3. 75.8 KN
4. 602.4 KN
- Answer 3 75 8 Kn 4 602 4 Kn 1 (31.46 KiB) Viewed 35 times
- Answer 3 75 8 Kn 4 602 4 Kn 2 (31.46 KiB) Viewed 35 times
Answer: 3. 75.8 KN 4. 602.4 KN
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Answer: 3. 75.8 KN 4. 602.4 KN
Answer:
3. 75.8 KN
4. 602.4 KN
Problem No 2 A simply supported W300 x 90 girder, em long carries a concentrated load P positioned at midpoint point. The beam also camies a uniform load of KN/m (including its own weight) and live load of 7-2 kN/m. Use Fy - 250 MPa and E 200 GPa. Properties of W350 x 90 64 d = 350 mm x 206.4 x 10 mm # mm tu 10 mm Sx 1510 x 102 = 250 mm bf= tf 16.5 mm 4 Allowable stresses Allowable deflection = 400 flexure, Fb0.auty Shear, Fv 0.40 Fy 3 Determine the value of P based on flexure. Determine the value of P based on shear. Determine the value of P based on deflection.
3. 75.8 KN
4. 602.4 KN
Problem No 2 A simply supported W300 x 90 girder, em long carries a concentrated load P positioned at midpoint point. The beam also camies a uniform load of KN/m (including its own weight) and live load of 7-2 kN/m. Use Fy - 250 MPa and E 200 GPa. Properties of W350 x 90 64 d = 350 mm x 206.4 x 10 mm # mm tu 10 mm Sx 1510 x 102 = 250 mm bf= tf 16.5 mm 4 Allowable stresses Allowable deflection = 400 flexure, Fb0.auty Shear, Fv 0.40 Fy 3 Determine the value of P based on flexure. Determine the value of P based on shear. Determine the value of P based on deflection.
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