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Calculation of Reactions, Internal Shears and Internal Moments ...

Calculation of Reactions, Internal Shears and Internal Moments ...

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Thus the maximum possible reaction on a simply supported beam due to a uniformlydistributed load <strong>of</strong> any length can be determined using the influence line <strong>of</strong> Figure 1.First, the influence line tells us where to place the uniform load for maximum positiveeffect, namely across the entire length <strong>of</strong> the beam where the influence line is positive.WxNext, the value <strong>of</strong> the reaction can be computed by multiplying the magnitude <strong>of</strong> theuniform load w by the area under the influence line over which the uniform load isapplied:Ra = Height <strong>of</strong> uniform load * Area under the influence line triangleRa = w kips/ft * (L ft * (1.0+0)/2) = wL/2 kipsNow admittedly this result is rather obvious, <strong>and</strong> most people would not create aninfluence line for such a simple case. However, we will later show that determiningwhere to place these loads for maximum effect can be quite confusing unless influencelines are available.<strong>Calculation</strong> <strong>of</strong> maximum positive shear at a point in a beam: Previously we showedthe generation <strong>of</strong> an influence line for shear at the quarter point in a simply supportedoverhanging beam, with critical placement <strong>of</strong> a set <strong>of</strong> concentrated <strong>and</strong> uniform loads,which will be repeated here. Note that for maximum positive shear, the uniform live loadis placed only in the ranges where the influence line is positive.wL/4w0.250.750.250.50Assuming a uniform load <strong>of</strong> 2,000 pounds/foot, <strong>and</strong> L = 10 feet, we can calculate themaximum positive shear at the quarter point by:V ¼ positive =Concentrated Load*height + Concentrated Load*height+ Uniform Load*Area under diagram+ Uniform Load*Area under diagramV ¼ positive = 20k * 0.75 + 10k * (2/3)*0.75+ 2k/ft * 0.25 * (10ft/4)/2 + 2k/ft * 0.75 * (3*10ft/4)/2= 26.25 kips<strong>Calculation</strong> <strong>of</strong> maximum negative shear at a point in a beam: Previously we showedthe generation <strong>of</strong> an influence line for shear at the quarter point in a simply supported


M ¼ positive = Concentrated Load*height + Concentrated Load*height+ Uniform Load*Area under influence line+ Uniform Load*Area under influence line= 20kip * 0.1875*10ft + 10kip * 0.1875*10ft*2/3+ 2kip/ft * (0.1875*10ft*(10ft/2))= 406.25 kip ft<strong>Calculation</strong> <strong>of</strong> maximum negative moment at a point in a beam: Previously weshowed the generation <strong>of</strong> an influence line for moment at the quarter point in a simplysupported overhanging beam, with critical placement <strong>of</strong> a set <strong>of</strong> concentrated <strong>and</strong>uniform loads, which will be repeated here:wwAssuming a uniform load <strong>of</strong> 2,000 pounds/foot, <strong>and</strong> L = 10 feet, we can calculate themaximum negative moment at the quarter point by:M ¼ negative =Concentrated Load*height + Concentrated Load*height+ Uniform Load*Area under influence line+ Uniform Load*Area under influence line= 20kip * -0.375*10ft + 10kip * (-0.375*10ft)/2+ 2kip/ft * (-0.375*10ft * (10ft/2)/2)+ 2kip/ft * (-0.03125*10ft)*(10ft/8)/2= 131.64

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