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- W 72 EI УБ A For the cantilever beam and loading shown, determine Part 1 (a) the equation of the elastic curve for portion AB of the beam, Part 2 out of 3 (b) the deflection at B, 10 wa 3x² W -2a- B - 16 ax down CSelect an A992 (E= 345 MPa and Fu= 450 MPa) W-shape for the floor beam AB of the floor system shown. In addition to the weight of the beam, the service dead load consists of 125 mm thick reinforced concrete slab (we= 24 KN/m³). The service live load is 3.80 KPa and there is 0.95 KPa movable partition load. The total deflection (DL + LL) must not exceed L/240, where L is the length between supports. Use ASD tairder Flear Beam - Hax column Im overhang 4@1.5 m = 6 mH.W-4: For the beam of sections and detail shown in Fig. find the maximum live load (PL) that the beam can be carry it. Neglect self-weight? Ans. PL=195.74 kN Use f 35 = - MPa fy 420 -2.0m- 70mm 2PL 2020mm 5025mm 350 mm sct.a-a 600 mm- a a -2.0m- -LHHH] .... .... 8025mm Soemy 3025mm 350 mm sct.b-b b b -2.0 m- 600 mm PL
- Q3) The beam shown in Figure below has lateral support at the ends only. The concentrated loads are live loads. Use A992 steel and select a W shape. (Do not check deflections. Use C-1). 23 k 25 k ttstA simply supported beam is subjected to a uniform service dead load of 15kN/m(including the weight of the beam), a uniform service live load of 30kN/m. the beam is 12 meters long and is laterally supported at the midspan, and A572 Gr.50 steel is used. Is W30x108 adequate? Assume Cb=1 by 4 d fore k h h Z S Ly 267 TH mm 19.3 757 13.8 min m 35.8 685.4 54.6 737.7 5670 x10^3 4900 x103 mm 60.8 x10 6 mm 2080 x10 3 8300 x10^9 mm wwwQ4 For the simply supported beam shown in Figure below, design the cross section against the maximum bending moment using Working Stress Method. The width of beam is 300 mm. Use 025 mm, fe 10 Mpa, f P-40 kN WD L) 14.32 kN/m (Including self-weight) L-8m
- For the beam below, determine the following, assume EI is constant. What is FEMAB in k-ft? -18 --96 Select the correct response: -69 2 k/ft 18 24ft4 - For the beam loading configuration and cross-section shown below, what is the maximum allowable load P given that the maximum allowable stress (in tension and compression) is 120 MPa? PO Р 1.6m PN ż 0.8m 0.8m P80 1.6m Р 30 30 T 10 dimensions in mm beam cross-section (same as problem 3, but inverted)The cross-sectison of a simply supported plate girder is shown in figure. The loading on the grider is symmetrical. The bearing stiffeners at supports are the sole means of providing restraint against torsion. Design the bearing stiffeners at supports. with minimum moment of inertia about the center line of web palte only as the sole design criterion. The flat section available are: 250 x 25, 250 x 32, 200 x 28, and 200 x 32 mm. Draw a sketch 500 25 8 .1445 Dimensions in mm 20 +425
- Q4 For the simply supported beam shown in Figure below, design the cross section against the maximum bending moment using Working Stress Method. The width of beam is 300 mm. Use 025 mm, fe 10 Mpa, f, = 140 Mpa and n-10. %3D P-40 kN WD L) 14.32 kN/m (Including self-weight) L-8mPart 1 The simply supported beam below is initially designed to undertake a concentrated factored live load of 45 kips (215 kN). Assume that the beam is laterally supported and the self weight neglected. Use: A992 steel (Fy= 50 ksi or 345 MPa) E = 29 x 10® psi (200 GPa), the maximum allowable deflection Aall = L/360 45 kips 16 ft 16 ft 1/ Determine the most economical W shape beam required Part 2 The most economical selected beam from Part 1 ends up being subjected to a concentrated live load of 20 kip. Determine the corresponding maximum live load deflection. Part 3 Assuming that the beam will only be braced laterally at the supports and every quarter spam points (8ft from each support), verify whether the chosen cross section still checks under the live load of 20 kips only. If not, what are your design recommendations?Acantilever beam AB, loaded by a uniform load (see figure), is applied on a channel section. If the cross section has the dimensions indicated and the moment of inertia about the z axis (the neutral axis) is I= 3.36 in" Find the maximum tensile stress fhtMAX and maximum compressive stress facMAX (Note: Draw the shear and moment diagram). 2.5 lb/ft 0.617 in. Z- C 2.269 in. 6.0 ft