The cross section shown is loaded by Vu = 180 kN and a torque of Tu = 25 kN-m. fc' = 28 MPa and fy = 275 MPa. Use 60 mm

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answerhappygod
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The cross section shown is loaded by Vu = 180 kN and a torque of Tu = 25 kN-m. fc' = 28 MPa and fy = 275 MPa. Use 60 mm

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The Cross Section Shown Is Loaded By Vu 180 Kn And A Torque Of Tu 25 Kn M Fc 28 Mpa And Fy 275 Mpa Use 60 Mm 1
The Cross Section Shown Is Loaded By Vu 180 Kn And A Torque Of Tu 25 Kn M Fc 28 Mpa And Fy 275 Mpa Use 60 Mm 1 (30.51 KiB) Viewed 36 times
The cross section shown is loaded by Vu = 180 kN and a torque of Tu = 25 kN-m. fc' = 28 MPa and fy = 275 MPa. Use 60 mm covering to the center of reinforcements. a. Determine the threshold torque. b. Calculate the ultimate developed strength of the beam in MPa c. Determine the spacing of two legs of 10 mm diameter stirrups to resist combined shear and torsion. d. Determine the maximum spacing of torsion reinforcement. e. Determine the number of 12 mm longitudinal reinforcement required to resist torsion. 500 250 600 125
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