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. 50 kips20 kips/ft560 kip-ftВDEF6 ft10 ft6 ft12 ft6 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

Answered: Image /qna-images/answer/9fa6fae3-0bf5-4fbf-a2a5-c00baf069c4d.jpg

Answered: 50 kips 20 kips/ft 560 kip-ft В D | F 6…

Engineering

Civil Engineering

50 kips 20 kips/ft 560 kip-ft В D | F 6 ft 10 ft 6 ft 12 ft 6 ft

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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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.

50 kips
20 kips/ft
560 kip-ft
В
D
E
F
6 ft
10 ft
6 ft
12 ft
6 ft

If the shear-force diagram is positive and looks like this ...
VE
VD
Linear
Linear
(first order)
(first order)
Constant
(zero order)
VB
VF
Vc.
then the bending-moment diagram looks like this:
..
MF
MB
MD
Quadratic
(second order)
Linear
(first order)
Quadratic
(second order)
ME
Mc
MA
Slope becomes
more positive
Slope becomes
less positive
Constant slope
(an arch)
(a hill that
gets steeper)
(an upward ramp)
V.
F
Vc
VA
VB
Linear
Constant
Linear
(zero order)
(first order)
(first order)
VD
... then the bending-moment diagram looks like this:
ME
Mc
Quadratic
(second order)
MA
Quadratic
(second order)
MD
Linear
(first order)
MB
Slope becomes
more negative
Slope becomes
less negative
Constant slope
(a valley)
(a waterfall)
|(a downward ramp)
Shear-Force Diagram V
Bending-Moment Diagram M
Load Diagram w
Equation
Rule 1: Concentrated loads create discontinuities in the shear-force diagram. [Equation (7.5)]
Slope = VB
%3D
Positive jump
in shear force V
MB
Slope = VA
MB
MA
VB
Po
AV = Po
V
MA
XA
XB
XB
XA
XB
XA
Rule 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) dx
w(x) dx
MB
w(x)
XA
XA
M
MB
B
M
Δν
Vg - VA = [" w(x) dx
XA
XB
XA
XB
XB
XA
Rule 3: The slope of the V diagram is equal to the intensity of the distributed load w. [Equation (7.1)]
WB
Slope = WB
MB
WB
Slope = w
"A
M
MB
MA
dV
= w(x)
dx
XA
XB
XB
XA
XB
Rule 4: The change in bending moment is equal to the area under the shear-force diagram. [Equation (7.4)]
AM=M;-M= | "V(x) dx
V(x) dx
XA
XA
VA
MB
AM
MB - MA = |" V dx
M
XA
XA
B VB
XB
XA
XB
XA
Rule 5: The slope of the M diagram is equal to the intensity of the shear force V. [Equation (7.2)]
Slope = V
Slope =w
-Slope = VA
MB
MB
%3D
dM
= V
dx
XB
XA
XB
XA
XB
Rule 6: Concentrated moments create discontinuities in the bending-moment diagram. [Equation (7.6)]
Negative jump in
No effect on
shear force V
bending moment
M
MB
VB
MA
AM = -Mo
XB
Mo
XB
XA
XB
A
XA

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