Determine/derive the Shear and Moment Equations of each segment of the beam (kindly include the FBD). Compute the Reactions and Draw the detailed Shear and Moment Diagram indicating values. y9 kips/ft5 kips/ftAВCD6 ft10 ft11 ft If the shear-force diagram is positive and looks like this ...VEVDLinearLinear(first order)(first order)Constant(zero order)VBVFVc.then the bending-moment diagram looks like this:..MFMBMDQuadratic(second order)Linear(first order)Quadratic(second order)MEMcMASlope becomesmore positiveSlope becomesless positiveConstant slope(an arch)(a hill thatgets steeper)(an upward ramp)V.FVcVAVBLinearConstantLinear(zero order)(first order)(first order)VD... then the bending-moment diagram looks like this:MEMcQuadratic(second order)MAQuadratic(second order)MDLinear(first order)MBSlope becomesmore negativeSlope becomesless negativeConstant slope(a valley)(a waterfall)|(a downward ramp)Shear-Force Diagram VBending-Moment Diagram MLoad Diagram wEquationRule 1: Concentrated loads create discontinuities in the shear-force diagram. [Equation (7.5)]Slope = VB%3DPositive jumpin shear force VMBSlope = VAMBMAVBPoAV = PoVMAXAXBXBXAXBXARule 2: The change in shear force is equal to the area under the distributed-load curve. [Equation (7.3)]AV= V½ – V¼ = J",w(x) dxw(x) dxMBw(x)XAXAMMBBMΔνVg - VA = [" w(x) dxXAXBXAXBXBXARule 3: The slope of the V diagram is equal to the intensity of the distributed load w. [Equation (7.1)]WBSlope = WBMBWBSlope = w"AMMBMAdV= w(x)dxXAXBXBXAXBRule 4: The change in bending moment is equal to the area under the shear-force diagram. [Equation (7.4)]AM=M;-M= | "V(x) dxV(x) dxXAXAVAMBAMMB - MA = |" V dxMXAXAB VBXBXAXBXARule 5: The slope of the M diagram is equal to the intensity of the shear force V. [Equation (7.2)]Slope = VSlope =w-Slope = VAMBMB%3DdM= VdxXBXAXBXAXBRule 6: Concentrated moments create discontinuities in the bending-moment diagram. [Equation (7.6)]Negative jump inNo effect onshear force Vbending momentMMBVBMAAM = -MoXBMoXBXAXBAXA

Question

Determine/derive the Shear and Moment Equations of each segment of the beam (kindly include the FBD). Compute the Reactions and Draw the detailed Shear and Moment Diagram indicating values. y9 kips/ft5 kips/ftAВCD6 ft10 ft11 ft If the shear-force diagram is positive and looks like this ...VEVDLinearLinear(first order)(first order)Constant(zero order)VBVFVc.then the bending-moment diagram looks like this:..MFMBMDQuadratic(second order)Linear(first order)Quadratic(second order)MEMcMASlope becomesmore positiveSlope becomesless positiveConstant slope(an arch)(a hill thatgets steeper)(an upward ramp)V.FVcVAVBLinearConstantLinear(zero order)(first order)(first order)VD... then the bending-moment diagram looks like this:MEMcQuadratic(second order)MAQuadratic(second order)MDLinear(first order)MBSlope becomesmore negativeSlope becomesless negativeConstant slope(a valley)(a waterfall)|(a downward ramp)Shear-Force Diagram VBending-Moment Diagram MLoad Diagram wEquationRule 1: Concentrated loads create discontinuities in the shear-force diagram. [Equation (7.5)]Slope = VB%3DPositive jumpin shear force VMBSlope = VAMBMAVBPoAV = PoVMAXAXBXBXAXBXARule 2: The change in shear force is equal to the area under the distributed-load curve. [Equation (7.3)]AV= V½ – V¼ = J&#34;,w(x) dxw(x) dxMBw(x)XAXAMMBBMΔνVg - VA = [&#34; w(x) dxXAXBXAXBXBXARule 3: The slope of the V diagram is equal to the intensity of the distributed load w. [Equation (7.1)]WBSlope = WBMBWBSlope = w&#34;AMMBMAdV= w(x)dxXAXBXBXAXBRule 4: The change in bending moment is equal to the area under the shear-force diagram. [Equation (7.4)]AM=M;-M= | &#34;V(x) dxV(x) dxXAXAVAMBAMMB - MA = |&#34; V dxMXAXAB VBXBXAXBXARule 5: The slope of the M diagram is equal to the intensity of the shear force V. [Equation (7.2)]Slope = VSlope =w-Slope = VAMBMB%3DdM= VdxXBXAXBXAXBRule 6: Concentrated moments create discontinuities in the bending-moment diagram. [Equation (7.6)]Negative jump inNo effect onshear force Vbending momentMMBVBMAAM = -MoXBMoXBXAXBAXA

Accepted Answer

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y 9 kips/ft 5 kips/ft A В C D 6 ft 10 ft 11 ft