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Determine the location of the shear center for the beams having the cross-sectional dimensions shown in the figures. All members are to be considered thin walled, and calculations should be based on the centerline dimensions. the expected solution is provided in the picture below.

Determine the location of the shear center for the beams having the cross-sectional dimensions shown in the figures. All members are to be considered thin walled, and calculations should be based on the centerline dimensions. the expected solution is provided in the picture below.

Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter6: Stresses In Beams (advanced Topics)
Section: Chapter Questions
Problem 6.10.1P: Determine the shape factor f for a cross section in the shape of a double trapezoid having the...
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Determine the location of the shear center for the beams having the cross-sectional dimensions shown in the figures. All members are to be considered thin walled, and calculations should be based on the centerline dimensions.

the expected solution is provided in the picture below.

 

Determine the location of the shear center for the beams having the cross-sectional dimensions shown in the figures. All members are to be considered thin walled, and calculations should be based on the centerline dimensions 10 80 mm Fig. P10-42
Transcribed Image Text:Determine the location of the shear center for the beams having the cross-sectional dimensions shown in the figures. All members are to be considered thin walled, and calculations should be based on the centerline dimensions 10 80 mm Fig. P10-42
10-41 I= (.002)(-1)'+2(.002 X-04)(.07.5 = 8.23 x 10? m² Ve = hi Fi +ha F> , A1= tb= Az. Fi= VA,, Fz= VA2 &= Az ZIt RIt e = 41 芸(hh . 002 (. 04)* 4 18.23x10°9) = . OIsg m = IS. 9 mm ( trom ¢ of rert. member) 10 -42 from problem 10-41 , I- 8.23×1C|| e- (0.002)(0.04)* (0.12_0.08²) 4 (2.23X10*") • 0. 00 35 m - 3.5 mm (from € of vertical member)
Transcribed Image Text:10-41 I= (.002)(-1)'+2(.002 X-04)(.07.5 = 8.23 x 10? m² Ve = hi Fi +ha F> , A1= tb= Az. Fi= VA,, Fz= VA2 &= Az ZIt RIt e = 41 芸(hh . 002 (. 04)* 4 18.23x10°9) = . OIsg m = IS. 9 mm ( trom ¢ of rert. member) 10 -42 from problem 10-41 , I- 8.23×1C|| e- (0.002)(0.04)* (0.12_0.08²) 4 (2.23X10*") • 0. 00 35 m - 3.5 mm (from € of vertical member)
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