8. A 9.00 m span simply supported beam carries a uniformly distributed load of 100 kN/m and point load of 400 kN. The point load is applied at 6.00 m from one end of the beam. as shown below. 400 kN...


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8. A 9.00 m span simply supported beam carries a uniformly distributed load of<br>100 kN/m and point load of 400 kN. The point load is applied at 6.00 m from one<br>end of the beam. as shown below.<br>400 kN<br>200<br>100 kN/m<br>700<br>dia = 50<br>150<br>6.00 m<br>100<br>150<br>400<br>9.00 m<br>CROSS - SECTION<br>Determine:<br>The highest shear force<br>At this section, compute the average shear stress<br>At this section, identify the maximum shear stress<br>i.<br>ii.<br>ii.<br>Answers:<br>Note: Student should draw the SFD, BMD and Deflected shape. Use the coordinates given below to check answers.<br>(i)<br>SFD: (0.00m, 583KN): (5.83m, 0.00KN); (6.00m, -16.7kN), (6.00m, -417KN); (9.00m, -717KN)<br>(ii)<br>fv, average = V/A = 717 x 103 N/(243.04 x 103 mm2) = 2.95 MPa<br>(iii)<br>y = 432.2 mm from base; or 567.8 mm from top.<br>At N.A., Q = 200 x 567.8 x 567.8/2 = 32.24 x 106 mm3.<br>fy = VQ/(It) where V = 717 kN, Q = 32.24 x 106 mm³; Ix = 22.08 x 10° mm“ and t = 200 mm<br>which yields fy = = 5.23 MPa<br>Screenshot<br>

Extracted text: 8. A 9.00 m span simply supported beam carries a uniformly distributed load of 100 kN/m and point load of 400 kN. The point load is applied at 6.00 m from one end of the beam. as shown below. 400 kN 200 100 kN/m 700 dia = 50 150 6.00 m 100 150 400 9.00 m CROSS - SECTION Determine: The highest shear force At this section, compute the average shear stress At this section, identify the maximum shear stress i. ii. ii. Answers: Note: Student should draw the SFD, BMD and Deflected shape. Use the coordinates given below to check answers. (i) SFD: (0.00m, 583KN): (5.83m, 0.00KN); (6.00m, -16.7kN), (6.00m, -417KN); (9.00m, -717KN) (ii) fv, average = V/A = 717 x 103 N/(243.04 x 103 mm2) = 2.95 MPa (iii) y = 432.2 mm from base; or 567.8 mm from top. At N.A., Q = 200 x 567.8 x 567.8/2 = 32.24 x 106 mm3. fy = VQ/(It) where V = 717 kN, Q = 32.24 x 106 mm³; Ix = 22.08 x 10° mm“ and t = 200 mm which yields fy = = 5.23 MPa Screenshot

Jun 10, 2022
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