Workshop Homework Problems (metric) based on SAP2000.pdf by Wolfgang Schueller

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

Workshop Homework Problems (metric) 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) EXAMPLE 6.1: Analysis of basic truss forms Investigate some simple, basic truss forms 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, that is, contours. Load arrangement, load direction, and load location: symmetrical and asymmetrical, vertical and horizontal, top-loaded and bottom-loaded. 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 1x 1-m square grid, as shown in the drawing below. Apply vertical, single loads of 5 kN (e.g. P vertical) at the top chord joints, as indicated in the drawing; treat the horizontal forces of 2 kN (e.g. P horizontal) as a 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 the deflection conceptually. 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.

1 1

The results of SAP2000 are checked for the particular truss layout of a parallel chord, modified Warren truss in case (a). For the manual check of some of the computer results, the following process is used. Check some of the SAP2000 results manually Check Warren truss: case a,


Cable-Supported Structures: stayed bridges

Problem 3 : Stayed bridges Investigate several basic stayed bridge structures shown below under a uniform gravity load of 40 kN/m; but assign 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 steel  250 MPa) as an initial input into SAP2000. Keep in mind that the beams in reality may consist of multi-cell roadway box girders. Use as a first trial nominal 50-mm diameter strands for the inner stays and 40-mm diameter strands for the outer stays, using A416 Gr. 250 (≈ 1500 N/mm2, MPa). Neglect axial strains in the towers by allocating large values to the cross-sectional area (just assume 1x1-m concrete sections using 6000 psi ≈ 40 N/mm2 for this conceptual study). Draw the bridge shapes on a 5x5-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 that 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


FLEXURAL SYSTEMS: Load Types and Boundary Conditions Problem 4: Beam load types Investigate for the beam cases shown below,  Boundary conditions, including cantilever action  Load types, a counterclockwise moment of 42 kNm acts at the left support of case (k), clockwise uniform torsional loads of 4 kNm/m act along the beam span of case (e), a clockwise moment of 63 kNm acts at the right reaction of case (d), and a counterclockwise moment of 16 kNm at the left support of the same case, etc.  Load distribution  Indeterminate action The beams are drawn on a 2-m grid; in other words, the beams span 6 m. The uniform load is 14 kN/m unless shown. For beam selection use SAP default setting (i.e. FSEC1). Show input (geometry and loading), moment, and shear diagrams with numerical values at critical locations, and deflections. The critical moments of the cases are checked (by referring to TABLES B.10 and B.11, or other references) to make sure that the computer results are meaningful; check the indeterminate beams approximately.

14 kN/m

A. 14 kN/m

14 kN/m

14 kN/m

14 kN/m

B.

C. 42 kNm D.

42 kNm

63 kNm

14 kN/m

I.

J. 14 kN/m 84 kN

4 kNm/m

K.

E. F.

G.

H.

42 kN

28 kN/m

28 kN/m

42 kN

L.

28 kN 28 kN 28 kN

M.

21 kN/m

N.

28 kN/m

28 kN/m

7 kN/m

O.


FLEXURAL SYSTEMS: Floor Framing 1

Problem 11: Design of steel floor beams The floor framing for a typical interior bay of a multi-story braced steel skeleton structure is shown below. The composite deck distributes dead and live load of 4 kN/m2 each to the beams; the live-load reduction factors are 0.96 for the beams and 0.8 for the interior girders; however, for this approximation, 0.96 also is used conservatively for the girders. Use A36 (Fy = 36 ksi ≈ 250 MPa or N/mm2= 25 kN/cm2) steel or Q235 (Fy = 235 N/mm2) in China 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 or the Chinese 2002 design code). For the automatic section selection in SAP2000, assume W18 (W460) beams or GB-HN300 to GB-HN400 in China which, however, is not included in the SAP2000Ed version. .

GI

3 Sp @ 2.5 m = 7.5 m

8m

BM

BM

BM

BM

GI


FLEXURAL SYSTEMS: Floor Framing 2

Problem 12: Design of steel floor framing Investigate the simple, hinged floor framing for the three 6 x 6-m bays shown below. The floor load consists of a dead load of 3.40 kN/m2 where beam weight is included, and a live load of 2.90 kN/m2; 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 (Fy = 36 ksi ≈ 250 MPa or N/mm2= 25 kN/cm2) steel and flexible connections; use working stress approach (AISC-ASD89). For the automatic section selection in SAP2000, try for the filler beams W10 (W250) sections and for the girders W16 (W410) sections. Check manually beams BM1, G2 and G4 to see whether the computer output makes sense.

G1

6m

BM2

BM1

BM1

BM2

G2

BM2 BM5

Bm5

BM1

G1

G4

BM5

BM5

G3

BM3

BM1 BM5

BM5

BM4 3 Sp @ 2 m = 6 m

BM2 6m

3 Sp @ 2 m = 6 m

BM3


Surface Structures: slabs Problem 15: square concrete slabs Investigate a square 15 cm concrete slab, 4 x 4 m in size shown below that carries a uniform load of 5.75 kN/m2 (COMB1), that is a dead load of 0.15(23.56) = 3.53 kN/m2 for its own weight (SLABDL taken care by self weight) and an additional dead load 0.24 kN/m2 (SUPERD), and a live load of 1.92 kN/m2 (LIVE). Use the default concrete material properties in SAP2000 (i.e. 4000 psi ≈ 28 MPa or N/mm2 = 2.8 kN/cm2) or in China C30 (fc’ = 30 N/mm2) and A615 Grade 60 rebars (Fy = 60 ksi ≈ 420 N/mm2 = 42 kN/cm2) or Chinese HRB400 (Fy = 360 N/mm2). Solve the problem by using 0.5 x 0.5 m plate elements using SAP. Use a concrete cover of reinforcement of 19 mm. 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 3 x 3 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

b 0.5 m

4m

3m

c

0.5 m

4m

0.5 m 2m 3m 2m 0.5 m d

e

f


FLEXURAL SYSTEMS: Floor Framing

Problem 16: Design of concrete floor beams A 6-story concrete frame office building consists of 9x10-m bays with a typical interior floor framing bay shown below. The 16.5-cm concrete slab weighs 0.165 m [24 kN/m3 (4.50 m – 0.30 m)/4.5 m] = 3.70 kN/m2, while the self weight of the floor beams is automatically taken into account by SAP2000. The slab supports 0.24 kN/m2 for ceiling and floor finish, a partition of 1.00 kN/m2, as well as a live load of 4.00 kN/m2. Hence, the superimposed dead load is: 3.70 + 0.24 + 1.00 = 4.94 kN/m2. Disregard the critical pattern live load moments in SAP2000 for this first investigation. The girders are 610 mm high and 410 mm wide, whereas the beams have the same depth but are 300 mm wide. The 3.70-m high columns have a size of 460x460 mm for the lower floors. A typical interior intermediate floor beam (i.e., beam between column lines) is investigated briefly assuming the ends of the columns fixed at the floors above and below (modeling 6 bays). Use a concrete strength of fc' = 4000 psi (28 N/mm2, MPa) or in China C30 (fc’ = 30 N/mm2), and A615 Grade 60 rebars, fy = fys = 60 ksi ≈ 420 N/mm2 = 42 kN/cm2, or Chinese HRB400 (Fy = 360 N/mm2), and assume a concrete cover to rebar center of 65 mm. Use ACI 318-05 or the Chinese 2002 design code (which however is not included in the SAP2000Ed version). The net spans of typical beams, girders, and columns respectively are, lbm1 = 10.00 – 0.41 = 9.59 m lbm2 = 10.00 – 0.46 = 9.54 m lg = 9.00 −0.46 = 8.54 m lcn= 3.70 − 0.610 = 3.09 m


Frame Structures: folded beams

Problem 17: Folded beam systems Simple-span 6-m span, folded beam systems as shown below, are drawn on a 1.50 x 1.50-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 48 kN for each case at mid-span (SINGLEP). b) Use a uniform load of 16 kN/m on global z-projection (UNIFORM). For the determinate structures the frame elements may be modeled by using A36 (Fy = 36 ksi ≈ 250 MPa or N/mm2= 25 kN/cm2) steel and the default setting (FSEC1 section), since member stiffness has no effect on the magnitude of internal member forces; but, do not use deflection results. 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


Frame Structures: three-hinged frames

Problem 18: Three-hinged frames Investigate the following 12-m span, three-hinged frame structure systems drawn on a 1.50 x 1.50 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 7.30 kN/m and a live load (L) of 10.95 kN/m on global zprojection and a lateral wind load of 7.20 kN/m on global x-projection (i.e. 0.80 kN/m2 for 9-m 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)]. For the design the steel frames use the AISC-ASD 89 working stress approach as based on A36 steel (Fy = 36 ksi ≈ 250 MPa or N/mm2= 25 kN/cm2) and W21 (i.e. W530) sections (AUTOW21) but use W12 (i.e. W310) 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 g h


Frame Structures: basic arches Problem 21: Basic arches Investigate the simple circular three-hinged, 12-m span arch systems and half-arch 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 2.4 m high consisting of ten linear segments and the steep arch is a semicircular arch consisting of 12 linear segments. Use for dead load wD = 7.30 kN/m (D) applied along the arch, for live load wL = 7.30 kN/m (LFULL for full loading and LHALF for loading half span) on the horizontal roof projection, and for wind wW = 5.80 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 1.20 x 1.20-m grid. Select W10 (i.e. W250) sections using Auto Select and AISC-ASD 89 working stress approach as based on A36 steel (Fy = 36 ksi ≈ 250 MPa or N/mm2= 25 kN/cm2). 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.

2.4 m

90 m 12

a

b

c

d 6m


FLEXURAL SYSTEMS: Beam Design

Problem 5: steel beam design Design a continuous 3-span steel beam with spans of 6 m, which carries a uniform dead load and live load each of 22 kN/m, using A36 (Fy = 36 ksi ≈ 250 MPa or N/mm2= 25 kN/cm2) or Q235 (Fy = 235 N/mm2 ) in China and a W18(W460) section or GB-HN300 to GB-HN400 in China; 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 unbraced length ratio about the minor axis of say 0.1. Use the AISC-ASD 89 working stress approach or the Chinese 2002 design code, which however is not included in the SAP2000Ed version. (I cannot check the Chinese solution because I do not know the Chinese 2002 design code! SAP2000 does not seem to auto select the most economical member but uses the largest section of the list)

Manual check of SAP2000 results: a) ALLOWABLE STRESS DESIGN (ASD): fb = M/S ≤ Fb = 0.66 Fy, or Sx  Mx /Fb = Mx/0.66Fy The critical moment occurs at the interior support and may be roughly approximated as, M  wL2/10 = 2(22)(6)2/10 = 158.4 kN-m = 15840 kN-cm Sx Mx/Fb = Mx/0.66Fy = 15840/0.66(25) = 960 cm3

Try W18x40 , (W460 x60), Sx = 1121 cm3 (from SAP2000). The SAP2000 reports the same section with a stress ratio fb /Fb = 0.925. b) LOAD AND RESISTANT FACTOR DESIGN (LRFD): Mu ≤ фbMn = фbMp = 0.9FyZx, or Zx ≥ Mu/0.9Fy, Mu ≈ (1.2wD +1.6wL)L2/10 = (1.2(22) + 1.6(22)62/10 = 222 kNm = 22176 kNcm Zx ≥ Mu/0.9Fy = 22176/0.9(25) = 986 cm3, try W18 x 35,, (460x52), Zx = 1090 cm3, using AISCLRFD99 The SAP2000 reports the same section with a stress ratio fb /Fb = 0.988.

Problem 6: concrete beam design Do a preliminary design of the steel beam in Problem 5 as a concrete beam disregarding the change in loading, using the default concrete material properties in SAP2000 (i.e. 4000 psi ≈ 28 MPa or N/mm2 = 2.8 kN/cm2) or in China C30 (fc’ = 30 N/mm2) and trying b/h = 25/50 cm (9.84/19.69 in) section; disregard the difference in loading conditions. Use ACI 318-05 or the Chinese 2002 design code (which however is not included in the SAP2000Ed version) and A615 Grade 60 rebars (Fy = 60 ksi ≈ 420 N/mm2 = 42 kN/cm2) or Chinese HRB400 (Fy = 360 N/mm2) and a bar cover of 6 cm. The typical, interior continuous beam to be investigated is supported by 30x30-cm columns, hence the beam has a net span of, ln = 6.00 – 0.30 = 5.70 m, or the end length offsets are 15 cm. Check the REBAR PERCENTAGE to see whether the assumed section makes sense. (I cannot check the Chinese solution because I do not know the Chinese 2002 design code!)


FLEXURAL SYSTEMS: Beam Types

Problem 7: the effect of beam: span, continuity, and live load arrangement Investigate for the multi-span beam types shown below, the effect of span, continuity, live-load arrangement, and hinging. The beams are shown on a 1.00-m grid, in other words, the top beam spans 12 m, while the 3-span beams each span 4 m. The beams carry dead and live loads of 7 kN/m each; investigate the various live-load arrangements and determine the critical ones. Design the laterally supported beams using W12 (W310) sections or GB-HN200 to 400 in China; use A36 (Fy= 36 ksi ≈ 250 MPa or N/mm2=25 kN/cm2) steel or Q235 (Fy = 235 N/mm2 ) in China as based on the AISC-ASD 89 working stress approach or the Chinese 2002 design code, which however is not included in the SAP2000Ed version. 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 SAP2000.

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


FLEXURAL SYSTEMS: Column Design

Problem 8: Steel column design Determine the preliminary size of a W14 (W360) column using A36 (Fy = 36 ksi ≈ 250 MPa or N/mm2=25kN/cm2), that is 3.66 m long and is not braced about its strong axis, but is braced about its weak axis and hence does not sway in that direction. The column carries an axial load of 2224 kN at the top. As a first trial, assume Ky = 1.0 and Kx = 1.5. Use the AISC-ASD 89 working stress approach.

Manual check of SAP2000 results: A typical column concentrically loaded by an axial force, P, causes a uniform compressive stress that should not exceed the allowable axial compressive stress, Fa that is the flexural buckling stress (i.e. critical stress), Fcr divided by the safety factor Ωc. Hence the required cross-sectional area, Ag, of the section is as follows, fa = P/Ag ≤ Fa = Fcr/Ωc or 

Ag ≥ P/Fa

Short columns are prevented from buckling below a slenderness of roughly, Kl/r = 30, for preliminary design purposes, Fa = Fcr / Ωc = Fcr/1.67 = 0.6Fcr = 0.6Fy = 0.6(25) = 15 kN/cm2



Long columns fail in elastic buckling above a slenderness of roughly, Kl/r = 126 ≤ 200, for A36 steel, Fa = Fcr / Ωc = [π 2E/(Kl/r)2]/1.92 = 102808/(Kl/r)2, where, E = 20000 kN/cm2



Intermediate columns fail in inelastic buckling; for preliminary design purposes assume for for Fy = 25 kN/cm2:

Fa ≈ 15.0 − 0.06Kl/r

For preliminary estimation purposes, it may be assumed that the braced weak axis controls the design: For W14, rrmin ≈ 10 cm (4 in),

hence, Fa ≈ 15 – 0.06(1(366/10) = 12.80 kN/cm2

Hence the required cross-sectional area, Ag, of the section is as follows, A = P/Ag = 2224/12.80 = 173.75 cm2

Try W14x99 (W360 x147), A = 187.74 cm2, rmin = 9.44 cm, rmax = 15.69 cm SAP2000 reports a W14x90 , A = 170.97 cm2, Fa = 13.29 kN/cm2 Check assumption: (KL/r)x = 1.5(366)/15.69 = 34.99, (KL/r)y = 1(366)/9.44 = 38.77 > 34.99 O.K.


Surface Structures: beam membrane

Problem 13: Beam membrane A simply supported, 12-m span glulam beam is 1.00 m deep and 15 cm wide. It has the following material properties for wood:        

Weight per unit volume: γ = 5.50 kN/m3 Mass per unit volume: m =  /g = 5.5/9,81 = 0.5606 kg/m3 Modulus of elasticity: E =11000 MPa (N/mm2) = 1100 kN/cm2 Poisson's ratio: ν = 0.3 Coefficient of thermal expansion: ε t = 3.78 (10)−6 (1/°C) = 3.780E-06 Shear modulus: G = 4231 MPa (N/mm2) = 423kN/cm2 Allowable bending stress: Fb = 12.4 MPa = 1.24 kN/cm2 Allowable shear stress: Fv = 1.14MPa (N/mm2) = 0.11 kN/cm2

The beam is modeled using membrane elements as indicated below. Determine how many elements are required for a sufficiently accurate solution of the stresses. Try, (a) 24 elements (n = 6 x 4 elements, each one 2.00 x 0.25 m), with an aspect ratio of 8. (b) 32 elements (n = 8 x 4 elements, each one 1.50 x 0.25 m), with an aspect ratio of 6. (c) 40 elements (n = 10 x 4 elements, each one 1.20 x 0.25 m), with an aspect ratio of 4.8. A load of 14 kN/m on top of the beam, which includes the self weight, is transformed into a surface load applied along the beam membrane: 14/1(1) = 14 kN/m2 = 1.40 N/cm2. 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 section properties for glulam timber.

1m

14 kN/m

12 m

Allowable stresses: wood: Fb ≈ 0.8 kN/cm2 = 8 N/mm2, Fv ≈ 0.1 kN/cm2 = 1 N/mm2, concrete: Fb ≈ 1.2 kN/cm2 = 12 N/mm2, Fv ≈ 0.05 kN/cm2 = 0.5 N/mm2, Ft ≈ 0.07kN/cm2 = 0.7 N/mm2 Problem 14: Deep beam behavior Study the beam in Problem 13 further by investigating the following features: (a) Put holes into the web and study the stress distribution. (b) Move the right roller support to midspan to obtain a cantilever beam by using the Model–Alive feature.


Frame Structures: statically indeterminate portals

Problem 19: Introduction to indeterminate portal frames Investigate a 12-m span, 4.50-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. The frames are spaced 12 m on center. For the design check of the steel frames use the AISC-ASD 89 working stress approach as based on A36 steel (Fy = 36 ksi ≈ 250 MPa or N/mm2= 25 kN/cm2). The following roof loads must be supported: 1.20 kN/m2 dead load, 1.44 kN/m2 live load, and 0.81 kN/m2 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)]. wD = 1.2(12) = 14.40 kN/m, wL = 1.44(12) = 17.28 kN/m, PW = 0.81(12 x 2.25) = 21.87 kN 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 (W610x113) beams together with W24x94 (W610x140) columns, where Ib = 0.78Ic, ψ = (Ib/Ic)(Lc/Lb) = 0.78(15/40) = 0.29  0.3 (b) Use a W24x68 (W610x101) beam together with W14x99 (W360x147) columns, where Ib = 1.65Ic: ψ = 0.62  0.6 (c) Use a W24x84 (W610x125) beam together with W14x90 (W360x134) columns, where Ib = 2.37Ic: ψ = 0.89  0.9 (d) Use a W24x94 (W610x140) beam together with W14x90 (W360x134) columns, where Ib = 2.70Ic: ψ = 1.01  1.0

EXPLANATIONS: The support moment, Ms, in the two-hinged frame is dependent on the stiffness of the beam (EI/L)b and the column (EI/L)c or for one material (Ec = Eb) on the relative stiffness factors Ib /Lb and Ic /Lc , that is on a distribution factor, ψ, and is equal to, MS = −(wL2/12)[3/(3+2ψ)] where, ψ = (I/L)b /(I/L)c = (Ib /Lb)Lc /Ic = n(Lc /Lb),

where Ib = nIc.

For some typical conditions, the support or column moment Ms is, ψ = 1.0 (i.e., equal beam and column stiffness): Ms =−wL2/20 ψ = 0.5:

Ms = −wL2/16

ψ = Lc /Lb , say = 0.3: Ms ≈ −wL2/14 (e.g., for same column and beam section, possibly braced frame)


Lateral Stability of Building Structures

Problem 22: Eccentrically braced, single-story building A simple single-story, 4.50-m high building is investigated with respect to lateral load flow only, disregarding gravity load action. The building consists of six 6.00 x 7.50 m bays as shown below. A lateral uniform wind pressure of 1 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 steel frame structure in case (a) 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, 1.00(4.5/2) = 2.25 kN/m. In other words, use a horizontal uniform line load along the spandrel beams of, 2.25 kN/m. For the determinate frame structure the frame elements may be modeled by using the default setting (FSEC1 section), since member stiffness has no effect on the magnitude of internal member forces; but, do not use deflection results.

4.5 m

a. 7.5 m

7.5 m

6m 6m 6m

b.

c.

d.

b) Investigate the asymmetrical lateral-force resisting braced, hinged frame structure in case (a), by using a 15-cm thick concrete slab for the roof structure rather than a rigid plane as based on diaphragm constraint.

Workshop Homework Problems (metric) based on SAP2000.pdf by Wolfgang Schueller

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