05-03ChapGere.0008

05-03ChapGere.0008 - s max in a simply supported wood beam...

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Shear Stresses in Rectangular Beams Problem 5.8-1 The shear stresses t in a rectangular beam are given by Eq. (5-39): in which V is the shear force, I is the moment of inertia of the cross-sectional area, h is the height of the beam, and y 1 is the distance from the neutral axis to the point where the shear stress is being determined (Fig. 5-30). By integrating over the cross-sectional area, show that the resultant of the shear stresses is equal to the shear force V . Solution 5.8-1 Resultant of the shear stresses t 5 V 2 I ¢ h 2 4 2 y 2 1 328 CHAPTER 5 Stresses in Beams (Basic Topics) t 5 V 2 I ¢ h 2 4 2 y 2 1 I 5 bh 3 12 V 5 shear force acting on the cross section R 5 resultant of shear stresses t [ R 5 V Q.E.D. 5 12 V h 3 ¢ 2 h 3 24 5 V 5 12 V bh 3 ( b ) # h / 2 0 ¢ h 2 4 2 y 2 1 dy 1 R 5 # h / 2 2 h / 2 t bdy 1 5 2 # h / 2 0 V 2 I ¢ h 2 4 2 y 1 2 bdy 1 N.A. V b dy 1 y 1 t h 2 h 2 Problem 5.8-2 Calculate the maximum shear stress t max and the maximum bending stress
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Unformatted text preview: s max in a simply supported wood beam (see figure) carrying a uniform load of 18.0 kN/m (which includes the weight of the beam) if the length is 1.75 m and the cross section is rectangular with width 150 mm and height 250 mm. Solution 5.8-2 Wood beam with a uniform load 1.75 m 18.0 kN/m 250 mm 150 mm L 5 1.75 m q 5 18 kN/m h 5 250 mm b 5 150 mm M AXIMUM SHEAR STRESS 5 630 kPa t max 5 3 V 2 A 5 3 qL 4 bh 5 3(18 kN / m)(1.75 m) 4(150 mm)(250 mm) V 5 qL 2 Ê A 5 bh M AXIMUM BENDING STRESS 5 4.41 MPa s max 5 M S 5 3 qL 2 4 bh 2 5 3(18 kN / m)(1.75 m) 2 4(150 mm)(250 mm) 2 M 5 qL 2 8 Ê S 5 bh 2 6 A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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This note was uploaded on 09/20/2009 for the course COE 3001 taught by Professor Armanios during the Spring '08 term at Georgia Institute of Technology.

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