A compression member is subjected to service loads of 165 kips dead load and 535 kips of live load. The member is 26 ft. long and pinned at each end. Use A992 steel and select the lightest W section. by utilizing AISC equations and table 4-1 only.
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- If the beam in Problem 5.5-9 i5 braced at A, B, and C, compute for the unbr Cb aced length AC (same as Cb for unbraced length CB). Do not include the beam weight in the loading. a. Use the unfactored service loads. b. Use factored loads.4.3-4 Determine the available strength of the compression member shown in Figure P4.3-4. in each of the following ways: a. Use AISC Equation E3-2 or E3-3. Compute both the design strength for LRFD and the allowable strength for ASD. 15 HSS 10x6x ASTM A500, Grade B steel (Fy=46 ksi) 2/3A W14X120 is used as a tension member in atruss. The flanges of the member are connected to a gusset plate by 3/4 inch boltas shown below. Use A36 steel with Fy-36 ksi and Fu=58 ksi Determine the Yielding Capacity of the section based on LRFD (kips) Determine the Tensile Rupture capacity of the section based on LRFD Determine the Demand to Governing Capacity Ratio (based on yielding and rupture only) if the Demand load carried by the section are DL=200 kips LL=400 kips use LRFD Properties and Dimension Ag=35.30 in^2 x = 6.24 in ry= 3.74 in d=14.5 in tf=0.94 in bf=14.7 in tw=0.59 in k=1.54 d=14.5 Y k1=1.5 bf=14.7 tf-0.94 X -tw=0.59 H
- Question 4: A W12 x 65 of A572 Grade 60 steel is used as a compression member. It is 26 feet long, pinned at each end, and has additional support in the weak direction at a point 12 feet from the top. Can this member resist a service dead load of 180 kips and a service live load of 320 kips? a. Use LRFD. b. Use ASD.snipA beam must be designed to the following specifications: Span length = 35 ft Beam spacing = 10 ft 2-in. deck with 3 in. of lightweight concrete fill (wc=115 pcf) for a total depth of t=5 in. Total weight of deck and slab = 51 psf Construction load = 20 psf Partition load = 20 psf Miscellaneous dead load = 10 psf Live load = 80 psf Fy=50 ksi, fc=4 ksi Assume continuous lateral support and use LRFD. a. Design a noncomposite beam. Compute the total deflection (there is no limit to be checked). b. Design a composite beam and specify the size and number of stud anchors required. Assume one stud at each beam location. Compute the maximum total deflection as follows: 1. Use the transformed section. 2. Use the lower-bound moment of inertia.
- A plate girder must be designed for the conditions shown in Figure P10.7-4. The given loads are factored, and the uniformly distributed load includes a conservative estimate of the girder weight. Lateral support is provided at the ands and at the load points. Use LRFD for that following: a. Select the, flange and web dimensions so that intermediate stiffeners will he required. Use Fy=50 ksi and a total depth of 50 inches. Bearing stiffeners will be used at the ends and at the load points, but do not proportion them. b. Determine the locations of the intermediate stiffeners, but do not proportion them.The beam shown in Figure is a two-span beam with a pin (hinge) in the center of the left span,making the beam statically determinate. There is continuous lateral support. The concentratedloads are service live loads. Determine whether a W12 × 79 of A992 steel is adequate.a. Use LRFD.b. Use ASD.Compute the maximum acceptable tensile SERVICE LOAD that may act on a single tee section that is connected to a gusset plate using welds 12 inches long, as shown in the figure. The service live load is three times the dead load. Use A992 steel. USE LRFD ONLY, no block shear will occur. What is the Maximum Tensile SERVICE Load Round your answer to 3 decimal places.