Select the lightest W shape for the beam shown in Figure1.Consider the following cases: (6) A36 steel , L_6ft; (6) A36 steel , L_14ft; A36 steel , L_24ft; Note: Consider moment only. 2. (c)
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- Q2) The members of the truss structure shown below is plain concrete. The compressive strength of the concrete is 25 MPa. Compute the maximum load P that can be carried by the structure. (Cross section of each member of the truss is 200 x 200 mm and don't use material factors and do not consider slenderness) Comment on your results briefly. P A& 2m SC 2 m 1380 2m DVerify the adequacy of column AB, part of a sway frame structure, to carry the loads shown in the figure. The column is made of ASTM A36 steel (Fy = 250 MPa). Use ASD specifications. For determining its effective length factor, neglect inelastic effect. %3D W21X62 C W21X62 L=5 m L=5 m Flanges of Columns and web of girders are in the plane of the frame NOTE: Take note of the moment of inertia to be used (ly) for columns. W18X50 W18X50 L=5 m L=5 m P(DL) = 40 kN NM 001 = (10)d NM 00S = (1)d NY SZI = (1)d W21X62 L-4.2m W21X62 L=4.2 mQuestion 6 of 6 Fill in the blanks: Determine the design moment capacity of the beam with given properties below. bf = 400mm bw = 350mm tf = 120mm d = 370mm d' = 100mm fc' 34.5 MPa fy = 420 MPa As = 8- 28mm bars USE NSCP 2015, Mu = KN-m
- Use A 992 Steel and Select the most economical W shape for the beam shown in the figure below. The beam weight is not included in the service loads shown. Do not check deflection. Assume continuous Lateral support. K 6 IPD = lok PL=20k * 161 1 WD= 3-33 K/F+ WL=6-67 k/Ft -xhulDetermine the design moment capacity of the beam with given properties below. bf = 650mm bw = 350mm tf = 120mm d = 370mm d' = 100mm fc' = 34.5 MPa fy = 400 MPa As = 8 - 28mm bars USE NSCP 2015, Mu = kN-mSituation 3: A W12x22 simply supported beam carries a uniformly distributed load. The properties of the section relevant to the problem are as follow: Use A50 steel and Cb - 1.14 d=312 mm bf - 102 mm tf 10.8 mm tw = 6.60 mm kdes 18.4 mm ry-21.5 mm Ix- 64.5 x 10^6 mm^4 ly- 1.94 x 10^6 mm^4 Sx416 x 10^3 mm^3 Zx 480 x 10^3 mm^3 J-122 x 10^3 mm^3 Cw-44.0 x 10^9 mm^6 1. Which of the following nearly gives the Design Strength Moment (LRFD) if the span of the beam is 2.5m? 89 KN-m O 114 KN-m O 123 KN-m O 101 KN-m
- For the beam shown below: a) Calculate the elastic curve equation O Ely Mo A R³L 6 D Mox³L + x + C₂ L- BThe 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 +425Situation 3: A W12x22 simply supported beam carries a uniformly distributed load. The properties of the section relevant to the problem are as follow: Use A50 steel and Cb - 1.14 d-312 mm bf - 102 mm tf = 10.8 mm tw = 6.60 mm kdes - 18.4 mm ry-21.5 mm Ix - 64.5 x 10^6 mm^4 ly-1.94 x 10^6 mm^4 Sx - 416 x 10^3 mm^3 Zx 480 x 10^3 mm^3 J-122 x 10^3 mm^3 Cw-44.0 x 10^9 mm^6 1. Which of the following nearly gives the maximum safe uniform load (Wu) that the beam could if the span of the beam is 0.800m? Neglect weight of the beam. 960 KN/m 933 KN/m 922 KN/m 975 KN/m
- Situation 1: A W8x31 simply supported beam carries a uniformly distributed load. The properties of the section relevant to the problem are as follow: Use A50 steel and Cb - 1.14 d=203 mm bf = 203 mm tf-11.0 mm tw = 7.24 mm kdes - 21.1 mm ry = 51.3 mm Ix = 45.8 x 10^6 mm^4 ly- 15.4 x 10^6 mm^4 Sx-451 x 10^3 mm^3 Zx = 498 x 10^3 mm^3 J-223 x 10^3 mm^3 Cw-142 x 10^9 mm^6 1. Which of the following nearly gives the maximum safe uniform load (Wu) that the beam could carry if the span of the beam is 5m? Neglect weight of the beam. 46 KN/m 41 KN/m 51 KN/m 33 KN/mW3D 25 Y= 155 %31 1oow= 2500 KN 27=310mm X= 475 Z= 60 %3D 102=600mmA built-up section shown, 30 feet long, is fixed at both ends and braced in the weak direction at the midheight. A992 steel is used. Determine the following: 11. Ix 12. Ky 13. effective slenderness ratio 14. Fe 15. LRFD design strength 16. ASD allowable strength 18 4 W10 × 49 A 14.4 in² bf 10 in d. 10 in tw 0.340 in tf 0.560 in Ix= 272 in¹ Iy=93 4in¹ W10×54 A 15-3 in² bf 10 in d = 10 lin tw0-370 in tf 0.615 in Ix= 303 in 4 Ty = 103 in 1