Lecture 9 part III

Lecture 9 part III - PART III SHEAR FORCE AND BENDING...

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18 PART III SHEAR FORCE AND BENDING MOMENT DIAGRAMS (CONT’)
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19 Considering the FBD with differential length dx shown in Fig. 8-11, the element is in equilibrium, we have the resultant force R= y F in y direction as Fig. 8-11 FBD for a differential length of beam with distributed load () 0 y FV w d xV d V =+ − + = or dV dV wdx w dx == (5) This indicates that in any section of the beam, the slope of the shear diagram is equal to the intensity of loading . When w is known as a function of x , the equation can be integrated between definite limits to yield 22 11 21 wdx dV V V V = Δ ∫∫ (6) This indicates that change in shear between two sections is equal to the a rea under the load diagram between these two sections, provided there are no concentrated forces in between . When the distributed load w exists, 2 1 x x wdx should be used by intuition. However, when only point force, V Δ should be used by ease. But V Δ is applied to all circumstances since it denoted the increment of the shear force. Applying the equilibrium equation for moment with respect to point O to the element shown in Fig. 8-11, we have
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20 0 () 0 2 dx MM V d x w d x M d M =+ + + = or dM dM Vdx V dx == (7) This indicates that at any section in the beam the slope of the moment diagram is equal to the shear . Similarly 22 11 21 xV Vdx dM M M M = Δ ∫∫ (8) This indicates the change in moment between two sections is equal to the area under the shear diagram between these two sections, provided there are NO concentrated couples in between .
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This note was uploaded on 04/20/2009 for the course CVEN 221 taught by Professor - during the Summer '08 term at Texas A&M.

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Lecture 9 part III - PART III SHEAR FORCE AND BENDING...

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