Shear and Moment Diagrams
Some Basic Definitions
Learning Objectives:
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Apply overall equilibrium to calculate the reactions and distributed internal forces (axial, shear, moment) for various beam configurations.
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The Shearing Force at any point is the algebraic sum of all the vertical forces to one side of the point considered.
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The Bending Moment at any point is the algebraic sum of the moments of all the external forces to one side of the point about that point.
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The Point of Contraflexure is a point, where the Bending Moment is zero or changes sign.
Solved Example: 9890-01
The point of a contraflexure is the point where: (UPPSC AE Mechanical 2013 Official Paper I)
A. B.M. changes sign
B. B.M. is maximum
C. B.M. is minimum
D. S.F. is zero
Correct Answer: A
Solved Example: 9890-02
Shear span is called the zone where: (KSSB JE CE Oct 2021- Shift I)
A. Shear force is constant
B. Shear force is zero
C. Bending moment is constant
D. Bending moment changes
Correct Answer: A
Some Basic Rules and Observations
Learning Objectives:
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Use equilibrium to derive the formal relationships between loading, shear, and moment (q, S, M) and apply these for various beam configurations.
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While drawing Shear Force and Bending Moment Diagrams, all positive values are plotted above the base line and negative values below it.
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The maximum Bending Moment occurs where the Shear Force is either zero or changes sign.
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If a beam is subjected to couple, the Shear Force does not change, but the Bending Moment suddenly changes in magnitude equal to that couple.
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If a beam is subjected to inclined loads their components at right angles to the axis of the beam will cause Shear Force and Bending Moments but the components along the axis of the beam will cause axial thrust.
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The slope of the shear diagram at a point is equal to the intensity of the distributed loading w(x) at that point.
\[w(x) = - \dfrac{dV}{dx}\]
where V is the shear force acting at position x of the point.
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The change in shear between any two points is equal to the area under the loading curve between the points.
\[V_2 - V_1 = \int_{x_1}^{x_2} - w(x) dx\]
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The change in moment between any two points is equal to the area under the shear diagram between the points.
\[M_2 - M_1 = \int_{x_1}^{x_2} V(x) dx\]
Solved Example: 39-2-01
Which of the following statements is true?
A. Shear force is constant when Bending moment is parabolic
B. Bending moment is maximum when shear force is zero.
C. Shear force is maximum when bending moment is zero.
D. When you know only shear force diagram, bending moment diagram can not be evaluated unless you know the loading diagram.
Correct Answer: B
Solved Example: 39-2-02
When the shear force diagram is a parabolic curve between two points, it indicates that there is a:
A. Point load at the two points
B. No loading between the two points
C. Uniformly distributed load between the two points
D. Uniformly varying load between the two points
Correct Answer: D
Solved Example: 39-2-03
For a loaded cantilever beam of uniform cross section, the bending moment (in N-mm) along the length is: $M(x) = 5x^2 + 10x$ where x is the distance (in mm) measured from the free end of the beam. The magnitude of shear force (in N) in the cross-section at x = 10 mm is:
A. 90
B. 95
C. 105
D. 110
\[M(x) = 5x^2+10x\] Shear force \[SF = \dfrac{dM}{dx} = 10x +10\]
At x = 10 mm, SF = 100 + 10 = 110 N
Correct Answer: D
Procedure to Draw Shear and Moment Diagrams
Learning Objectives:
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Determination of expressions for internal shear forces and internal bending moment at a general location along a multisegmented beam, using method of sections.
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To find the maximum moment/shear and their locations.
The following is the procedure for constructing the shear and moment diagrams for a beam.
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Determine the support reactions for the structure.
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To construct the shear diagram, first, establish the V and x axes and plot the value of the shear at each end of the beam.
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Since the \(\dfrac{dV}{dx}\) = w, the slope of the shear diagram at any point is equal to the intensity of the applied distributed loading.
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The change in the shear force is equal to the area under the distributed loading. If the distributed loading is a curve of degree n, the shear will be a curve of degree n+1.
Solved Example: 39-3-01
The best way to get the Shear Force Diagram (SFD) and Bending Moment Diagram (BMD) for a beam containing internal hinge is:
A. Take shear force at that hinge to be zero
B. Neglect the hinge and continue calculating SFD and BMD
C. Take deflection at that hinge to be zero
D. Separate the portions left and right of that hinge
Correct Answer: D
Simply Supported Beam with a Point Load at the Center
Learning Objectives:
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To draw shear and bending moment diagram for a simply supported beam with a point load at the center.
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To find Maximum shear force and maximum bending moment.
Simply Supported Beam with a Point Load Anywhere
Learning Objectives:
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To draw shear and bending moment diagram for a simply supported beam with a point load anywhere on the span.
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To find Maximum shear force and maximum bending moment.
Solved Example: 39-5-01
A sudden increase or decrease in shear force diagram between any two points indicates that there is:
A. No loading between the two points
B. Point loads between the two points
C. U.D.L. between the two points
D. None of these.
Correct Answer: B
Solved Example: 39-5-02
In the following figure, at point B, the value of B.M. will be:
A. 0.55 kN.m
B. 1 kN.m
C. 4 kN.m
D. None of these.
Correct Answer: C
Simply Supported Beam with Uniformly Distributed Load
Learning Objectives:
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To draw shear and bending moment diagram for a simply supported beam with Uniformly Distributed Load (UDL).
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To find Maximum shear force and maximum bending moment.
Solved Example: 39-6-01
The bending moment in the centre of a simply supported beam carrying a uniformly distributed load of w per unit length is:
A. 0
B. $\dfrac{wl^2}{2}$
C. $\dfrac{wl^2}{4}$
D. $\dfrac{wl^2}{8}$
Correct Answer: D
Solved Example: 39-6-02
A simply supported beam, 10 m long carries a uniform distributed load of 20 kN/m. What is the value of the maximum moment of the beam due to the load?
A. 10,000 kN-m
B. 5,000 kN-m
C. 2,000 kN-m
D. 250 kN-m
\[ M = \dfrac{wl^2}{8} = \dfrac{20\times 10^2}{8} = 250 KN.m\]
Correct Answer: D
Solved Example: 39-6-03
The maximum shear force for the following simply supported beam:
A. $\dfrac{WL}{6}$
B. $\dfrac{WL}{3}$
C. $WL$
D. $\dfrac{2WL}{3}$
The maximum shear force will occur at point A.
\[R_A + R_B = \dfrac{WL}{2}\] Taking moments about the point A = 0 (Clockwise is positive) \begin{align*} \Sigma M_A &= 0 \\ \left(\dfrac{WL}{2} \times \dfrac{L}{3}\right) - \left(R_B \times L\right) &= 0\\ R_B &= \dfrac{WL}{6} \end{align*} \[R_A = \dfrac{WL}{2} - R_B = \dfrac{WL}{2} - \dfrac{WL}{6} = \dfrac{WL}{3}\]Correct Answer: B
Solved Example: 39-6-04
In a simply supported beam carrying a uniformly distributed load w kN/m over the left half span, the point of contraflexure will occur in:
A. Left half span of the beam
B. Right half span of the beam.
C. Quarter points of the beam
D. Does not exist
Correct Answer: D
Solved Example: 39-6-05
When the bending moment is parabolic curve between two points, it indicates that there is:
A. No loading between the two points
B. Point loads between the two points
C. U.D.L. between the two points
D. Uniformly varying load between the two points
Correct Answer: C