C C leg श * Pleese see an ac. 105. A member ABCD is subjected to a force system as shown in the figure A C D 365 450 45+ 130 The resistive forče in the part BC is (a) 365 (compressive) (b) 450 (tensile) (c) 85 (compressive) (d) 320 (compressive)
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- Compute the nominal shear strength of an M107.5 of A572 Grad 65 steel.Channel sections are used as a purlin. The top chords of the truss are sloped at 5H to 2V. Trusses are spaced 5m on centers and the purlins are spaced 1.4m on centers. Use the properties of the channel section given below. Loads: (Consider all loads pass thru the centroid of the section.) Dead load = 750 Pa Live load = 1,000 Pa Wind load = 1,400Pa Wind coefficients: Windward = 0.3 Leeward = -0.5 Use Fbx=Fby=250 MPa. Determine the maximum value of the interaction equation using the load combination of 0.75(D+L+W)A steel bracket of solid circular crosssection is subjected to two loads, each of which isP = 4.5 kN at D (see figure). Let the dimension variablebe b = 240 mm.(a) Find the minimum permissible diameter dminof the bracket if the allowable normal stress is110 MPa.(b) Repeat part (a), including the weight ofthe bracket. The weight density of steel is77.0 kN/m3.
- Determine the nominal strength of a A36 steel column (Fy = 36 ksi), W 14 X 68 with a total length of 50 feet (both pin-end connection). W14x68 A = 20 in.^2 d = 14 in. tw = 0.415 in. bf = 10 in. tf = 0.72 in. T = 10-7/8 in. k = 1.31 in. k1 = 1.0625 in. gage = 5-1/2 in. rt = 2.71 in. d/Af = 1.94 Ix = 722 in.^4 Sx = 103 in.^3 rx = 6.01 in. Iy = 121 in.^4 Sy = 24.2 in.^3 ry = 2.46 in.pls help! Write the complete solutions and legibly. Answer in 2 decimal places. UPVOTE WILL BE GIVEN! MECHANICS OF DEFORMABLE BODIES Let Q = 0 and T = 8 (Example: 12Q°C = 120°C). Just substitute the number in letter. Thank you! For the frame shown, a 4QQT-N load is acting on member ABD at D. If the allowable material shear stress for is 4Q MPa, determine the required diameter (rounded off to the nearest 2.5 mm) of the pins at C and D. Pin C and pin D are subjected to double shear and single shear, respectively. If the thickness of member BC is 12 mm and that of member DE is 16 mm, determine the maximum bearing stress at C.Channel sections are used as a purlin. The top chords of the truss are sloped at 5H to 2V. Trusses are spaced 5m on centers and the purlins are spaced 1.4m on centers. Use the properties of the channel section BELOW. Loads: (Consider all loads pass thru the centroid of the section.) Dead load = 750 Pa Live load = 1,000 Pa Wind load = 1,400Pa Wind coefficients: Windward = 0.3 Leeward = -0.5 Use Fbx=Fby=250 MPa. Determine the maximum value of the interaction equation using the load combination of 0.75(D+L+W)
- Given the following allowable stresses:• 50 MPa for shear in rivets• 100 MPa for bearing between a plate and a rivet• 80 MPa for tension in the platesNOTE: Draw FBD(a) If we increase the diameter of each bar to 20 mm, determine the maximum load Pthat can be applied to the lap joint.(b) Using the value computed for maximum load P from (a) compute for the over-allfactor of safety given that the actual P= 20kN.A flanged bolt coupling consists of eight ½ -in. steelbolts evenly spaced around a bolt circle 12 in. indiameter, and six ¾ -in. aluminum bolts on aconcentric bolt circle 8 in. in diameter. What torquecan be applied without exceeding 15000 psi in thesteel or 10000 psi in the aluminum? Assume Gst =12 × 106 psi and Gal = 4 × 106 psi.The Channel section is used as a purlin placed on top of trusses with spacing of 4.1 m and slope 1V to 4H. The purling are supporting a total uniform load of 20.2 kN/m.Properties of Channel:Sx = 119 × 10^3 mm^3Sy = 87 × 10^3 mm^3 Questions:Compute the maximum bending stress. Note: The answers 346.05 Mpa for this problem is not acceptable answers. kindly double check your answers again.
- Channel sections are used as a purlin. The top chords of the truss are sloped at 6H to 1V. Trusses are spaced 4m on centers. Use the properties of the channel section in the picture. Loads: Dead load = 600 Pa Live load = 800 Pa Wind load = 1,300Pa Wind coefficients: Windward = 0.25 Leeward = -0.4 Use Fbx=Fby=248 MPa. Determine the safe spacing of purlins (Use load combination of 0.75(D+L+W)A flanged bolt coupling consists of six ½-in. steel bolts evenly spaced around a bolt circle 12 in. in diameter, and four ¾-in. aluminum bolts on a concentric bolt circle 8 in. in diameter. What torque can be applied (kip-in) without exceeding 9000 psi in the steel or 6000 psi in the aluminum? Assume Gst = 12 × 106 psi and Gal = 4 × 106 psi.Prob 01: An 18” depth beam is bolted to a 24” depth girder is connected similar to that in figure shown below. Thediameter of the rivet is 20mm and the angles are each 100mm x 100mm x 12mm thick. For each bolt, assume thatthe allowable bearing stress is 220 MPa. Determine the allowable load P in KN on the connection. *show all theload P in kN for bearing stressBeam web thickness = 5mmGirder web thickness =