Lecture 9 part IV

Lecture 9 part IV - PART IV SHEAR FORCE AND BENDING MOMENT...

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28 PART IV SHEAR FORCE AND BENDING MOMENT DIAGRAMS (CONT’)
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29 Example 8-7 For 0<x<4 21 0 5500 y VV V A l b =+ Δ=+ = 2 0 (4) 0 (4) 0 5500(4) 22,000 y M M M V A lb ft Δ = + = + = + = x y V 2 o A y 1 2 M 2 V 1 M 1
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30 For 4=<x<8, note now section 2 in previous interval becomes section 1 in this interval 21 5500 ( 2000) 3500 VV V l b =+ Δ= + = 2 22,000 (4) 22,000 3500(4) 36,000 M MM V l b f t Δ = + = + = For 8=<x<12 3500 ( 6000) 2500 V l b + = 2 36,000 (4) 36,000 ( 2500)(4) 26,000 M V l b f t Δ = + = + = x y V 2 o 2 6 000 lb V 1 1 M 1 M 2 x y V 2 o V 1 1 2 2000 lb M 1 M 2
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31 For 12=<x<16 21 2500 ( 4000) 6500 VV V l b =+ Δ= + = 2 26,000 (4) 26,000 ( 65000)(4) 0.0 M MM V l b f t Δ = + = + = This is a good check since there is no moment at E. Finally the shear and moment diagram read as the follows The best way to get these two diagrams is to draw the diagrams during the computing procedures above. You are suggested to draw the diagrams for V and M and do the computation at the x y V o 2 4 000 lb V 1 1 M 1 M 2
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32 same time. This is because drawing the V diagram will help you in drawing the M diagram correspondingly. For instance, the area under the V diagram between sections 1 and 2 is the moment increment M Δ at section 2 with respect to the moment at section 1.
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33 Example 8-8 Alternatively , instead of using the mathematical expressions, we have the following Set up a consistent x-axis for all different zones as the reference along the length of the beam
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34 For 0<x<2 21 08 y VV V A k N =+ Δ=+ = 2 0 (2) 0 (2) 8(2) 16 y M MM V A k N m Δ = + = + = = For 2=<x< inflection point Since V 1 and the distributed load 10kN/m have opposite direction and V 1
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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 IV - PART IV SHEAR FORCE AND BENDING MOMENT...

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