Workshop Homework Problems Based on SAP2000 by Wolfgang Schueller

February 1, 2018 | Author: wolfschueller | Category: Beam (Structure), Bending, Truss, Stress (Mechanics), Structural Engineering
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Refer to: Building Support Structures, Analysis and Design with SAP2000 Software, 2nd ed., eBook by Wolfgang Schueller,...

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WORKSHOP The Building Support Structure in Architecture: a visual analysis and design of structures with computers using SAP2000.

Workshop Homework Problems as presented by the following Examples/Problems reference

Reference: Building Support Structures, Analysis and Design with SAP2000 Software, 2nd ed., eBook by Wolfgang Schueller, 2015 The SAP2000V15 Examples and Problems SDB files are available on the Computers & Structures, Inc. (CSI) website: csiamerica.com/go/schueller

Prof. Wolfgang Schueller

AXIAL SYSTEMS: Trusses Problem 1 (EX. 6.1) Investigate some simple, basic truss forms as based on the Howe-type of member layout (a similar approach can be used for other common layouts such as Pratt, Warren, K-truss, and lattice). Then make the following changes as indicated in the drawing below by reshaping the truss configuration that is play with the truss object, using the Set Reshape Element Mode in SAP2000, by considering:  

Profile: rectangular, triangular, curved, trapezoidal, and other asymmetrical shapes, i.e. contours Load arrangement, load direction, and load location : symmetrical and asymmetrical, vertical and horizontal



Support location and orientation: simple beams, cantilever beams, overhanging beams, frames, etc.

For determinate structures disregard the effect of material and member sizes, i.e. the frame elements may be modeled with zero moments of inertia, or the default setting may be used, since member stiffness has no effect on the magnitude of internal member forces; however, do not use deflection results. Generate at least four different truss shapes on a 4 x 4-ft (1.22 x 1.22-m) square grid as shown in the drawing. Apply vertical, single loads of 1k = 4.45 kN (D = DEAD) at the top chord joints as indicated below; treat the horizontal forces of 0.5 k = 2.22 kN (W = WIND) as separate load case. Show the axial force flow with numerical values, and show the reaction forces. Study the character of the given trusses and also the relationship of member tension and compression, so you can develop a feeling for the structure and predict the direction of the force flow; try also to predict conceptually the deflection. Check some of your results manually (graphically or analytically):  Check the reactions of two trusses with asymmetrical support or loading conditions.  Check the member forces of two joints for the two trusses.

a. b. c. d.

4' 4'

Problem 2 (Problem 6.1): Indeterminate rectangular trusses •Investigate some typical rectangular indeterminate trusses such as the cross-braced truss (6.3g) and the lattice truss (Fig. 6.8c). Use the ModelAlive feature in SAP2000 to derive the respective truss responses instantly by rearranging the modified Warren truss configuration in EXAMPLE 6.1 (Fig. 6.9a). Check manually some of the results approximately. •Use the Model-Alive feature and modify the geometry of the rectangular truss using the Set Reshape Element Mode button.

Prof. Wolfgang Schueller

Cable-Supported Structures: stayed bridges

Problem 2 (EX. 11.6): Stayed bridges Investigate several basic stayed bridge structures in Fig. 11.12 under a uniform gravity load of 2.75 k/ft (40 kN/m), but assign 1.5 k/ft (22 kN/m) to the outer spans of the stayed bridge in case f); assign zero to self-weight. Disregard in this oversimplified first approach live load arrangement, wind loads, thermal loads and the prestressing of the cables and the eccentric application of the stays to the girders. The beams are simply supported at the towers. Determine the approximate beam size using W30 (W760) or W36 (W920) sections (A36 ≈ 250 MPa) as an initial input into SAP. Keep in mind that the beams in reality may consist of multi-cell roadway box girders. Use as a first trial nominal 2-in-diameter (51 mm) strands for the inner stays and 1.5-in-diameter (38 mm) strands for the outer stays using Fu = 200 ksi (1380 MPa). Neglect axial strains in the towers by allocating large values of the cross-sectional area, A. Draw bridge shapes on an 15 x 15-ft (4.57 x 4.57-m) grid as shown in the drawing. Consider P-Delta analysis plus large displacements. Investigate conceptually the axial force flow and bending moments in the beams together with the reaction forces and deformations. Study the character of the given structures and also the relationship of member tension and compression, so you can develop a feeling for the structure and predict the direction of the force flow; try also to predict conceptually the deflection.

a

c

b

e

d

f

Prof. Wolfgang Schueller

FLEXURAL SYSTEMS: Load Types and Boundary Conditions Problem 3 (EX. 7.1) Investigate for the beam cases below and study the effect of:  Boundary conditions including Cantilever action  Load types, a counterclockwise moment of 18 k-ft (24.40 kNm) acts at the left support of case (k), clockwise uniform torsional loads of 1 k-f/ft (4.45 kNm/m) act along the beam span of case (e), a clockwise moment of 12 k-ft acts at the right reaction of case (d) and a counterclockwise moment of 12 k-ft at the left support of the same case, etc.  Load distribution  Indeterminate action The beams are drawn on a 4-ft (1.22-m) grid, in other words the beams span 12 ft (3.66 m). The uniform load is 1 k/ft (14.59 kN/m) unless shown. Show input (geometry and loading), moment and shear diagrams with numerical values at critical locations, and deflections. Check the maximum moments of the cases by referring to tables to make sure that the computer results are meaningful; check the indeterminate beams approximately.

1 k/ft

A. 1 k/ft

1 k/ft

1 k/ft

1 k/ft

B. I. C. 12 kft D.

12 kft

1 k/ft

J.

18 kft

1 k/ft K.

12 k

1 kft/ft E. F.

6k

2 k/ft

4k G.

H.

6k

4k

4k

L.

M.

2 k/ft

2 k/ ft

0.5 k/ft

1.5k/ft

N. O.

Prof. Wolfgang Schueller

FLEXURAL SYSTEMS: Floor Framing 1

Problem 7 (EX. 4.3) The floor framing for a typical interior bay of a multistory braced steel skeleton structure is shown below. The composite deck distributes dead and live load of 80 psf (3.83 kPa) each to the beams; the live load reduction factors are 0.96 for the beams and 0.8 for the interior girders (EXAMPLE 2.2); however, for this approximation also 0.96 is conservatively used also for the girders. Use A36 (250 MPa) steel and flexible connections. The compression flanges of the filler beams and girders are assumed fully laterally supported by the floor slab. Design the floor beams using SAP for working stress approach (AISC-ASD89). AssumeW18 (W460) beams for the automatic section selection.

GI

3 Sp @ 8' = 24'

25'

BM

BM

BM

BM

GI

Prof. Wolfgang Schueller

FLEXURAL SYSTEMS: Floor Framing 2

Problem 8 (EX. 7.6) Investigate the simple hinged floor framing for the three 21 x 21 ft (6.40 x 6.40 m) bays shown below using SAP. The floor load consists of a dead load of 70 psf (3.35 kPa or kN/m2) where beam weight is included, and a live load of 60 psf (2.87 kPa); ignore live load reduction for this preliminary investigation. Select the most economical sections for the beams as well as girders assuming full lateral support of the compression flanges. Design the floor beams using A36 (250 MPa) steel and working stress approach. For the automatic section selection, try for the filler beams W10 sections (W250) and for the girders W16 sections (W410). Check manually beams BM1, BM3and G4 to see whether the computer output makes sense.

G1

21'

BM2

BM1

BM1

BM2

BM2 G2 BM5

BM5

G1

G4

BM5

BM5

G3

BM3

BM1 BM5

BM5

BM4 3 Sp @ 7' = 21'

BM2 21'

3 Sp @ 7' = 21'

BM1

BM3

Prof. Wolfgang Schueller

Surface Structures: slabs

Problem 11 (EX. 12.5): square concrete slabs Investigate a square 6-in. (15 cm) concrete slab, 12 x 12 ft (3.66 x 3.66 m) in size in Fig. 12.14 that carries a uniform load of 120 psf (5.75 kPa or kN/m2, COMB1), that is a dead load of 75 psf (3.59 kPa) for its own weight (SLABDL taken care by self weight) and an additional dead load 5 psf (0.24 kPa, SUPERD), and a live load of 40 psf.(1.92 kPa, LIVE). The concrete strength is 4000 psi (28 MPa) and the yield strength of the reinforcing bars is 60 ksi (414 MPa). Solve the problem by using 2 x 2 ft (0.61 x 0.61 m) plate elements using SAP. a) Assume one-way, simply supported slab action. b) Assume a two-way slab, simply supported along the perimeter. c) Assume the slab is clamped along the edges to approximate a continuous interior two-way slab. d) Assume flat plate action where the slab is simply supported by small columns at the four corners. e) Assume cantilever plate action with four corner supports for a center bay of 8 x 8 ft (2.44 x 2.44 m). f) Assume umbrella action of the plate with the center column fixed to the plate. Check the answers manually using approximations. Compare the various slab systems that is study the effect of support location on force flow.

a 2'

8'

c

2'

12'

6'

6'

12'

b

d

e

f

Prof. Wolfgang Schueller

Frame Structures: folded beams

Problem 12 (EX. 8.1): Folded beam systems Simple-span 20-ft (6.10-m) span, folded beam systems as shown below, are drawn on a 5 x 5 ft (1.52 x 1.52 m) grid, are investigated with respect to the effect of geometry on the bending moment distributions, shear distribution, axial force flow, and reactions. a) Use a single load of 10 k (44.48 kN) for each case at mid-span (SINGLEP). b) Use a uniform load of 1 klf (14.59 kN/m) on global z-projection (UNIFORM). Show the magnitude of the reactions and moment diagrams as well as axial force diagrams with their maximum values. Check some of your answers manually to be sure that the computer solutions are all right.

b a

c

d

f

e

g

Prof. Wolfgang Schueller

Frame Structures: three-hinged frames

Problem 13 (EX. 8.2): Three-hinged frames Investigate the following 40-ft (12.19-m) span, three-hinged frame structure systems drawn on a 5 x 5 ft (1.52 x 1.52 m) grid, with respect to the effect of geometry (e.g. column and beam inclination) on force flow in statically determinate structures by studying conceptually the bending moment distribution, axial force flow, the reactions, and the deflected shapes. Use a uniform dead load (D) of 0.5 k/ft = 7.30 kN/m and a live load (L) of 0.75 k/ft = 1.95 kN/m on global z-projection and a lateral wind load of 0.5 klf = 7.30 kN/m on global x-projection (i.e. 17 psf = 0.82 kPa for 30-ft spacing of frames). Consider the following load combinations for this preliminary investigation: COMB1 (D + L), COMB2 (D + W), and COMB3 [D + 0.75(L + W)]. Design the steel frames as based on SAP default material properties by using W21sections (AUTOW21) but use W12 sections (AUTOW12) for the arched shapes (b, d, and e). After the first design cycle, the structure has to be reanalyzed with the new member sections and be redesigned. Keep in mind that in the computer program design is an iterative process, where the analysis and design must be run multiple times to complete the design process. In other words check: Design > Steel Frame Design > Verify Analysis vs. Design Section. For the design of the frame beams assume an unbraced length ratio of Lb/L = 0.1 about the minor axis for preliminary design purposes, and consider the columns laterally braced about their minor axis (Ky =1). Show the magnitude of the reactions and moment diagrams as well as axial force diagrams with their maximum values. Check several of your answers manually to be sure that the computer solutions make sense.

A. B.

C. D.

E. F.

Prof. Wolfgang Schueller

Frame Structures: basic arches Problem 16 (EX. 9.1): Basic arches Investigate the simple circular three-hinged, 40-ft (12.19-m) span arch systems and halfarch systems shown below, with respect to the effect of arch proportion and load arrangement on intensity of force flow using SAP. The shallow arch is 8 ft high consisting of ten linear segments and the steep arch is a semicircular arch consisting of 12 linear segments. Use for dead load wD = 0.5 k/ft = 7.30 kN/m (D) applied along the arch, for live load wL = 0.5 k/ft (LFULL for full loading and LHALF for loading half span) on the horizontal roof projection, and for wind wW = 0.4 k/ft = 5.84 kN/m (W) on the vertical roof projection. Consider the following load combinations for this preliminary investigation: COMB1 (D + LFULL), COMB2 (D + LHALF), COMB3 [D + 0.75(LFULL + W)] and COMB4 [D + 0.75(LHALF+ W)]. Draw the arch images on a 4 x 4-ft (1.22 x 1.22-m) grid. Select W10 sections (A36) using Auto Select. For the design of the arches use an unbraced length ratio of Lb/L = 0.1 about the minor axis for preliminary design purposes. Study the load combinations and determine which ones control the design. Show and study the magnitude of the reactions and bending moment distribution with critical values, as well as axial force flow with their maximum values. Check some of your answers manually to see whether the computer solutions make sense.

a

b

c

d

Prof. Wolfgang Schueller

FLEXURAL SYSTEMS: Beam Types

Problem 4 (EX. 4.2): steel beam design Design a continuous 3-span steel beam with spans of 20 ft (6.10 m), which caries a uniform dead load and live load each of 1.5 klf (21.89 kN/m) using A36 (250 MPa or N/mm2) steel and a W18 (W460) section using SAP. Consider the critical live load arrangement. Assume the weight of the beam is included in the dead load. Consider the beam laterally supported by the floor slab assuming an unsupported length ratio of say 0.1. Use AISC-ASD 89 working stress approach. Problem 5 (Pr. 4.1): concrete beam design Do a preliminary design of the steel beam in EXAMPLE 4.2 as concrete beam using the default concrete material properties in SAP and trying a b/h = 10/20 in (254/508 mm) section; disregard the difference in loading conditions. The typical, interior continuous beam to be investigated, is supported by 12x12-in (305x305-mm) columns, hence has a net span of, ln = 20 – 12/12 = 19 ft (5.79 m). Check the REBAR PERCENTAGE to see whether the assumed section makes sense. Problem 6 (EX. 7.4): the effect of beam: span, continuity, and live load arrangement Investigate for the multi-span beam types below the effect of span, continuity, live load arrangement, and hinging. The beams are shown on a 3-ft (0.914-m) grid, in other words the top beam spans 36 ft (10.97 m), while the bottom 3-span beams each span 12 ft (3.66 m). The beams carry a dead and live loads of 0.5 k/ft (7.30 kN/m) each; investigate the various live load arrangements and determine the critical ones. Design the laterally supported beams using W12 (W310) sections and A36 (250 MPa) steel using working stress approach. Set the self weight of beams equal to zero. Show input (geometry and loading), shear and moment diagrams with numerical values at critical locations, deflections and member sections. Check the design of the beams and make sure that the critical load combinations are used by SAP.

A.

SIMPLE BEAMS

B.

OVERHANGING BEAMS: SINGLE-CANTILEVER BEAMS

C.

OVERHANGING BEAMS: DOUBLE-CANTILEVER BEAMS

2-SPAN CONTINUOUS BEAMS D.

3-SPAN CONTINUOUS BEAMS E.

F.

HINGE-CONNECTED BEAMS

G FIXED BEAMS

Prof. Wolfgang Schueller

Surface Structures: beam membrane

Problem 9 (EX. 12.1): Beam membrane A simply supported, 40-ft (12.19 m) span glulam beam 4 ft (1.22 m) deep, 6 in. (15.24 cm) wide with allowable stresses of, Fv = 165 psi and Fb = 1800 psi, is modeled with membrane elements. Determine how many elements are required for a sufficient accurate solution of the stresses. Try, a) 24 elements, (n = 6 x 4 elements, each one 1 ft x 6.67 ft) with an aspect ratio of 6.67. b) 32 elements, (n = 8 x 4 elements, each one 1 ft x 5 ft) with an aspect ratio of 5. c) 40 elements, (n = 10 x 4 elements, each one 1ft x 4 ft (0.31 x 1.22 m) with an aspect ratio of 4. A load of 1 k/ft (14.59 kN/m) on top of the beam, which includes the self weight, is transformed into a surface load applied along the beam membrane:1/4(1) = 0.25 ksf ≈ 12 kPa (kN/m2). Check the maximum bending and shear stresses manually and compare the values with the S11 (SMAX, SMIN) and S12 computer stress diagrams. The critical computer results are checked disregarding the precise properties for glulam timber.

4'

1 K/ft

40'

Problem 10 (Pr. 12.1): Deep beam behavior Study the beam in EXAMPLE 12.1 further by investigating the following features: 1) Move the right roller support to mid-span to obtain a cantilever beam. 2) Put holes into the web and study the stress distribution. 3) To the simply supported beam add a roller support at mid-span bottom. Study the stress distribution. Then move the center support to the top. What does change? 4) Add two supports at the bottom, i.e. three-span beam with 4-ft cantilever. Compare the stress distribution of the beams as their height-to-span ratio (h/L = 0.33) increases. For further discussion of deep beam behavior refer to Fig. 3.16.

Prof. Wolfgang Schueller

Frame Structures: statically indeterminate portals Problem 14 (EX. 8.3): Introduction to indeterminate portal frames Investigate a 40-ft (12.19-m) span, 15-ft (4.57-m) high, rectangular, single-bay, two-hinged portal steel frame with respect to the effect of indeterminacy, that is, the change of relative member stiffness on the force flow, using A36 (250 MPa). The frames are spaced 40 ft (12.19 m) on center. The following roof loads must be supported: 25 psf (1.20 kPa) dead load, 30 psf (1.44 kPa) live load, and 17 psf (0.81 kPa) wind load against the curtain walls. Consider the following load combinations to check the given sections: COMB1 (D + L), COMB2 (D + W), and COMB3 [D + 0.75(L + W)]. The columns do not sway about their weak axes (Ky = 1) because the building is laterally braced in the long direction. The frame naturally sways in the cross direction, where effective length factor Kx  1.0 is determined by SAP2000 from the stiffness of the members. For the design of the beams, an unbraced length ratio of Lb /L = 0.1 about the minor axis is used for preliminary design purposes. (a) Treat the frame as a beam; use W24x76 beams together with W24x94 columns, where Ib = 0.78Ic, ψ = (Ib/Ic)(Lc/Lb) = 0.78(15/40) = 0.29  0.3 (b) Use a W24x68 beam together with W14x99 columns, where Ib = 1.65Ic: ψ = 0.62  0.6 (c) Use a W24x84 beam together with W14x90 columns, where Ib = 2.37Ic: ψ = 0.89  0.9 (d) Use a W24x94 beam together with W14x90 columns, where Ib = 2.70Ic: ψ = 1.01  1.0 Problem 15 (Pr. 8.5): The effect of member sizes on force flow in indeterminate frames Do a general investigation of the effect of member stiffness on force distribution of the portal frame in EXAMPLE 8.3 under gravity loading only, using modification factors in SAP2000 for, ψ = nLc /Lb = n(15/40) = 0.375n, where n = Ib /Ic, by assuming the following, (a) (b) (c) (d)

for the beam, I, and column, 1.3I, (ψ = 0.29) for the beam, 1.6I, and column, I, (ψ = 0.6) for the beam, 2.1I, and column, I, (ψ = 0.79) for the beam, 2.7I, and column, I, (ψ = 1.01)

Since the relative, not actual, stiffness controls the force distribution, the default section in SAP2000 or any other section may be used; i.e., after you have defined the material (e.g., A36), go to Define > Frame Sections > Add New Property > select e.g., Steel > click on Import a Steel Section > give Section Name, e.g., I > click OK > follow the same procedure for the next section, naming it 1.3I > but click on Property Modifiers > specify Moment of Inertia about 3 axis: 1.3 > click OK > follow the same procedure for the other sections.

Prof. Wolfgang Schueller

Lateral Stability of Building Structures

Problem 17 (EX. 13.1): Eccentrically braced, single-story building A simple single-story, 15-ft (4.57-m) high building is investigated with respect to lateral load flow only, disregarding gravity load action. The building consists of six 20 x 25 ft (6.10 x 7.62 m) bays as shown in Fig.13.7. A lateral uniform wind pressure of 20 psf (0.96 kPa or kN/m2) is assumed. Do a preliminary investigation of the lateral force distribution to the vertical resisting shear walls or braced frames using default sections in SAP a) Investigate the asymmetrical lateral-force resisting braced, hinged frame structure in Fig.13.7a by using the diaphragm constraint to model the concrete floor. . A uniform wind pressure is assumed against the short building façade (WINDY) and against the long façade (WINDX); but the loads are not treated in combination with each other. The wind load at the roof level is equal to, 0.0020(15/2) = 0.15 klf = 2.19 kN/m. In other words, use a horizontal uniform line load along the spandrel beams of, 0.15 k/ft. The total wind pressure against the narrow and broad façades respectively, which the rigid roof diaphragm must support, is Py = 0.15(50) = 7.5 k,

Px = 0.15(60) = 9.00 k

15'

a. 25'

25'

20' 20' 20'

b.

c.

d.

b) Investigate the asymmetrical lateral-force resisting braced, hinged frame structure in Fig. 13.7a by using a 6-inch (152-mm) thick concrete slab for the roof structure rather than a rigid plane as based on diaphragm constraint.

Problem 18 (EX. 13.2): Concrete slab diaphragms with concrete shear walls Assume four 8-in (203-mm) thick concrete shear walls as shown in Fig.13.7b and Fig. 13.10, to resist wind action against the long face of the building. The 6-in (152-mm) thick concrete roof membrane distributes the lateral forces to the shear walls. Determine the lateral force distribution. Use the default setting of the 4000 psi (28 MPa) concrete for the material properties. Disregard the self weight of the members, only consider the wind load. A uniform wind pressure is assumed against the long building façade (WINDX). The wind load at the roof level is equal to, 20(15/2) = 150 plf. The horizontal uniform line load along the spandrel beams, in turn, is replaced by, 150/50 = 3 psf, distributed along the roof membrane. Problem 19 (EX. 13.3): Lateral force distribution for cross-wall structure A simple single-story, 15-ft (4.57-m) high building is investigated with respect to lateral load flow. The building consists of six 20 x 25 ft (6.10 x 7.62 m) bays as shown in Fig.13.7c. A lateral uniform wind pressure of 20 psf (0.96 kPa or kN/m2) is assumed in the y-direction (i.e. short building side). Do a preliminary investigation of the lateral force distribution to the vertical resisting braced frames assuming a 6-inch roof concrete slab. Use 1-in (25-mm) -diameter steel rods for the cross bracing and the default material properties in SAP. Model the rods as tension-only frame elements using a nonlinear analysis. Assume three equally braced, identical, parallel frames as shown in Fig.11.7c, but using also a brace in the other direction so the structure is stable in general, and not just for the symmetrical condition. In this case, the lateral force resisting structures are arranged symmetrically so that there is no torsion under symmetrical load action. •Model the roof as a rigid diaphragm using the diaphragm constraint. •Model the roof as a relatively rigid membrane by assuming a 6-in (152 mm) concrete membrane. •Model the roof as an unfilled deck, say a 1-in (25-mm) thick concrete membrane, and replace the braced bays with 8-in (203-mm) concrete shear walls. Here the roof diaphragm is flexible in comparison to the stiff vertical supports. Problem 20 (EX. 13.4): Lateral force flow to asymmetrically arranged cross frames Change the width of the interior x-braced panel in EXAMPLE 13.3b, to 10 ft (3.05 m). Do a preliminary investigation of the lateral force distribution to the vertical resisting braced frames. Problem 21 (EXAMPLE: 13.5): Lateral force flow in eccentric core structure A simple single-story, 15-ft (4.57-m) high building is investigated with respect to lateral load flow. The building is braced to form a core, a/b = 25/20, as shown in Fig.13.7d. A lateral uniform wind pressure of 20 psf (0.96 kPa or kN/m2) against the short building side is assumed. Do a preliminary investigation of the lateral force distribution to the vertical resisting braced frames assuming a 6-inch (152-mm) roof concrete slab. Use 1-in (25-mm) diameter steel rods for the cross bracing as tension-only members and the default material properties in SAP. A uniform wind pressure is assumed against the short building façade (WINDY). The wind load at the roof level is equal to, 0.0020(15/2) = 0.15 klf = 2.19 kN/m. In other words use a horizontal uniform line load along the spandrel beams of, 0.15 k/ft.

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