Lecture on Strength of Materials August 2014

STRENGTH OF MATERIALS

Mechanics of Deformable Bodies

BY: DIVINA R. GONZALES

•MECHANICS OF DEFORMABLE BODIES- Study of the relationship between externally applied loads and their internal effects on deformable bodies. RIGID BODY – Bodies which neither change in shape and size after the application of forces FREE BODY DIAGRAM – Sketch of the isolated body showing all the forces acting on it.

THREE MAJOR DIVISIONS OF MECHANICS •Mechanics of Rigid Bodies – Engineering Mechanics •Mechanics of Deformable Bodies – Strength of Materials •Mechanics of Fluids – Hydraulics

SIMPLE STRESS TENSILE STRESS AND COMPRESSIVE STRESS

STRESS – unit strength of the body

P S A

Where: S – Uniform internal stress Perpendicular P – Axial force Passing through the centroid A – Uniform cross-sectional area

Compressive stress P Tensile stress P

Which bar is stronger A or B? Assume that the given loads are the maximum loads each can carry.

BAR A

BAR B

A=20mm2

A=50mm2

1000N

500N

PROBLEMS ON SIMPLE STRESS 1.Determine the weight of the heaviest traffic lighting system that can be carried by the two wires shown if the allowable stress on wire AB is 90MPa and on wire AC is 110MPa given that the cross sectional areas of wire AB is 50mm2 and that of AC is 80 mm2.

B

C

70 A

35

2. Determine the required cross sectional areas of members BE, CD and CE of the given truss shown, if the allowable stress in tension is 120MPa while in compression is 105MPa. A reduced allowable stress in compression is given to reduce the danger of buckling. D C

G 4m

3m B

A 3m

E 3m

3m 50KN

75KN

3m F 3m

50KN

H

3. An bronze rod is rigidly attached between a aluminum rod and a steel rod as shown. Axial loads are applied at the positions indicated. Find the maximum value of P that will not exceed a stress in steel of 140MPa, in aluminum of 90MPa or in bronze of 100MPa.

Aluminum Bronze Steel 2 A= 500mm A= 200mm2 A= 150mm2 2P 3P P L alum= 3.5m

L br = 2m L st = 1.2m

C

4. Determine the weight of the heaviest cylinder that can be supported by the structure shown if the cross sectional area of the cable is 120mm2 and its allowable stress is 80MPa.

8m

B

9m

A

3m

5. The figure shows the landing gear of a light airplane. Determine the compressive stress in strut AB caused by the landing reaction R=30KN. Neglect the weights of the members. The strut is a hollow tube, with 40mm outside diameter and of uniform thickness of 7mm.

A B 200mm R

55 600mm

A

5m

C 9m

B

3m

In the recently opened World Trade Center, a showcase of designer’s work is being featured. There is a center plate of which is just supported by three bars. If the weight of the triangular plate is 500N, find the stress in each leg if its cross-sectional section is 50mmx50mm.

The tripod shown supports the total station which weighs 10N. Find the required diameter of the leg if the maximum allowable stress in each leg is 25MPa.

A

8m

C 4m 3m

3m

B 6m

D

SHEARING STRESS – This arises whenever the applied loads cause one section of the body to slide past its adjacent section. The force acts parallel to the area. Where: Ss – Shearing stress P – Force acting parallel to the area A – Surface area

P Ss  A P

P

Shearing area parallel to the load

TYPES OF SHEARING STRESS 1. SINGLE SHEAR LAP JOINT P 130mm Width of plate

t

Rivet under Single shear

P

P

P

2. DOUBLE SHEAR

Rivet under Double Shear

BUTT JOINT P/2

tSPLI CE

P/2

tSPLI CE

t MAIN PLATE

P

3. PUNCHING SHEAR

P

4. INDUCED SHEAR SHEARED AREA

P

P

P SHEARED AREA

BEARING STRESS – Contact pressure exerted by one body upon another body. Also know as end stress. The force acts perpendicular to the area. LAP JOINT P

t

P

130mm

Width of plate

P

P ENLARGEMENT OF THE RIVET HOLES

DIAMETER t

OF THE PLATE

PROBLEMS ON SHEARING STRESS AND BEARING STRESS A circular hole is to be punched out of a plate that has a shear strength of 40ksi The working compressive stress in the punch is 50ksi. Compute the maximum thickness of a plate in which a hole 2.5in in diameter can be punched. B) If the plate is 0.25in thick, determine the smallest diameter that can be punched.

The lap joint is connected by three 20mm diameter rivets. Assuming that the axial load P = 50KN is distributed equally among the three rivets, find a) the shearing stress in each rivet; b) bearing stress between the plate and a rivet and c) the maximum average tensile stress in each plate. t=25mm

130mm

Determine the maximum force P that the top chord can carry if the allow shearing stress is 50MPa, bearing stress is 60MPa and tensile stress is 85MPa of the connection shown P

b a

c



a b c d 

200mm 120mm 40mm 75mm 30

A 25.6 mm diameter bolt having a diameter at the root of the threads of 21.7mm is used to faster two timbers as shown. The nut is tightened to cause a tensile force of 34KN in the bolt. Determine the shearing stress developed in the head of the bolt and the threads. Also determine the outside diameter of the washer if the inside diameter is 30mm and the bearing stress is limited to 0.9MPa.

15mm

17mm

The bracket is supported by ½ inch diameter pins at A and B (the pin at B fits in the 45 slot in the bracket). Neglecting friction, determine the shear stresses in the pins, assuming single shear.

45

A

200lb 12in B 36in

Compute the maximum force P that can be applied to the foot pedal. The ¼ inch diameter pin at B is in single shear, and its working shear stress is 40000psi. The cable attached at C has a diameter of 1/8 inch and a working normal stress of 20,000psi. P

A

B

2”

C 10

6”

2”

T

The figure shows a roof truss and the detail of the connection at joint B. Members BC and BE are angle sections with thickness shown in the figure. The working stresses are 70MPa for shear in rivet and 140MPa for bearing stress due to the rivets. How many 19-mm diameter rivets are required to fasten the said members to the gusset plate?

D

B

E

C

A

4m

F

6m

4m

4m 96KN

G

200KN

H 4m

96KN DETAIL OF JOINT B 10mm GUSSET PLATE

75X75X13

75X75X6

P BE

P BC

THIN WALLED CYLINDERS –

LONGITUDINAL JOINT

CIRCUMFERENTIAL JOINT

SCIRCUMFERENTIAL 

 D 2t

S LONGITUDINAL 

 D 4t

1.A cylindrical tank 8m in diameter is 12m high. If the tank is completely filled with water, determine the required thickness of tank plating if the allowable stress is 40MPa.

The tank shown in the figure is fabricated from 400mm

600mm

Determine the minimum thickness of the tank plating if the internal pressure is 1.5MPa and the allowable stress is 40MPa.

A large pipe called a penstock in hydraulic work is 1.5m in diameter. Here it is composed of wooden staves bound together by steel hoops, each 300mm2 in cross-sectional area, and is used to conduct water from a reservoir to a power house. If the maximum tensile stress permitted in the hoops is 130MPa, what is the spacing between hoops under a head of water of 30m?

A spiral-riveted penstock 1.5m in diameter is made of steel plate 10mm thick. The pitch of the spiral or helix is 3m. the spiral seam is a singleriveted lap joint consisting of 20-mm diameter rivets. Using SS=70Mpa and Sb=140MPa, determine the spacing of the rivets along the seam from a water pressure of 1.25MPa. Neglect end thrust. What is the circumferential stress?

STRESS- STRAIN DIAGRAM Actual Rupture Strength STRESS

Ultimate Strength

Yield Point

Rupture Strength

Elastic limit Proportional limit

STRAIN

The strength of the material is not only the criterion that must be considered in designing a structure. The stiffness of a material is frequently of equal importance. Hooke’s Law states that up to the proportional limit, the stress is proportional to strain. The constant of proportionality based from experiment is the modulus of elasticity.

STRESS      STRAIN  P   S S     A L  AXIAL

PL  AE

•During a stress-strain test, the unit deformation at a stress of 35MPa was observed to be 167x10-6 m/m and at a stress of 140MPa it was 667x10-6. If the proportional limit was 200MPa, what is the modulus of elasticity? Would these results be valid if the proportional limit were 150MPa? Explain.

The rigid bars AB and CD are supported by pins at A and D. The vertical rods are made of aluminum and bronze. Determine the vertical displacement of the point where the force P =10kips is applied. Neglect the weight so the member.

Aluminum E=10x10 6 psi L=3ft

A 2ft

B

A= 0.75 in2 3m

Bronze

E=12x10 6 psi L=4ft A= 0.25 in2

D C

2ft

2ft P

•The compound bar containing steel bronze and aluminum segments carries the axial loads shown in the figure. The properties of the segments and the working stresses are listed in the table. Determine the maximum allowable value of P if the change in length of the entire bar is limited to 0.08in and the working stresses are not to be exceeded. A (in2)

E (psi)

S (psi)

Steel

0.75

30 x 106

20000

Bronze

1.00

12 x 106

18000

Aluminm

0.50

10 x 106

12000

Steel 2 ft P

Bronze 4 ft 3P

4P

Aluminum 3 ft 2P

•A round bar of length L tapers uniformly from a diameter D at one end to a smaller diameter d at the other end. Determine the elongation caused by an axial tensile load P if E is its modulus of elasticity.

4P L   E Dd

•The rigid bars shown are separated by a roller at C and pinned at A and D. A steel rod at B helps support the load of 50KN. Compute the vertical displacement of the roller at C. Answer: 2.82mm STEEL E=200x10 6 N/m2 L=3m A= 300 mm2 A

P= 30KN D B C

The rigid bars AB and CD are supported by pins at A and D. The vertical rods are made of aluminum and bronze. Determine the vertical displacement of the point where the force P=10kips is applied. Neglect the weights of the member. Answer: 0.115 in

Aluminum L=3ft A=0.75in2 E=10x106psi

2 ft

3ft Bronze L=4ft A=0.25in2 E=12x106ps i

2ft P

2ft

STATICALLY INDETERMINATE MEMBERS

Static indeterminacy does not imply that the problem cannot be solved; it simply means that the solution cannot be obtained from the equilibrium equations alone. A statically indeterminate problem always has geometric restrictions imposed on its deformation.

The mathematical expressions of these restrictions, known as the compatibility equations, provide us with the additional equations needed to solve the problem (the term compatibility refers to the geometric compatibility between deformation and the imposed constraints).

Because the source of the compatibility equations is deformation, these equations contain as unknowns either strains or elongations. We can, however, use Hooke’s law to express the deformation measures in terms of stresses or forces. The equations of equilibrium and compatibility can then be solved for the unknown forces.

The figure shows a copper rod that is placed in an aluminum sleeve. The rod is 0.005 inch longer than the sleeve. Find the maximum safe load P that can be applied to the bearing plate, using the following data: Answer: 60.3kips

COPPER ALUMINUM Area (in2) 2 3 E (psi) 17x106 10x106 S (ksi) 20 10 10 in

Bearing plate 0.005” alum copper

•A reinforced concrete column 250mm in diameter is designed to carry an axial compressive load of 400KN. Using the allowable stress in concrete of Sconc =6MPa and S steel = 120MPa, determine the required area of reinforcing steel. Assume Econc=14GPa and Esteel = 200GPa. Answer: 1320mm2

A rigid block of mass M is supported by three symmetrically spaced rods as shown. Each copper rod has an area of 900mm2 Ecopper =120GPa and the allowable stress is 70MPa. The steel rod has an area of 1200mm2, Esteel=200GPa and allowable stress of 140MPa. Determine the largest mass M which can be supported. Answer: 22.3x103Kg

Copper 160mm

Steel 240mm

Copper 160mm

Before the 400KN load is applied, the rigid platform rests on two steel bars each of cross-sectional area of 1200mm2, as shown. The cross-sectional area of copper is 2400mm2. . Compute the stress in each rod after the 400KN load is applied. Neglect the weight of the platform. Esteel =200GPa Ealuminum =70GPa. 400KN 0.1mm 250mm

steel

aluminum

steel

•The composite bar is firmly attached to unyielding supports. Compute the stress in each material caused by the application of the axial load P=50kips. b) If the maximum allowable stress in each material is Salum=22psi and Ssteel=40psi, find the maximum P that the structure can support. Aluminum A=3.25 in2 E=10x106psi

Steel A=5.5 in2 E= 29x106psi

P 12in

25in

•The rigid beam is supported by the two bars shown in a horizontal position before the load P is applied. If P=200KN, determine the stress in each rod after its application. B) Find the vertical movement of P. c) If the allowable stress in aluminum is 80MPa and steel is 120MPa, find the maximum load P that the system can carry L=6 m

steel

aluminum L=4.5 m 3m

Steel A in mm2 E in GPa

Aluminum 600 70

800 200

3m

3m •P

THERMAL STRESS It is well known that changes in temperature cause dimensional changes in a body. An increase in temperature results in expansion, whereas a temperature decrease produces contraction.

The thermal stress is:

TS    E  T The thermal deformation is:

    L  T Where  = coefficient of thermal expansion and change in temperature.

T=

PROBLEMS ON THERMAL STRESS AND THERMAL DEFOMATION

•A steel rod with a cross-sectional area of 150mm2 is stretched between two fixed points. The tensile load at 20C is 5000N. What will be the stress at -20C? At what temperature will the stress be zero? steel = 11.7x10-6/C and E=200GPa. Answer=127MPa ; T= 34.2C

•Two identical steel bars 500mm long support the rigid beam shown. An aluminum bar is placed exactly in between them, that is 0.1mm shorter. a) If the rigid beam is weightless determine the change in temperature for the middle bar to just touch the beam. b) If the beam weighs 300KN, find the stress in each bar. c) If the beam weighs 250KN, determine the stress in each bar when the temperature raises 35C d) If the beam weighs 320KN, determine the stress in each bar when the temperature drops 25C.

A in mm2  in x 10-6 /C E in GPa

Aluminum 2400 23 70

Steel 1200 11.7 200

0.1mm

500mm

steel

aluminum

steel

•The composite bar is firmly attached to unyielding supports. The initial temperature is 80F when the load P = 20kips is applied, compute the stress in when the temperature is 150F and when the temperature is 5F.

Aluminum A=3.25

in2

E=10x106psi

= 12.8x10-6/F 12in

Steel A=5.5 in2 E= 29x106psi = 12.8x10-6/F P 25in

•The rigid beam is supported by the two bars shown in a horizontal position before the load P is applied. If P=200KN, determine the stress in each rod after an increase in temperature of 40C B) drop of 65C.

aluminum 3m

L=6 m

steel L=4.5 m

3m

3m •P Steel

A in mm2  in x 10-6 /C E in GPa

Aluminum 600 23 70

800 11.7 200

•BEAMS It is a structure usually horizontal acted upon by transverse loads (forces that acts perpendicular to the plane containing the longitudinal axis of the beam) W P

N/m

W

N/m

axis beam

Statically Determinate Beams supported such that the number of reacting forces equals the number of available equations static equilibrium conditions, P

Simply Supported W

P

N/m

Cantilever Beams

Beams with Overhang

W

N/m

P

•TYPES OF LOADS P

Concentrated Loads

W

F

N/m

Uniformly Distributed

•Uniformly Varying Loads Triangular W

N/m

•Parabolic Loads

Trapezoidal W1

N/m

W2

N/m W

N/m

•Moving Loads

The fundamental definitions of shear and bending moments are expressed by

V   Fy left and

M   Mleft    M right 

in which upward acting forces or loads cause positive effects. The shearing force V should be computed only in terms of the forces to the left of the section being considered: the bending moment M may be computed in terms of the forces to either the left or the right of the section depending on which requires less arithmetical work.

Relations between load, shear moment are given by:

dv w dx dM v dx

=

slope of the shear diagram

= slope of the moment diagram

These relations are amplified to provide a semi graphical method of computing shear and moment which supplements the equations

V   Fy left

and

M   Mleft    M right 

V2  V1  Aload  diagram

M 2  M 1  AShear  daigram

A summary of the principles presented suggests the following procedure for the construction of shear and moment diagrams 1. Compute the reactions 2. Compute the values of shear at the change of load points using 3. Sketch the shear diagram, drawing the correct shape and concavity of the shear diagram.

4. Locate the points of zero shear. 5. Compute values of bending moment at the change of load points and the points of zero shear using 6. Sketch the moment diagram

PROPERTIES OF THE SHEAR DIAGRAM 1. At every change of loading we have to investigate our shear. 2. For concentrated loads or reactions the left and right portion of the point where they are acting must be investigated 3. Whenever we have a concentrated load or reaction there will always be a vertical line in the shear diagram 4. The shear diagram is one degree higher than the load diagram

Reference of the Concavity of the Shear Diagram

Concavity of the circle corresponds to the concavity of the shear diagram Positively decreasing load

Negatively decreasing load

Negatively increasing load

Positively increasing load

Note: When M=0 ( point of inflection) there is a change of the concavity of the elastic curve of beam

PROPERTIES OF THE MOMENT DIAGRAM 1. For every change in shear diagram, the moment must be investigate 2. Consider only the moment at any point and not the left and the right portion of the point except when we have a moment load or reaction 3. Analyze the point where the shear intersects the reference line (V=0) since when shear is zero moment is maximum or minimum 4. Vertical line will only be observed in the moment diagram whenever we have a moment load or reaction 5. The concavity of the moment diagram depends upon the load: if the load is downward, moment diagram is downward 6. The moment diagram is one degree higher than the shear diagram

DESIGN FOR FLEXURAL STRESS

1. A cantilever beam, 75mm wide by 200 mm high and 6m long carries a load that varies uniformly from zero at the free end to 1500 N/m at the wall. (a) Compute the magnitude and location of the maximum flexural stress (b) Determine the type and magnitude of the stress in a fiber 25mm from the top of the beam at section 2.5m from the free end.

2. Determine the minimum width “b” of the beam shown if the flexural stress is not to exceed 10Mpa.

b

3m

5000N 2000N/m

1m

3m

200mm

3. A 2” diameter bronze bar is used as a simply supported beam 8ft long. Determine the largest uniformly distributed load, which can be, applied over the right half of the beam if the flexural stress is limited to go 10ksi?

4. A simply supported rectangular “yacal” beam, 75mm wide by 150mm deep, carries a uniformly distributed load of 2250 N / m over its entire length. What is the maximum length of the beam if the flexural stress is limited to 18 Mpa?

5. A simply supported beam 6m long is composed of two C 200x28 channels riveted back to back. What uniformly distributed load can be carried, in addition to the weight of the beam, without exceeding a flexural stress of 125 MN/m2 if (a) the webs are vertical and (b) the webs are horizontal. Refer to Appendix B for channel properties.

6. A beam with a S380 x 74 section is simply supported at the ends. It supports a central concentrated load of 40 kN and a uniformly distributed load of 15 kN/m over its entire length, including the weight of the beam. Determine the maximum length of the beam if the flexural stress is not to exceed 130 Mpa. Refer to Appendix B for properties of S shapes.

7. A beam 15 m long is simply supported 2 m from each end. It is a built–up made of four angle bars 100x75x13, with long legs horizontal (see Table B-6) welded to a flat bar 25mmx300mm as shown in the figure. Determine the total uniformly distributed load that can be carried along its entire length without exceeding a flexural stress of 120 MPa.

8. A beam with a W2360x33section (see Table B-2) is used as a cantilever beam 7.5 m long. Find the maximum uniformly distributed load which can be applied over the entire length of the beam, in addition to the weight of the beam, if the flexural stress is not to exceed 140 MN / m2

9. A 12-m beam simply supported at the ends carries a uniformly distributed load of 20kN / m over its entire length. What is the lightest W shape beam that will not exceed a flexural stress of 120MPa ? What is the actual stress in the beam selected?

10. A simply supported steel beam 10m long carries a uniformly distributed load of 18kn/m load over the entire length and a central concentrated load of 25Kn. Determine the lightest Wide flanged section that can be used to support the load. What is the actual resulting stress in the beam selected.

11. A cantilever wooden beam is composed of two segments with rectangular cross sections. The width of each section is 75mm but their depths (150mm and 250mm) are different, as shown in the figure. Determine the maximum bending stress in each beam. 50KN 1.5m

2m

UNSYMMETRICAL SECTION DESIGN FOR BENDING STRESS

The previous discussions are all about beams symmetric with respect to the neutral axis. Because flexural stress vary directly with distance from the neutral axis of symmetric beams, which is the centroidal axis, such beam sections are desirable for materials that are equally strong in tension and compression.

However, for materials relatively weak in tension and strong in compression such as cast iron, it is desirable to use beams that are unsymmetrical with respect to the neutral axis. With such a cross section, the stronger fibers can be located at a greater distance from the neutral axis than the weaker fibers. The ideal treatment for such materials is to locate the centroidal or neutral axis in such a position that the ratio of the distances from it to the fivers in tension and in compression is exactly the same as the ratio of the allowable stresses in tension and in compression. The allowable stresses thus reach their permitted values simultaneously.

1. A cast-iron beam carries a uniformly distributed load on a simple span. Compute the flange width “b” of the inverted T section so that the allowable stresses fb(tension)=30MPa and fb(compression)=90MPa reach their limits simultaneously. 20mm W N/m L

yc

120mm

NA

20mm “b”

yt

2. Compute the maximum tensile and compressive stresses developed in the beam that is loaded and has the crosssectional properties shown. 8KN

30mm

10KN/m 4m

1m

125mm 25mm

100mm

3. Determine the maximum safe value of W that can be carried by the beam shown if given the following allowable stresses fb(tension)=60MPa and fb(compression)=90MPa

W

2m

W

6W KN

8m

80mm 20mm 20mm 35mm

2m

DESIGN FOR SHEARING STRESS The vertical shear sets up numerically equal shearing stresses on longitudinal and transverse sections, which are determined from:

VQ fv  Ib

Q=A’B in which A’ is the partial area of the cross section above a line drawn through the point at which the shearing stress is desired. Q=A’y is the static moment about the NA of this area (or of the area below this line).

.

Maximum shearing stresses occur at the section of maximum V and usually at the NA. For rectangular beams, the maximum shearing stress is

3V Max. f v  2bh

1. Draw the shearing stress distribution for a rectangular beam 75mmx200mm which is simply supported on a 10m beam with a load of 20KN/m over the entire length.

2. Determine the maximum and minimum shearing stress in the web of the wide flange section if V=120KN. 180mm

30mm

300mm 220mm

3. The distributed load shown is supported by a box beam shown using four pieces of 50mm x 250mm plank of Molave timber. Determine the maximum safe value of w that will not exceed a flexural stress of 10 MPa or a shearing stress of 1MPa.

4. The distributed load shown is supported by a wideflange section W 360x45 of the given dimensions. Determine the maximum safe w that will not exceed a flexural stress of 140 MPa or a shearing stress of 75 MPa.

2m

4m

DESIGN FOR FLEXURE AND SHEAR