1.Determine the force in members AB, BD, BE, and DE of the Howe roof truss shown in figure below. 600lb, 1000lb, 400lb
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Q: FIGURE 1 60 kN Height 0.75 :1.00 1.00. 1.00 C00 25 kN
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A: ∈Fx=0HA=0∈MH=0RA×18-1×18-2×14-2×10-6×1.75-1.5×3=0RA×18=81RA=4.5KN∈Fy=0RA+RH=1+2+2+1.75+1.5+0.75RH=4.…
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1.Determine the force in members AB, BD, BE, and DE of the Howe roof truss shown in figure below. 600lb, 1000lb, 400lb
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- A howe truss has six panels and each panel is 1m. The span is 6m and height 1.50m. Joints U1, U2, U3, U4, and U5 each carry a vertical full panel load PL = 2600lb and joints L0 and L6 each carry a vertical half panel load PL/2 = 1300lb. Find the value of maximum tensile stress of such member.Calculate the area A of the A36 steel truss member by , where Fmax is the highest load in the truss members,For the given Bollman Bridge truss assembly shown in Figure 1and using the information given in Table 1, determine the internal forces acting along BC, EC and EHusing sectioning method if the point A is a roller support and point G is pin support. Also mention if the members are in tension or in compression.
- A Mansard roof truss is loaded as shown. Determing the force in members EG in kN. P = 6.5 kN and x = 4.7 m. Write numerical value and in two decimal places.What is ϕMn for W8x24 - A36 steel in lb-ft for: a) 15 ft b) 30 ftA Howe scissors roof truss is shown. A Howe scissors roof truss is loaded as shown. Determine the force in members Gl, Hl, and HJ when P1 = 1 kips and P2 = 2 kips. Note: please double check your answers!. Thanks
- 22 A concrete floor slab 100 mm thick is cast monolithic with concrete beams 2.0 m on centers. The beams have a span of 4 m and have a web width of 250 mm, an effective depth of 400 mm and overall depth of 500 mm. The tensile reinforcement consists of 6-ϕ32 mm bars in two rows. Use material strengths f’c = 21 MPa and fy = 415 MPa. Calculate the ultimate bending moment strength of the T-beam in kN·m. 640 778 711 701A howe truss has six panels and each panel is 1m. The span is 6m and height 1.50m. Joints U1, U2, U3, U4, and U5 each carry a vertical full panel load PL = 2600lb and joints L0 and L6 each carry a vertical half panel load PL/2 = 1300lb. Find the value of the stress of member L2L3.Three bars, AB, AC, and AD, are pinned together to support a load P = 20 kN as shown in Fig. P-256. Horizontal movement is prevented at joint A by the short horizontal strut AE. Determine the stress in each bar and the force in the strut AE. For the steel bar, A = 200 mm and E=200 GPa. For each aluminum bar, A=400 mm^2 and E= 70 GPa.
- W 14 x 142 is used as a column having a length of 8.48 m. long. It is hinged at the r end and fixed at the lower end but there is a lateral bracing perpendicular to minor axis of the W section at a point 4.78 m. above the bottom support. It is assumed to be pinned connected at the bracing point. Using A 36 steel Fy= 248 MPa the NSCP Specificatioris. E, = 200000 MPa. Properties of W 14 x 142 A = 26967.69 mm2 d= 374.65 mm bf= 393.70 mm tf= 27.00 mm tw= 17.27 mm4 Ix= 695.11 x 106 mm4 Iy = 274.71 x 106 mm4 rx = 160.53 mm rY = 100.84 mm Assume: Kx = 0.80 for 8.48 m. length KY = 1.9 for the length on the upper support KY = 0.80 for the length on the bottom support Compute the slenderness ratio along y-axis.W 14 x 142 is used as a column having a length of 8.48 m. long. It is hinged at the r end and fixed at the lower end but there is a lateral bracing perpendicular to minor axis of the W section at a point 4.78 m. above the bottom support. It is assumed to be pinned connected at the bracing point. Using A 36 steel Fy= 248 MPa the NSCP Specificatioris. E, = 200000 MPa. Properties of W 14 x 142 A = 26967.69 mm2 d= 374.65 mm bf= 393.70 mm tf= 27.00 mm tw= 17.27 mm4 Ix= 695.11 x 106 mm4 Iy = 274.71 x 106 mm4 rx = 160.53 mm rY = 100.84 mm Assume: Kx = 0.80 for 8.48 m. length KY = 1.9 for the length on the upper support KY = 0.80 for the length on the bottom support Compute the slenderness ratio along y-axis. Please show clear solution thank youCalculate for the cross-section of members AG, BC, and CE for the truss assembly shown in the figure below, considering that the stresses are not to exceed 20 ksi categorized in tension and 14 ksi categorized in compression. NOTE: A reduced stress in compression is specified to reduce the danger of buckling.