Chapter 4 Shear Forces and Bending Moments

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for a < x < a + bV = R A - q (x - a)M = R A x - q (x - a) 2 / 2for a + b < x < LV = - R B M = R B (L - x)maximum moment occurs where V = 0i.e.x 1 = a + b (b + 2c) / 2LM max = q b (b + 2c) (4 a L + 2 b c + b 2 ) / 8L 2for a = c, x 1 = L / 2M max = q b (2L - b) / 8for b = L, a = c = 0(uniform loading over the entire span)M max = q L 2 / 8Example 4-5construct the V- and M-dia for thecantilever beam supported to P 1 and P 2R B = P 1 + P 2M B = P 1 L + P 2 bfor 0 < x < aV = - P 1M = - P 1 xfora < x < LV = - P 1 - P 2 M= - P 1 x - P 2 (x - a)10
Example 4-6construct the V- and M-dia for thecantilever beam supporting to a constant uniformload of intensity qR B = q L M B = q L 2 / 2then V = - q x M = - q x 2 / 2alternative methodV max = - q L M max = - q L 2 / 2xV - V A = V - 0 = V = -∫ q dx = - q x0M - M A = M - 0 = M =x-∫ V dx = - q x 2 / 20Example 4-7an overhanging beam is subjected to auniform load of q = 1 kN/m on ABand a couple M 0 = 12 kN-m on midpointof BC, construct the V- and M-dia forthe beamR B = 5.25 kNR C = 1.25 kNshear force diagramV = - q xV = constanton ABon BC11
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Page 9: R A = R B = q L / 2V = R A - q x =

Chapter 4 Shear Forces and Bending Moments

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