Please show all steps 8. A 9.00 m span simply supported beam carries a uniformly distributed load of100 kN/m and point load of 400 kN. The point load is applied at 6.00 m from oneend of the beam. as shown below.400 kN200100 kN/m700dia = 501506.00 m1001504009.00 mCROSS - SECTIONDetermine:The highest shear forceAt this section, compute the average shear stressAt this section, identify the maximum shear stressi.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 mmwhich yields fy = = 5.23 MPaScreenshot

Given:- The length of the beam = 9 m The magnitude of the uniformly distributed load = 100 kN/m The…

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: i. The highest shear force ii. At this section, compute the average shear stress ii. At this section, identify the maximum shear stress 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) tv, 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

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: i. The highest shear force ii. At this section, compute the average shear stress ii. At this section, identify the maximum shear stress 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) tv, 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

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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Question

Please show all steps

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

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