Structural Analysis
6th Edition
ISBN: 9781337630931
Author: KASSIMALI, Aslam.
Publisher: Cengage,
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- Problem 2) Determine the maximum compressive force developed in member HG of the side truss due to the wheel loads of the car and trailer.arrow_forwardSOLVE FOR THE FORCE IN ALL MEMBERS OF THE TRUSS. SHOW SOLUTIONS CLEARLY USING METHOD OF JOINTS AND WRITE WELL. 0.8m 2.4m A BD BD 3 KN MEMBER AB AC BC BE BF EF GE FG 2m Av(REACTION @ hinge) Gv(REACTION @ roller) C 1m 10KN B 1m 8KN 3.5m SUMMARY OF RESULTS FORCE E 8KN 1m F 2.4m 6KN G NATURE(T or C)arrow_forwardLight-grade steel channel was used as a purlin of a truss. The top chord of the truss is inclined I V : 4 H and distance between trusses is equal to 6 m. The purlin has a weight of 79 N/m and spaced at 1.2 m. on centers. The dead load including the roof materials is 720 Pa, live load of 1000 Pa and wind load of 1.2 1.2 1440 Pa. Coefficient of Purlins pressure at leeward and windward are 0.6 and 0.2 respectively. Assume all loads passes through the centroid of the section. Properties of C 200 x 76 mm Sx = 6.19 x 104 mm Sy = 1.38 x 104 mm3 W = 79 N/m 1.2 Truss 1.2 12 12 1.2 Allowable bending stress Fbr = Fby = 207 MPa Truss %3D 6m O Calculate the bending stress, fox, for dead load and live load combination (D + L). Calculate the bending stress, foy, for dead load and live load combination (D + L). O Calculate the maximum ratio of actual to the allowable bending stress for load combination 0.75 (D + L + W) at the windward side. fbx = 151.14 MPa fby = 169.6 MPa Interaction = 1.25arrow_forward
- A 66-foot long plate girder must support a uniformly distributed load and concentrated loads at the one-third points. The uniform load consists of a 1.3-kipyft dead load and a 2.3 kip/ft live load. Each concentrated load consists of a 28-kip dead load and a 49-kip live load. There is lateral support at the ends and at the one-third points. Use A572 Grade 50 steel, a total depth of 80 inches, and LRFD. a. Select the girder cross section and the required spacing of intermediate stiffeners. b. Determine the size of intermediate and bearing stiffeners. c. Design all weldsarrow_forward2. Determine the force in each member of the truss. IHG F D 2a L6@a²6aarrow_forwardc) The three suspender bars are made of A-36 steel and have equal cross-sectional areas of 450 mm² .Determine the average normal stress in each bar if the rigid beam is subjected to the loading shown. A B C 80 kN 2 m 50 kN D E F -1m--1m--1 m--1m-|arrow_forward
- .arrow_forward4/47 Compute the force in member HN of the loaded truss. BE 2 m A 4/40 n L L D R Q L L F 6 m L P O N -8 panels at 3 m Problem 4/47 age to L BODJEN H M L I L L 2- J [2 m EM Karrow_forwardPROBLEM SET #5 . Situation 1 Determine the force in each member of the truss and state whether it is in tension or compression. . Situation 2 -2 m PROBLEM SET #5 12 ft H B 600 N 2k 2 m -2 m B 800 N Determine the force in members GH, CH, BC, FG, and CG. State whether the it is tension or compression. -2 m 600 N 3k 3 k 2. C D 2 m 10 ft 10 ft 10 ft 10 ft Earrow_forward
- Learning Goal: To design a plate girder for bending, confirm the shear stress limit, and determine the required weld stress limit for the web/flange joints. A half-through bridge is a design that uses two l-section plate girders to support the roadway as shown in the cross section below. Suppose this kind of design is to be used bridge with a span of 80 ft . Each plate girder can be modeled as a simply supported beam with a uniform load of 6 kip/ft. The web thickness tw = 0.5 in. , flange thickne tf = 2 in. , and web depth d = 60 in. have all been determined (Figure 1). The allowable stresses for the plate girder are oallow = 30 ksi and Tallow = 18 ksi Beams support loads perpendicular to their long axes. For beams with long spans, the bending loads typically cause the highest stresses, so the beam section is designed based on the maximum moment. Then the chosen design can be checked for adequate shear stress limit. If the beam has a short span and large loads, chen the shear stress…arrow_forward1. Girders AC and DF have a width of 350 mm and a total depth of 500 mm. Given: Total dead load - 4.9 kPa (including wt. of slab and beam) Concrete f'e = 20.7 MPa Longitudinal bars fy = 415 MPa Live Load = 4.8 kPa Shear bars fyv = 275 MPa Distance on center of girders: Concrete cover = 70 mm L= 6 m, S = 2.8 m, Column = 0.35x0.35 a. ) For beam BE, calculate the factored shear force (kN) at the critical section. Assume a simply supported span. b.) Determine the spacing (mm) of two legs of 10 mm ø shear bars at the critical section. c. ) In accordance with NSCP provisions, what should be the maximum spacing (mm) of stirrups at the critical section of the shear. A 2.80 m E 2.80 m 6.00 m Girders 350x500 mm Beams 250x400 mmarrow_forwardRefer to figure CD and DE of the gable frame shown are subjected only to vertical loads only along their lengths. Compute the maximum shear at CD (kN)arrow_forward
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