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Problem 1-91/1-92/1-93/ Engineering Mechanics Materials
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Engineering mechanics problem with solution.
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1–91. If the allowable bearing stress for the material under the supports at A and B is 1sb2 allow = 1.5 MPa, determine the size of square bearing plates A! and B! required to support the load. Dimension the plates to the nearest mm. The reactions at the supports are vertical. Take P = 100 kN.
1–92. If the allowable bearing stress for the material under the supports at A and B is 1sb2allow = 1.5 MPa, determine
the maximum load P that can be applied to the beam. The bearing plates A! and B! have square cross sections of 150 mm * 150 mm and 250 mm * 250 mm, respectively.
1–93. The rods AB and CD are made of steel. Determine their smallest diameter so that they can support the dead
loads shown. The beam is assumed to be pin connected at A
and C . Use the LRFD method, where the resistance factor for steel in tension is f = 0.9, and the dead load factor is gD = 1.4. The failure stress is sfail = 345MPa.
Go to my playlist to get more specific topics
1–91. If the allowable bearing stress for the material under the supports at A and B is 1sb2 allow = 1.5 MPa, determine the size of square bearing plates A! and B! required to support the load. Dimension the plates to the nearest mm. The reactions at the supports are vertical. Take P = 100 kN.
1–92. If the allowable bearing stress for the material under the supports at A and B is 1sb2allow = 1.5 MPa, determine
the maximum load P that can be applied to the beam. The bearing plates A! and B! have square cross sections of 150 mm * 150 mm and 250 mm * 250 mm, respectively.
1–93. The rods AB and CD are made of steel. Determine their smallest diameter so that they can support the dead
loads shown. The beam is assumed to be pin connected at A
and C . Use the LRFD method, where the resistance factor for steel in tension is f = 0.9, and the dead load factor is gD = 1.4. The failure stress is sfail = 345MPa.
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