bolt diameter, which is normally slightly larger than the nut thickness. For example two 50 mm members joined by a set of M12 bolts will require at least (2 . 50) + 12 = 112 mm long bolts, but the standard sizes are 110 and 120 mm so the joint will need 120 mm long bolts. Bolts and washers should be corrosion resistant, hot-dip galvanising is the most commonly used method and in certain circumstances stainless steel may be needed. Do not mix types, or protection of metals or electrolytic corrosion may occur. Hankison’s Formula may be used to calculate load capacities of the bolts when loads are between 00 and 900 to the grain. SANS do not suggest a method of determining joint shear stress values but Breyer recommends this formula: t=

1,5 ⋅ V b ⋅ de

Where t = Shear stress in MPa. V = Shear stress in kN. 107

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b = Breadth or thickness of the member in mm. de = Distance from unloaded edge to the centre of the nearest bolt (or edge of split-ring or circular toothed connector) in mm. Stalnaker & Harris have a variation of this formula, adapted from a beam shear formula. However it gives different results to Breyer’s equation. Breyer’s formula is more conservative and therefore preferable.

3.3.7.1 Example of bolted joint design

A 152 x 38 mm V4 SAP joist is bolted to a 114 x 38 mm V4 SAP column. The permanent load is 3,6 kN and an imposed load of 6kN is applied for 8 hours every day in a sheltered area. Design the bolted joint.

Assume 4 m 12 bolts will be used then from Table 27 in SANS 10163-2 basic force per bolt = Pb = 1,7 kN (Perpendicular to grain) k1 = 3,6 . 6 / (3,6 / 1 + 0,66) = 1, 270 (Clause 6.3.3) k2 to k5 each equals 1 in this case Allowable force per bolt = P = Pb . k1 = 1,7 . 1,27 = 2,16 kN 108

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Assume steel side-plates are used then Clause 8.2.2 applies: P = P . 1,25 = 2,70 kN Allowable force on joint from Clause 8.1.5 SANS 10163-2: PN = n . P . α = 4 . 2,70 . 0,99 = 10,69 kN Total load on joint = W = 3,6 + 6 = 9,6 kN The design is satisfactory so far because PN > W Check for shear:

Grade shear stress parallel to grain = Pv = 0,7 MPa (Table 3) Allowable shear stress = σ v = k1 . Pv = 1,27 . 0,7 = 0,889 MPa. Shear force = V = 1 . 3,6 + 0,66 . 6 = 7,56 kN. de = 80 mm (See figure 3.3) and b – 38 mm Shear stress = t = 1,5 . 7,56 . 103 / (38 . 80) = 3,73 MPa.

The actual shear stress exceeds the allowable so the joint will fail because the column is loaded parallel to the grain and it will not withstand the applied load. In addition owing to the restrictions on bolt spacing, it is not possible to put any more bolts into the connection. Some other method of connection will have to be tried on this joint or the column and/or beam sizes and possibly the grade will have to be changed. Changing the grade alone would not provide a sufficiently high enough allowable shear stress for this load.

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3.4

TIMBER SECTION PROPERTIES

Table 3.3: Rough-Sawn (Minimum Dimensions) SA Pine Section Properties

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Table 3.4: Nominal SA Pine Stocklam Section Properties (Cont.)

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Table 3.4 (Continued): SA Pine Stocklam Section Properties

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Table 3.4 (Continued): SA Pine Stocklam Section Properties

These tables were generated on a computer spreadsheet. You can easily produce your own table in the same way for a number of different reasons. For example, round pole section tables, PAR tables, etc. The more preparatory work you do now, the less tiresome the exam will be.

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3.5

SPECIFIC OUTCOMES

At the end of this section the learner will be able to: 1. Determine the permissible stresses and size of timber section required to resist tension, compression, bending and shear effects. 2. Design bolted and nailed joints in timber elements.

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3.6

REFERENCES

1. BURDZIK, Prof. WMG: Personal communications. Pretoria: 1996. 2. SANS 10160: The general procedures and loadings to be adopted in the design of buildings. 3. SANS 10163-2: The structural use of timber. Part 2: Allowable stress design. 4. Baird JA & Ozelton EC: Timber designer’s manual. (2nd Edition). Blackwell Science Ltd, Oxford, 1995. 5. Stalnaker JJ & Harris EC: Structural design in wood. Van Nostrand reinhold, New York, 1989 (Imperial units) 6. South African lumber Millers Association: SALMA Timber Manual. SALMA, Isando, 1995.

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3.7

QUESTIONS FOR SELF-EVALUATION

Try these questions before turning the page to look at the solutions. 3.7.1

Determine a safe span for a laterally supported beam carrying a 22 mm thick floor in a house. The joints are spaced at 600 mm centres, 38 x 200 mm Grade 5 SAP is used, moisture content is 12%.

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3.8

SOLUTIONS TO THE SELF-EVALUATION QUESTIONS

3.8.1 From SANS 10160:

UDL = 0,6 . 1,5 = 0, 900 kN/m Flooring = 0,6. 9,81 . 0,5 . 22 / 1 000 = 0,065 kN/m Self weight = 470 . 9,81 . 0,036 . 0, 197 / 1 000 = 0,033 kN/m Total self weight = 0, 065 + 0,033 = 0,098 kN/m Total load = 0.098 + 0,9 = 0,998 kN/m Modification factors:

k1 = 1,15 k3 = k4 = k5 = 1,0 k1 = (0,098 + 0,9) / (1,0 . 0,098 + 0,66 . 0,66 . 0,9) = 1,4421 Referring to Table 3 SANS 10163-2:

Grade stress in bending = Pb = 5,2 MPa Allowable bending stress = fb = 5,2 . 1,4421 . 1,15 . 1 . 1 = 8,62 MPa Allowable shear stress = fv = 0,7 . 1,4421 . 1,15 . 1 . 1 = 1,16 MPa Allowable bearing stress = fcp = 2,1 . 1,4421 . 1,15 . 1 . 1 = 3,38 MPa Design for bending. Test for shear and bearing: Bending stress = σ b =

6⋅M ≤ fb b ⋅ d2

Moment M = 0, 998 ⋅ L2 / 8 kN .m

8 ⋅ 8, 62 ⋅ 36 ⋅ 197 2 so L 〈 4011 mm 0, 998 ⋅ 6 set L 〈 4 000 mm

therefore L2 =

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Required support width:

Design stress ≤ Allowable stress

R ≤ f cp b⋅w R = 0,998 . 4 / 2 = 1,996 kN R / w . b = 3,48 therefore w = 1,996 . 103 / 3,48 . 36 = 16 mm No problems here, use a 38 x 76 mm wallplate. Test for shear stress:

Design stress ≤ Allowable stress 1,5 ⋅ V ≤ fv b⋅d 1,5 ⋅ 1,996 ⋅ 10 3 ≤ 1,16 36 ⋅ 197 0,422 ≤ 1,16 The shear is therefore satisfactory. Calculate final deflection:

Design deflection = Elastic deflection . d1 . d2 ≤ ∆ ALLOWABLE ∆ ALLOWABLE

= L / 300 = 4 000 / 300 = 13, 3 mm (Table 1 SANS 1 0160)

/ = b . h 3 / 12 = 36 . 197 3 / 12 = 22 936 119 mm 4 5 ⋅ w ⋅ L4 5 ⋅ 0, 998 ⋅ 4000 4 = = 18, 6 mm 384 ⋅ E ⋅ I 384 ⋅ 7800 ⋅ 22936119 WD + WI + WW 0, 9 + 0, 098 = = 1, 041 d1 = 0, 6 ⋅ 0, 098 + 1, 0 0, 9 C dD ⋅ WD + C dI ⋅ WI + C dW ⋅ WW

∆ ELASTIC =

d 2 = 1, 0 (Table 9 SANS 1 0163 − 2)

Because the allowable deflection is already exceeded, take another look at the span, using 13 mm as deflection: 118

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Set L4 =

384 . 7800 . 22936119 . 13 then L = 3 621 mm say 3, 6 m 5 . 0, 998 . 1, 041 . 1,0

Safe span is 3,6 m

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3.9

TUTORIAL

3.9.1

A 38 x 228 mm M8 SAP simply supported beam spanning 4,5 m has lateral support on the ends and the centre of the top edge. It carries a central point load.

a) What is the moment resistance of the beam?

(5)

b) What is the maximum central point load W this beam can carry, based on bending strength only? 3.9.2

(5)

A 50 x 228 mm M6 SAP beam is supported by a 38 x 114 mm SAP wallplate. The reaction at the support is 20 kN. Show whether or not there is a possibility of bearing and shear failure.

3.9.3

(5)

A 52 x 114 mm M5 SAP has an axial tensile load of 19 kN. It has a bolted joint in the centre which has steel side-plates. Design the bolted joint to resist the axial force. (10)

3.9.4

A tension splice in a 38 x 152 mm V4 SAP member with a permanent load of 5 kN, a medium duration load of 2 kN and a short duration load of 1,5 kN is nailed together with 25 mm thick VS splice plates on both sides. Design the nailed joint, taking withdrawal forces into account.

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MODULE 2 STUDY NOTES CHAPTER 4 FORMWORK CONTENTS

PAGE

4.1

OBJECTIVES

122

4.2

FORMWORK

122

4.2.1 Loads acting on formwork

122

4.2.1.1 Example of formwork load calculation

124

4.2.2 Timber formwork materials 4.2.3

125

Timber formwork design

128

4.2.3.1 Wall formwork

128

4.2.3.1.1 Example of wall formwork design

128

4.3

FALSEWORK

136

4.4

SPECIFIC OUTCOMES

139

4.5

REFERENCES

140

4.6

QUESTIONS FOR SELF-EVALUATION

141

4.7

SOLUTIONS TO THE SELF-EVALUATION QUESTIONS

142

4.8

TUTORIAL

148

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4.1

OBJECTIVES

The objective of this section is to study the loads & pressure acting on formwork and shutterboards.

4.2

FORMWORK

Shuttering is the part of the formwork which is in contact with the concrete. Formwork is the framework supporting the shuttering and falsework is the framework supporting the formwork. However, the terms formwork and shuttering are generally used interchangeably to describe all three terms. Overseas publications refer to sheathing when referring to the sheathing used in formwork, but we call it shuttering or shutter-panels or shutter-boards in Southern Africa.

4.2.1

Loads acting on formwork

SANS 10160 does not make any recommendations for loads on formwork, however both Parts 1 and 2 of SANS 10163 do mention formwork and scaffolding in their respective duration-of-load modification factors. There seems to be little consensus on what design method to use for formwork in South Africa. A common trend is merely to use safe-load tables published by scaffolding and formwork suppliers, these tables were originally designed by empirical and allowable stress methods. There seems to be little choice but to follow established practice overseas and adapt these methods to local conditions. Meuwese produced a comprehensive book on formwork design, for in-house use by Murray & Roberts, using allowable stress design principles and drawing heavily on American Concrete Institute recommendations. Unfortunately this work has not been updated and timber grades referred to no longer exist.

The following factors affecting pressure on formwork are adapted from Hurst.

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Densities can vary from 10 kN/m3 for light-weight concrete to as much as 40 kN/m3 for heavy aggregate concretes but most concrete poured in Southern Africa is in the order of 25 kN/m3. Note that density has no influence on hydrostatic pressures. Height of discharge could have an influence on the formwork but if good concreting practice is adhered to this should not become an issue. Temperature of concrete should be in the range of 15 – 300C. When no precautions are taken to control concrete temperature (which is generally the case), the concrete pressure and the hydrostatic pressure must be increased by 3% for every 10C below 150C and decreased by 3% (with a maximum of 30%) for every 10C above 150C. Rates of placing concrete influence pressure on the shuttering and the pressure depend on the fluidity of the concrete where: Pressure on shuttering, Psh = 5 . Vp + 21 kN/m2 (Stiff, dry mix) Pressure on shuttering, Psh = 10 . Vp + 19 kN/m2 (Soft mix) Pressure on shuttering, Psh = 14 . Vp + 18kN/m2 (Pumping mix) Where Vp is the rate of placing in a wall in vertical m/hr, setting time is 5 hours, temperature is 150C, concrete density is 25 kN/m2 and internal vibration is used for compaction. See figure 4.1

Figure 4.1: Formwork pressure diagram (After Hurst) 123

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Workability of concrete has a similar influence on pressure to the rate of placing. Factors affecting workability include mix design, admixtures, fillers such as PFA, etc. Figure 4.1 shows some typical values. Imposed loads will include the 25 kN/m2 of the concrete, a 2,5 kN/m2 load for the workmen, the self-weight of the formwork, wind loads and any abnormal loads such as additional machinery or vehicles which may be used on the formwork during construction. Vibration should only be from internal poker vibrators used for compaction, external vibrators should not be used in the interests of good concreting practice. Figure 4.1 assumes that vibration ceases when compaction is achieved. Hydrostatic of concrete, Phs = Dc . hs kN/m2 Where: Dc = concrete density in kN/m3 hs = height from top surface of concrete downwards in m. Note: The height is NOT the top of the shutter! This is the highest possible value for hydrostatic pressure from the concrete and is only used when all other values are higher. The lowest value for pressure due to the concrete is always used in the design. See example below.

4.2.1.1 Example of formwork load calculation If a soft mix concrete, at 130C, is poured at a rate of 3 m/hr for a 4 m high wall, what hydrostatic pressure can be expected?

From Figure 4.1 (or by calculation): Psh = 10 . 3 + 19 = 49 kN/m2 (This is the Consistency Limit) hs = 49 / 25 = 1,96 m 124

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Phs = 25 . 4 = 100 kN/m2 (This is the Hydrostatic Pressure) The consistency Limit is the lowest value so it becomes the design pressure. Temperature adjustment of 6% increase is needed in this case, therefore the design pressure is 51, 94 kN/m2 and is distributed equally from the bottom of the shutter to a point 1, 96 m below the top of the concrete. From this point to the top of the concrete the value reduces to zero at the top surface.

4.2.2

Timber formwork materials

Although formwork can be made entirely of wood, in practice it is usual to have a combination. For example the shutterboard is a plywood, soldiers are fabricated proprietary steel units and walers are scaffold tubes. The plywood is mounted in a steel frame which is compatible with the soldiers using gutter-bolts (which are not subject to any forces of consequence), this makes it easy to replace the plywood when it is damaged. Usually five more pours are possible with each set of shutters, depending on how much care is taken when erecting and stripping the formwork. The steel units and scaffolding can be used for 30 or more pours. Obviously the industry gravitates towards the most economical solution. We will consider both the design of timber formwork and combinations of tubular scaffolding with timber formwork. The following definitions represent a small fraction of the terminology used in the formwork industry but they are the terms used in timber formwork:

Brace or strut – A diagonal prop stabilising the formwork. Button – A round mushroom-shaped piece of plastic used to blank off

unwanted

holes

that may have been drilled in shuttering in previous pours.

Deck or Decking – The horizontal shuttering boards used for the soffit of a slab. Double waler – Two walers used as a set where the members are not more than 30 mm apart between which ties are fixed. 125

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Joist – A horizontal member serving as a beam to support slab formwork. Kicker or sill – A strip of wood nailed or spiked to the blinding or foundation to hold the base of the shuttering in place.

Prop – A timber column used to support soffit formwork. Shutter – Boarding or planking used, at the interface with the concrete to be poured, to form the shape of the structure.

Soldier – A vertical member providing stiffening and bearing for shuttering. Spreader – A temporary piece of wood used in wall formwork to keep the shutters the correct distance apart.

Stud – A horizontal soldier. Tie – A metal device for holding shuttering in position and capable of withstanding hydrostatic pressures. Some ties are sleeved and recoverable with only the sleeve remaining in the concrete. Others remain cast in the concrete with only the bolt portion being recoverable.

Twin soldier – A double waler used in vertical position when studs rather than soldiers are used.

Waler or wailing – A horizontal member which supports the soldier and through which ties are fastened.

Shutterboards are usually plywood, the grade and thickness being a combination of personal choice and load requirements. Plywood or battenboard or similar board products are used for “off-shutter” finishes where appearances are important. Where work is not exposed to view such as the back of retaining walls, etc. 25 mm thick planking may be used. In practice however, this is quite an expensive option and contractors would be more inclined to use old damaged boards which still have strength but which are too warped or holed from use to be used for good finishes.

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The common shutterboard used to be 5-ply treated plywood, 22 mm thick (marine ply) but these tend to be expensive and there is a trend towards using lumberboards which is 19 mm thick with a core of 16 mm thick solid timber strips of wood glued together faced on both sides with a veneer, contractors seal the board with a polyurethane sealant for the first pour and then use shutter-oil for subsequent pours. There are other thicknesses of lumberboards which is also known by other names such as, battenboard and “VP Boards”. (See SANS 929: Plywood and composite board and Tables 35 – 37 SANS 10163-2). Another choice of material for shuttering is planed 25 x 228 mm SAP which is either buttjointed or tongued and grooved, but this also tends to be expensive and is only really viable in small work or for filling awkward areas. Soldiers are the vertical members directly behind the shutterboard, at spacings determined by design. Typically soldiers are made from V4 timber and dimensions start at 50 x 50 mm. Walers or waling are the horizontal members which carry the bulk of the load and ensure that the formwork stays in place during construction. Two (typically 50 x 114 mm) walers are placed 20 – 30 mm apart (double walers) and serve as a set through which ties are fastened. When a single waler is used its strength is compromised by having to drill holes through it for the ties. Ties are available in a greater variety of types but the simple threaded rod type with a hexagonal nut and a 50, 75 or 100 mm square washer (6 mm thick) is one of the most effective and easiest to use with timber shuttering. These ties come in different diameters from 10 to 30 mm but most commonly used is a 16 mm diameter high tensile rod which has a permissible tensile load of about 90 kN. The rod is separated from the concrete by placing it in a 20 mm diameter plastic tube which in turn acts as a control for the width of the panel being cast. It is advisable to use spreaders to prevent these plastic sleeves from buckling.

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4.2.3

Timber formwork design

4.2.3.1 Wall formwork According to Stalnaker & Harris the procedure (adapted for Southern African conditions) for designing formwork for walls is: i.

Calculate the design lateral pressure.

ii.

Check the shutterboard for bending, deflection and shear. (According to Hurd, shear normally does not need to be checked). Either the thickness of the shutterboard or the spacing of its soldiers may be adjusted. Availability of materials will determine which is adjusted. The deflection limit of l / 360 applies to both shutterboard and supporting members. If the shutterboard dimensions are fixed then the maximum allowable span of the shutterboard will equal the soldier spacing. If the soldiers are fixed then the bending check will need to determine the required section modulus of the shutterboard based on the bending stresses.

iii.

Design soldiers considering bending, shear and deflection.

iv.

Design walers for bending, shear and deflection.

v.

Choose ties and spacing.

vi.

Check bearing stresses of soldiers on walers and tie-holders on walers.

vii.

Design the lateral bracing of the formwork.

viii.

Design the falsework if required.

4.2.3.1.1

Example of wall formwork design

A 3,6 m high wall is to be cast at 1 m/hr with a soft mix of concrete at 15 0C, poker vibrators will be used to achieve compaction. There is no wind and only 3 labourers will be placing

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the 25 kN/m3. Tie used have a permissible tensile load of 80 kN, V4 SAP members are to be used for soldiers and walers. The shuttering will be sealed 19 mm lumberboard (battenboard) which has E = 4,6 kN/mm2 and allowable bending stress of 4,7 N/mm2 parallel to the grain. (Table 36, SANS 10163-2). Note: The dry strength (10 % moisture content) of plywood is about 30 % greater than its saturated strength, therefore if shuttering is not sealed the modulus of elasticity should be reduced by 30 %.

i.

Design lateral pressure: Psh = 10 . 1 + 19 = 29 kN/m2 (Consistency Limit) hs = 29 / 25 = 1,16 m Phs = 25 . 3,6 = 90 kN/m2 (Hydrostatic Pressure) Pmax = 29 kN / m2 (Design Lateral Pressure, the lowest value)

ii.

Check shutterboard for bending: Assume that bending stresses are parallel to the face grain and that the board is continuous over three or more spans.

M =

w ⋅ L2 = 0,1 ⋅ 29 ⋅ L2 = 2,9 L (Lateral Pressure Moment) 10

k1 = 29 / 0,66 . 29 = 1,515 fb = Pb . k1 . k2 . k3 . k4 = 4,7 . 1,515 . 1 . 1 . 1 = 7,12 MPa Z =

b ⋅ d2 1000 ⋅ 19 2 = = 60 167 mm 3 6 6

Mr = fb . Z (Allowable bending stress times section modulus) Mr = 7, 12 . 60 167 / 1 000 000 = 0, 428 389 kN.m Equate M = Mr to solve for L: 129

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2,9 . L2 = 0,428 389 Therefore L = 384 mm % (max. allowable span based on bending) Assume a 38 mm wide soldier will be used then the clear span will be Lc = L – 38 = 346 mm iii)

Check shutterboard for deflection: I =

b ⋅ d3 1000 ⋅ 19 3 = = 571 583 mm 4 12 12

Deflection is limited to 3 . L / 1000 ∆ allowable = 0,003 . L = 1,153 mm ∆ max =

0, 0064 . Pmax . Lc

4

E.I

=

0, 0064 . 29 . 346 4 = 1, 012 mm 4600 . 571583

Therefore deflection is satisfactory. iv)

Check shutterboard for shear: For Rolling or Interlaminar Shear Grade stress = Pv = 0,67 MPa (Table 36 SANS 10163-2) When used on the flat the induced shear in a board should not exceed the allowable interlaminar shear. Boards are treated as beams as far as formulae are concerned, therefore: Shear stress σ v = 1, 5

V ≤ Pv (Note: b = 1000 mm) b⋅d

Where shear for continuous beams (or panels) = V = 0,6 . w . Lc V = 0,6 . 29 . 0,346 = 6,02 kN

σ v = 1,5 . 6,02 . 103 / 1000 . 19 = 0,48 MPa ≤ Pv 130

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v)

Check soldiers for bending: A width of 38 mm has already been assumed for the soldier and the spacing L = 384 mm c

c

The imposed load on each soldier wis = 29. 0,384 = 11,147 kN/m

M =

wis ⋅ L2 = 0,1 . 11,147 . 0, 384 2 = 0,1644 kN .m 10

K1 = 11,147 / 0,66 . 11,147 = 1,515 fb = Pb . k1 . k2 . k3 . k4 = 4,0 . 1,515 . 1 . 1 . 1 = 6,06 MPa Assume a depth for the soldier of 114 mm Z =

b ⋅ d2 38 ⋅ 114 2 = = 82 308 mm 3 6 6

Mr = fb . Z (Allowable bending stress times section modulus) Mr = 6,06 . 82 308 / 1 000 000 = 0,4988 kN.m Therefore M ≤ Mr vi)

Check soldiers for deflection

I =

b ⋅ d 3 38 ⋅ 114 3 = = 4 691 556 mm 4 12 12

Assume that V4 walers are to be placed at 0,9 m c

c

which means that 5 sets of walers

are needed and 4 spans. Deflection is limited to 3 . Ls / 1000 ∆ allowable = 0,003 . Ls = 2,70 mm ∆ max = 131

0, 0064 . Pmax . Ls E .I

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Therefore deflection is satisfactory. vii)

Check soldiers for shear: Grade stress = Pv = 0,4 MPa (Table 3 SANS 10163-2) Hurd recommends that the allowable shear is multiplied by load-duration factor of 1,5 and a “two-beam” factor of 1,5. Therefore allowable shear stress = Pv = 0,4 . 1,25 . 1,5 = 0,75 MPa Shear stress σ v = 1, 5

V ≤ Pv (Note: b = 38 mm) b⋅d

Where shear for continuous beams = V = 0,6 . wis . Ls Ls = 900 – 130 = 770 mm (Assuming double waler with 50 mm wide members 30 mm apart) = 0,6 . 11,147 . 0,770 = 5,15 kN

σ v = 1,5 . 5,15 . 103 / 770 . 38 = 0,26 MPa ≤ Pv viii)

Check walers for bending: A width of 50 mm has already been assumed for the double waler, now assume a depth of 114 mm and a spacing of 30 mm between the two 50 mm wide members. The imposed load on each waler wiw = P max . Ls / 2 Wiw = 29 . 0,770 / 2 = 11,17 kN/m wiw . Ls 2 = 0,1 . 11,17 . 0, 770 2 = 0, 662 kN .m 10 k1 = 11,17 / 0, 66 . 11,17 = 1, 515 M =

f b = Pb . k1 . k 2 . k 3 . k 4 = 4, 0 . 1, 515 . 1 . 1 . 1 = 6, 06 MPa Z =

132

b . d 2 100 . 114 2 = = 216 600 mm 3 6 6

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Mr = fb . Z (Allowable bending stress times section modulus) Mr = 6,06 . 216 600 / 1 000 000 = 1,313 kN.m Therefore M ≤ Mr ix)

Check walers for deflection: I =

b ⋅ d 3 100 ⋅ 114 3 = = 12 346 200 mm 4 12 12

Deflection is limited to 3 . Ls / 1000

∆ allowable = 0,003 . Ls = 2,70 mm ∆ max =

0, 0064 . Pmax . Ls E .I

4

=

0, 0064 . 11,147 . 900 4 = 0, 63 mm 6000 . 12346200

Therefore deflection is satisfactory. x)

Check ties: Load tie = PTIE = Pmax . L . Ls = 29 . 0,384 . 0.900 = 10,03 kN Compressive strength perpendicular to the grain = Pcp = 1,6 MPa Required bearing area under washer = PTIE / Pcp = 6 270 mm2 A standard size washer of 100 mm square less the space gives a bearing area = 100 . (100 – 30) = 7 000 mm2, therefore it is OK. Tie spacing = Tie capacity / Waler load = 80 / 11,17 = 7,16 m. This is obviously out of the question because the wall is only 3,6 m high. The tie is therefore too strong for the application, either use a smaller tie or place ties at 900 mmc/c vertically and 1 152 c/c to suit the waler and soldier spacings.

xi)

Check bearing stresses: Compressive stress between waler and soldier 133

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= (half soldier load x Ls) / (Width of soldier x width of waler) = 0,5 . 11,147 . 900 / 38 . 100 = 1,32 MPa ≤ Pcp therefore OK. xii)

Lateral bracing: Adequate lateral bracing is very important for stability and safety considerations, it is better to have too much than too little. An unexpected gust of wind could easily blow the formwork down if it is not braced.

The following method (after Hurd) is based on a modified Euler formula and does not conform to SANS 10163. However it is a safe method. Slenderness value = L / d ≤ 50 Where L = Unsupported length in mm (ignoring Effective length) d = Least dimension of the bracing member in mm (Not the radius of gyration as given in SANS 10163) The allowable load P = PB / A =

0, 3 ⋅ E ≤ σ (Safety factor = 3) (L ÷ d ) c

Where P = Total load on the bracing in kN. A = Net cross-sectional area in mm2. E = Modulus of elasticity in MPa. (Table 3 SANS 10163-2)

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Table 4.1: Allowable load for formwork bracing based on unsupported length for v4 sap Size

50 x 114

76 x 114

114 x 114

50 x 152

76 x 152

A mm2

5700

8664

12 996

7600

11 552

L mm 100

P kN 25,6

L/d

L/d

20

P kN 90,0

L/d

13,1

P kN 304

L/d

8,7

P kN 34,2

L/d

20

P kN 120

1500 11,4

30

40,0

19,7

135

13,1

15,2

30

53,3

19,7

2000 6,4

40

22,5

26,3

76,0

17,5

8,5

40

30,0

26,3

2500 4,1

50

14,4

32,8

48,6

21,9

5,4

50

19,2

32,8

3000

10,4

39,4

33,7

26,3

13,3

39,4

3500

7,3

46,0

24,8

30,7

9,8

46,0

4000

19,0

35,7

4500

15,0

39,4

5000

12,1

43,8

5500

10,0

48,2

13,1

(ONLY VALUES OF L/d ≤ 50 SHOWN) Care must be taken when applying the values given in Table 4.1 to ensure that the bracing is adequately fixed to the formwork and in the ground (either to a concrete kicker-block or to a heavy stake). The allowable bearing stress must not be exceeded if there is a wood-to wood contact. If stresses are in this order of magnitude then the load may be transferred by means of other devices, such as metal brackets bolted to the brace and to the waling.

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4.3

FALSEWORK

Timber falsework is rarely used owing to the cost of timber and labour in Southern Africa. Proprietary scaffolding systems are generally cheaper to hire, quicker to erect and strip and have the added advantage of light-weight, durability and sets of design tables for quick accurate specification of the requirements of the project. Timber falsework does still seem to have a place in the USA for certain applications. In the Far East bamboo scaffolding is still commonplace. In 1996, bamboo scaffolding was erected to renovate a church in Singapore, the structure was about 50 m high. In South Africa the most common usage would be pine or gum pole bracing and props under small deck slabs, where empirical methods of sizing are used. It is unlikely that timber falsework will be used in any heavy construction. Some factors to bear in mind when designing falsework, irrespective of the material being used, are shown in Table 4.2 (After Ratay).

Table 4.2: interrelated falsework items ITEM

SIZE & SPACING

DEPENDS ON

Shutterboard

Optimum

Joist spacing

Joist spacing

Optimum

Joist pan between cap beams*

Cap beams*

Optimum

Span between props

Props #

Spacing

Finite maximum load / prop #

Sills

Size & length

Prop # loads

Footings

Size & length

Loads & bearing capacity of soil

Note: *Cap beams = Ledgers or Main bearers #Prop = Frame legs or trestles (Standards for scaffolding-tubing)

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Failure to optimise any one of the items in Table 4.2 could have an adverse effect on the design of other components of the system. For example thinner shutterboard would result in more closely spaced joists. The consequence of this is that joists are more expensive and additional handling increases both time of erection and stripping, which in turn costs the contractor and the developer more money. From this it may be concluded that the entire formwork and falsework system should be designed as an entity.

Figure 4.2: Falsework Trestle

Figure 4.2 shows the sort of standard used for heavy-duty falsework for slab construction subjected to heavy loadings. Only a single trestle set (without joist and shuttering) is shown for clarity. The joist would be perpendicular to the cap beams. Trestle sets would be spaced at centres determined by the loads and the joist design. Additional cross-bracing between trestle sets would be necessary to provide stability. Footings may need to be more closely spaced than illustrated to avoid settlement into the ground. For heavy traffic loads the members could be 300 x 300 mm except that footings, side plates and bracing could be as big as 300 x 100 mm sections. Decking of two layers perpendicular to each other with the bottom 100 mm planking and the top 50 mm planks.

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The considerations applicable to formwork design also apply to the falsework. The American Concrete Institude (ACI) recommendations (Hurd) include, an imposed load of 2,4 kN/m2 horizontally for the weight of labour, equipment and impact. This is increased to 3,6 kN/m2 when concrete dumpers are used. Some authorities in the USA specify a minimum combined load of 4,8 kN/m2 horizontally. The ACI also recommend that 1,46 kN/m or 2% of the slab total permanent load is taken as a UDL acting horizontally on the edge of the slab, whichever is the greater. A loaded 1360 kg concrete dumper travelling at 10 km/hr stopping in 2 seconds will impose a horizontal force of 1,824 kN and at 20 km/hr the force will be 3,65 kN. Retay recommends that, for lengths (spans) from 0 to 12 m the imposed loads should be increased by 30% to allow for impact loads.

Table 4.3: Some ohs and nosa requirements DESCRIPTION

LIMITS

Slope of scaffolding ramps

1: 1,5 max

Factor of safety for scaffolding ramps

2

Guardrail height for scaffolding > 2 m high

900 – 1 000 mm

Spacing wooden scaffolding standards

3 m c/c max

Dimension of any component of wooden scaffold

75 mm φ or equivalent

Height of wooden framework for scaffolding

10,0 m max

Dimensions of scaffold plank

38 x 275 mm min.

In addition to normal design considerations the requirements of the National Occupational Safety Association (NOSA) and the Occupational Health and Safety Act No. 85 of 1993 (OHS) should also be taken into consideration (See Table 4.3). The actual design is carried out using SANS 10163 (either Part 1 or Part 2), but bearing in mind the overseas practice mentioned above.

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SANS 10100: The structural use of concrete specifies that untreated timber must not be used for formwork and that cambers should be introduced to ensure that tolerances of the finished concrete are met. As far as falsework is concerned the SANS state that designs should be such that if one member breaks or is damaged, the adjacent one will be strong enough to carry the load thus preventing further damage or a domino effect collapse.

4.4

SPECIFIC OUTCOMES

At the end of this section the student will be able to follow the correct procedure for designing formwork and be capable of checking the strength of all components of the formwork system for slabs, beams, walls and columns.

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4.5

REFERENCES 1. SANS 1783: Sawn softwood timber 2. SANS 10100: The structural use of concrete. 3. SANS 10163-2: The structural use of timber. Part 2: Allowable stress design. 4. Baird JA & Ozelton EC: Timber designer’s manual. (2nd Edition). Blackwell Science Ltd, Oxford, 1995. 5. Breyer DE: Design of wood structure. McGraw-Hill, London, 1988. 6. Hurd MK: Formwork for concrete. American Concrete Institute, Detroit. 7. Hurst MP: Formwrk. Construction Press, Longman Group Ltd, Harlow, 1983. 8. Meuwese RFA: Form design. Murray & Roberts (Cape) Ltd. 9. Peurifoy RL: Construction planning, equipment and methods. (3rd edition). McGrawHill, 1979. 10. Ratay RT: Handbook of temporary structures in construction. McGraw-hill, 1984. 11. Richardson JG: Formwork construction and practice. Viepoint. 12. Stalnaker JJ & Harris EC: Structural design in wood. Van Nostrand Reinhold, New York, 1989 (Imperial units). 13. South African Lumber Millers Association: SALMA Timber Manual. SALMA, Isando, 1995.

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4.6

QUESTIONS FOR SELF-EVALUATION

Try these questions before turning the page to look at the solutions. 4.6.1

A 280 mm thick reinforce concrete slab 3,0 m above level ground is to be cast with a high slump mix of concrete at 150C, poker vibrators will be used to achieve compaction. The plan size of the slab is 20 x 15,5 m. There is no likelihood of wind and only labourers will be placing the concrete. V4 SAP members are to be used for the falsework. The shuttering will be sealed 19 mm lumberboard (battenboard) which has E = 4,6 kN/mm2 and an allowable bending stress of 4,7 N/mm2 parallel to the grain. (Table 36 SANS 10163-2) Design the formwork and falsework using timber for the soffit, joist and cap beams and steel scaffolding frames for the falsework. Scaffolding standard can carry a permissible load of 2 000 kg each and the system allows for either 1,2 m or 1,5 m horizontal bracing to form towers.

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4.7

SOLUTIONS TO THE SELF-EVALUATION QUESTIONS

4.7.1

Slab formwork design follows much the same set of rules as wall formwork, except that soldiers and walers would be replaced by cap beams (ledgers or bearers) and joists, ties would not be needed and lateral bracing might apply to props under the beams.

i)

Design load: Load due to concrete = Pc = 0,28 . 1 . 1 . 2 400 . 9,81 = 6,592 kN/m2 Imposed load due to workers = PI = 1,5 kN/m2 Pmax = PI + Pc = 1,5 + 6,6 = 8,1 kN/m2

ii)

Check shutterboard for bending: Assume that bending stresses are parallel to the face grain and that the board is continuous over three or more spans.

w ⋅ L2 = 0,1 . 8,1 . L2 = 0, 81 L2 10 k1 = 8,1 / (0, 66 . 8,1) = 1, 515 M =

f b = Pb . k1 . k 2 . k 3 . k 4 = 4, 7 . 1, 515 . 1 . 1 . 1 = 7,12 MPa Z =

b ⋅ d 2 1000 ⋅ 19 2 = = 60167 mm 3 6 6

Mr = fb . Z (Allowable bending stress times section modulus) Mr = 7,12 . 60 167 / 1 000 000 = 0,428 389 kN.m Equate M = Mr to solve for L: 0,81 . L2 = 0,428 389 Therefore L = 727 mm c/c (max. allowable span based on bending) Assume a 38 mm wide joist will be used then the clear span will be Lc = L – 38 = 689 mm

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iii)

Check shutterboard for deflection:

I =

b . d 3 1000 . 19 3 = = 571 583 mm 4 12 12

Deflection is limited to 3 . L / 1000

∆ allowable = 0,003 . L = 1,153 mm 4

∆ max =

0, 0064 . Pmax . Lc 0, 0064 . 8,1 . 689 4 = = 4, 44 mm E.I 4600 . 571583

This deflection is excessive. Try a shorter span Lc say 462 mm then L = 462 + 38 = 500 mm which is a module of a standard scaffold frame’s 1 500 mm width.

∆ max =

0, 0064 . 8,1 . 462 4 = 0, 898 mm 4600 . 571583

Therefore deflection is now satisfactory. iv)

Check shutterboard for shear: For Rolling or Interlaminar Shear Grade stress = Pv = 0,67 MPa (Table 36 SANS 10163-2) Shear stress σ v = 1,5

V ≤ Pv b.d

(Note: b = 1 000 mm)

Where shear for continuous beams (or panels) = V = 0,6 . w . Lc V = 0,6 . 8,1 . 0,462 = 2,25 kN

σ v = 1,5 . 2,25 . 103 / 1 000 . 19 = 0,18 MPa ≤ Pv v)

Check joists for bending: A width of 38 mm has already been assumed for the joist and the spacing 143

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L = 500 mm c/c The imposed load on each joist wIJ = 8,1 . 0,500 = 4,05 kN/m M =

wIJ . L2 = 0,1 . 4, 05 . 0, 500 2 = 0,1013 kN .m 10

k1 = 4,05 / (0,66 . 4,05) = 1,515 fb = Pb . k1 . k2 . k3 . k4 = 4,0 . 1,515 . 1 . 1 . 1 = 6,06 MPa Assume a depth for the joist of 114 mm

b ⋅ d2 38 ⋅ 114 2 Z = = = 82 308 mm 3 6 6 Mr = fb . Z (Allowable bending stress times section modulus) Mr = 6,06 . 82 308 / 1 000 000 = 0,4988 kN.m Therefore M ≤ Mr (A 38 x 76 mm would work but the minimum preferred size is 114 mm) vi)

Check joist for deflection: I =

b . d3 38.114 3 = = 4691556mm 4 12 12

Assume that cap beams are spaced at 1.5 m c/c to suit scaffold frames. Deflection is limited to 3 . LJ / 1000

∆ allowable = 0,003 . LJ = 4,50 mm

Note that this allowable deflection is excessive for the required tolerances of the finished concrete. In most cases this will happen because tolerances for offshutter concrete are quite small.

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Pmax = WIJ = 4,16 kN / m 4

0, 0064 . Pmax . LJ 0, 0064 . 4, 05 . 1500 4 ∆ max = = = 4, 66 mm E.I 6000 . 4691556 Once again deflection is excessive. Try a 38 × 152 mm joist. ∆ max =

0, 0064 . Pmax . LJ E.I

4

=

0, 00064 . 4, 05 . 1500 4 = 1, 97 mm 6000 . 11120725

This deflection is within acceptable tolerances for the slab soffit. vii)

Check joist for shear: Grade stress = Pv = 0,4 MPa (Table 3 SANS 10163-2) Hurd recommends that the allowable shear is multiplied by a load duration factor of 1,5 and a “two-beam” factor of 1,5. Therefore allowable shear stress = Pv = 0,4 . 1,25 . 1,5 = 0,75 MPa Shear stress for continuous beams = V = 0,6 . wIJ. LJ LJ = 1 500 mm V = 0,6 . 4,05 . 1,5 = 3,645 kN

σ V = 1, 5 . 3, 645 . 10 3 / 1500 . 38 = 0, 096 MPa ≤ PV viii)

Check cap beams (ledgers) for bending: The imposed load on each cap beam is in effect two point load acting at the third points of the length and a point load directly over each support. These loads are the total load per joist span for the length of the cap beam span plus the self-weight of the joists and shuttering. Wshutter = 1,5 . 0,500 . 0,019 . 460 .9,81 / 1 000 = 0,064 kN Wjoist = 1,5 . 0,152 . 0,038 . 460 . 9,81 / 1 000 = 0, 039 kN Wimposed = 1,5 . 0,5 . 8,1 = 6,075 kN 145

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Although the self weights are small in this case, it is a good practice to always include them. WICB = 0,064 + 0,039 + 6,075 = 6,18 kN (Point load imposed by joist) Max. bending moment: continuous beam point load at third points. Mmax = 0,119 . 2 . wICB . LJ = 0,119 . 2 . 6,18 . 1,5 = 2,206 kNm K1 = 22,06 / 0,66 . 22,06 = 1,515 fb = Pb . k1 . k2 . k3 . k4 = 4,0 . 1,515 . 1 . 1 . 1 = 6,06 MPa Assume cap beams are 50 x 225 mm members (This is a common size). Z =

b ⋅ d2 50 ⋅ 225 2 = = 421 875 mm 3 6 6

Mr = fb . Z (Allowable bending stress times section modulus) Mr = 6,06 . 421 875 / 1 000 000 = 2,557 kN.m Therefore M ≤ Mr ix)

Check cap beams for deflection:

I =

b . d3 50 . 225 3 = = 47 460 937 mm 4 12 12

Deflection is limited to 3 . LJ / 1000 ∆ allowable = 0,003 . LJ = 4,50 mm 4

∆ max =

0, 0064 . Pmax . LS 0, 00064 . 12, 36 . 1500 4 = = 1, 41 mm E.I 6000 . 47460937

Therefore deflection is satisfactory. x)

Check bearing stresses: Compressive stress between cap beam and joist =

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(joist load) / (width of joist x width of cap beam) = 6,18 . 1000 / 38 . 50 = 3,25 MPa ≥ 1,6 MPa Pcp for V4. Either increase grade to M8 or use two 50 x 225 mm members back-to back as cap beams. xi)

Load on scaffolding: Three of the four loads mentioned in viii) above plus the self weight of the cap beam will make up the load on each prop. wprop = 3 . 6,18 + 2 . 0,050 . 0,225 . 1,5 . 460 . 9,81 / 1 000 wprop = 18,7 kN per prop. Allowable load per leg of the scaffolding is: 2 000 . 9,81 / 1 000 = 19,62 kN ≥ wprop therefore acceptable.

The final solution is: Scaffolding 1,5 x 1,5 m braced towers with 1,5 m spacing in both directions between towers. Cap beams 2 No. 50 x 225 mm V4 SAP@1,5mc/c Joist

38 x 152 mm V4SAP@500mmc/c

SOFFIT

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4.8

TUTORIAL

4.8.1

Would a 50 x 50 mm member be acceptable as bracing for a wooden scaffolding? Qualify your answer.

4.8.2

(2)

What is the difference in strength between dry plywood and saturated plywood? (1)

4.8.3

Design a timber form for a 12 m long rectangular beam which is to cast on ground level, cured and then lifted into place. The beam will 1,2 m high and 300 mm wide, it will be reinforced and 40 MPa concrete with a 75 mm slump will be used. Include all bracing and ties.

4.8.4

(15)

Describe the principles you would use to design timber formwork for a column. (7)

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