Question 6 (Marks:20) A steel girder has the cross-section shown in the adjacent figure. The wall...

  

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Question 6 (Marks:20) A steel girder has the cross-section shown in the adjacent figure. The wall thickness is uniformly 1.25 cm. The stress due to twisting should not exceed 350 MPa. Neglect stress 12.5 cm concentrations. Find the maximum allowable torque, the twist per metre length for that torque and the shear stress in the middle web for that torque. Where necessary, use the symbol G for the shear modulus. Hint: Divide the cross-section into two parts of an a = on perimeter of A1, az = 4 ds ds on perimeter of Az and a = ds on the web. %3D %3D T = qA + q>Az and Aq 42º12 – 4342 – 91412 , where q, and q; are the shear flow Ay %3D ds 2.AG t over either section. Angle of twist per unit length is 0=- 4.4°G ion 6 (Marks:20) Igirder has the cross-section shown in the adjacent The wall thickness is uniformly 1.25 cm. The stress 125 cm twisting should not exceed 350 MPa. Neglect stress 12.5 cm trations. Find the maximum allowable torque, the er metre length for that torque and the shear stress in ddle web for that torque. Where necessary, use the IG for the shear modulus. Hint: Divide the cross-section into two parts of areas A, Ay md 25 cm ds ds on perimeter of A1, az =0 ds on perimeter of Az and a =0- on the web. Then, T-7+ %3D %3D Ai + a4, and 444– 92412 – 4242– 4je12 where q, and q, are the shear flows of Áy and Ay Az %3D f ds 96as over either section. 2.AG %3D of twist per unit length is 6= 4.4 G TED TEMPTED Hide Question 6 (Marks:20) A steel girder has the cross-section shown in the adjacent figure. The wall thickness is uniformly 1.25 cm. The stress due to twisting should not exceed 350 MPa. Neglect stress 12.5 cm concentrations. Find the maximum allowable torque, the twist per metre length for that torque and the shear stress in the middle web for that torque. Where necessary, use the symbol G for the shear modulus. Hint: Divide the cross-section into two parts of an a = on perimeter of A1, az = 4 ds ds on perimeter of Az and a = ds on the web. %3D %3D T = qA + q>Az and Aq 42º12 – 4342 – 91412 , where q, and q; are the shear flow Ay %3D ds 2.AG t over either section. Angle of twist per unit length is 0=- 4.4°G ion 6 (Marks:20) Igirder has the cross-section shown in the adjacent The wall thickness is uniformly 1.25 cm. The stress 125 cm twisting should not exceed 350 MPa. Neglect stress 12.5 cm trations. Find the maximum allowable torque, the er metre length for that torque and the shear stress in ddle web for that torque. Where necessary, use the IG for the shear modulus. Hint: Divide the cross-section into two parts of areas A, Ay md 25 cm ds ds on perimeter of A1, az =0 ds on perimeter of Az and a =0- on the web. Then, T-7+ %3D %3D Ai + a4, and 444– 92412 – 4242– 4je12 where q, and q, are the shear flows of Áy and Ay Az %3D f ds 96as over either section. 2.AG %3D of twist per unit length is 6= 4.4 G TED TEMPTED Hide

Answer rating: 100% (QA)

Sel 125 cm A Emax 350 M6 1 25 cm 25cm A A2 13 25cm Consider eputibrig conditi... View the full answer