EXAMPLE D-1 Given: Solution: EXAMPLE D-2 Given: Solution: Determine the design strength of a W200x35.9 as...

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EXAMPLE D-1 Given: Solution: EXAMPLE D-2 Given: Solution: Determine the design strength of a W200x35.9 as a tension member in 345 MPa steel. How much dead load can it support? If there are no holes in the member, A, A, and Equation 2-3 governs 311 MPax 4 580 mm² 1 000 N/KN P=311 MPa x A₂ = -=1 420 KN Assuming that dead load is the only load, the governing load combi- nation from Section A is 1.4D. Then, the required tensille strength P=1.4P, SOP,=1 420 kN P 1 420 kN/1.4 = 1 010 kN maximum dead load that can be sup- ported by the member. Repeat Example D-1 for a W20Cx35.9 in 345 MPa steel with four 24 mm diameter holes, two per flange, along the member (i.e., not at its ends) for miscellaneous attachments. See Figure D-1(a). a. For yielding in the gross section 6,P,= 1 420 kN, as in Example D-1. b. For fracture in the net section EXAMPLE D-1 Given: Solution: EXAMPLE D-2 Given: Solution: Determine the design strength of a W200x35.9 as a tension member in 345 MPa steel. How much dead load can it support? If there are no holes in the member, A, A, and Equation 2-3 governs 311 MPax 4 580 mm² 1 000 N/KN P=311 MPa x A₂ = -=1 420 KN Assuming that dead load is the only load, the governing load combi- nation from Section A is 1.4D. Then, the required tensille strength P=1.4P, SOP,=1 420 kN P 1 420 kN/1.4 = 1 010 kN maximum dead load that can be sup- ported by the member. Repeat Example D-1 for a W20Cx35.9 in 345 MPa steel with four 24 mm diameter holes, two per flange, along the member (i.e., not at its ends) for miscellaneous attachments. See Figure D-1(a). a. For yielding in the gross section 6,P,= 1 420 kN, as in Example D-1. b. For fracture in the net section

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Answer Solution for Example D1 The design strength for this example will be determined using Equatio... View the full answer