Please show all steps to reach answers provided. Final Test review 6. A simply supported beam carries two point-loads as shown below. If the maximum allowable shear stress is 40.0 MPa, and the maximum bending stress allowed is 100 MPa: Draw the deflected shape, SFD, and BMD ii. Locate the centroid above the base and the moment of inertia (I) of the cross-section What is the highest value of P allowed? 200 5.0 15.0 2.00 m 230 2.00 m 15.0 8.00 m 6.0번 150 Note: Self-weight may be ignored CROSS - SECTION Answers: Note: Student should draw the SFD, BMD and Deflected shape. Use the coordinates given below to check answers. SFD: (0.00m, P kN): (2.00m, P kN): (6.00m, -P kN), (8.00m, -P kN) BMD: (0.00m, 0.00 kNm): (2.00m, 2P kNm): (6.00m, 2P kNm), (8.00m, 0.00 kNm) Deflected shape: smile throughout! (i) = 157 mm from base; I, = 85.6 x 10° mm (ii) At N.A., Q = 397.6 x 10' mm. Using fy = VQ/(It) where fy = 40.0 N/mm, t = 6.00 mm Ix = 85.6 x 10° mm“ and V = P yields P = 51.7 kN Using fu = ("/1ly where fs = 100 N/mm, y = 157 mm, I, = 85.6 x 10° mm and M = 2P yields P = 27.2 kN. The Point load allowable is the less of the two values computed above. Therefore P = 27.2 kN

Please show all steps to reach answers provided. Final Test review 6. A simply supported beam carries two point-loads as shown below. If the maximum allowable shear stress is 40.0 MPa, and the maximum bending stress allowed is 100 MPa: Draw the deflected shape, SFD, and BMD ii. Locate the centroid above the base and the moment of inertia (I) of the cross-section What is the highest value of P allowed? 200 5.0 15.0 2.00 m 230 2.00 m 15.0 8.00 m 6.0번 150 Note: Self-weight may be ignored CROSS - SECTION Answers: Note: Student should draw the SFD, BMD and Deflected shape. Use the coordinates given below to check answers. SFD: (0.00m, P kN): (2.00m, P kN): (6.00m, -P kN), (8.00m, -P kN) BMD: (0.00m, 0.00 kNm): (2.00m, 2P kNm): (6.00m, 2P kNm), (8.00m, 0.00 kNm) Deflected shape: smile throughout! (i) = 157 mm from base; I, = 85.6 x 10° mm (ii) At N.A., Q = 397.6 x 10' mm. Using fy = VQ/(It) where fy = 40.0 N/mm, t = 6.00 mm Ix = 85.6 x 10° mm“ and V = P yields P = 51.7 kN Using fu = ("/1ly where fs = 100 N/mm, y = 157 mm, I, = 85.6 x 10° mm and M = 2P yields P = 27.2 kN. The Point load allowable is the less of the two values computed above. Therefore P = 27.2 kN

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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