Question

Shear and Bending Moment Diagrams

Learning Goal:

To determine the reactive forces and moments acting on a beam; express the shear and bending moment as functions of their positions along the beam; and construct shear and bending moment diagrams.

The cantilever beam shown is subjected to a moment at A and a distributed load that acts over segment BC, and is fixed at C. Determine the reactions at the support located at C. Then write expressions for shear and bending moment as a function of their positions along the beam. Finally, use these expressions to construct shear and bending moment diagrams.

$ME_1404314_Intro.jpg$

Part A - Reactions at support C

Draw a free-body diagram of the beam on paper. Use your free-body diagram to determine the reactions at support C. Assume positive moments act counterclockwise.

Express your answer as integers separated by commas.

Cx, Cy, MC =

0,30,-210

kips, kips, kip-ft

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Part B - Shear and moment expressions as a function of x

Six free-body diagrams and interval(s) are shown below. For the beam given, which one is correct? (Place the origin at the beam’s left end and assume positive internal shear and moment.) The term "interval" refers to the parts of the beam that will be cut in order to study the internal forces.

$ME_1404314_B-1.jpg$

$ME_1404314_B-2.jpg$

Six free-body diagrams and interval(s) are shown below. For the beam given, which one is correct? (Place the origin at the beam’s left end and assume positive internal shear and moment.) The term "interval" refers to the parts of the beam that will be cut in order to study the internal forces.

$ME_1404314_B-1.jpg$

$ME_1404314_B-2.jpg$

D

C

F

A

B

E

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Part C - Shear and moment values for segment 1

Draw a free-body diagram of the segment {0≤x<8 ft} on paper and calculate the internal shear, V, and moment, M, over this segment.

Express your answers as integers separated by a comma. Assume positive moments act counterclockwise.

V, M =

0,-120

kips, kip-ft

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Part D - Shear and moment expressions for segment 2

Draw a free-body diagram of the segment 8<x≤14 ft on paper. Write expressions for the internal shear, V, and moment, M, over this segment.

Express your answers in terms of x separated by comma. Assume positive moments act counterclockwise.

V, M =	30,210	kips, kip-ft

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Part E - Internal shear force diagram

Construct a shear force diagram using the expressions for the internal shear force. First, draw the lines of discontinuity. Then, plot the internal shear segments.

Draw the shear diagram on the grid provided. Begin by drawing the line(s) of discontinuity. A line of discontinuity is a vertical line through the point where the discontinuity occurs. Then, draw the functions between the lines of discontinuity.

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Part F - Internal bending moment diagram

Construct an internal bending moment diagram using the expressions for the bending moment that you developed earlier.

Begin by drawing the line(s) of discontinuity. NOTE - The curve you choose from the drop-down is only a pictorial representation of a real quadratic/cubic curve. The equation of this curve is not mathematically equivalent to the correct answer. Consequently, slopes at discontinuities and intercepts with the x-axis (if any) are not accurate.

Answer 1

Concepts and reason

Shear force:

When force acting on a body pushes two parts of the body in opposite direction then the force is referred as shear force. Shear force on an object acts normal to its section.

Equilibrium of a rigid body:

An object is said to be in equilibrium when the sum of external forces and couples are zero.

For a rigid body to be in equilibrium in three dimensions, the sum of external forces acting along , and directions have to zero.

For a rigid body to be in equilibrium in three dimensions, the sum of external couples about any point should be zero.

Moment:

It refers to the propensity of the force to cause rotation in a body about any fixed point. The moment’s magnitude can be obtained by multiplying force’s magnitude with the perpendicular distance at which the force acts. The moment is denoted by and its unit is.

Cantilever beam:

A beam which is fixed at one end and free at the other end is called cantilever beam.

Support reactions:

If there is a force acting by a support due to which, the translational movement of the object is restricted or prevented. Then the force developed is called as a support reaction.

Types of load:

The different types of Load are shown in Table (1):

Fundamentals

The static force equilibrium condition in the horizontal direction is written as,

Here, and are the sum of all the forces acting along x and y directions
‎in the system.

The expression for the calculation of the moment of force is written as,

Here, is the magnitude of the moment of force, is the applied force and is the distance from the reference point.

The moment equilibrium condition is written as,

Here, is the sum of moments due to all the forces in the system.

The steps required to plot the shear force and bending moment diagrams are:

1. Determine all the reactive forces and couple moments acting on the beam and resolve all of them into components acting perpendicular and parallel to the beam's axis.

2. Take the beam's left end as the origin and extend to regions of the beam between concentrated forces and couple moments, or where there is no discontinuity of distributed loading.

3. Take different beam sections and draw the free-body diagram of one of the segments. The shear force V and bending moment M should be shown acting in their positive sense, in accordance with the sign convention.

4. Calculate the shear force by summing forces perpendicular to the beam's axis.

5. Calculate the moment by summing moments about the sectioned end of the segment.

6. Plot the shear force diagram (V versus x) and the bending moment diagram (M versus x). If V and M in the diagram are positive, their values are plotted above the x axis, whereas negative values are plotted below the axis.

7. To enhance readability, show the shear force and moment diagrams below the free-body diagram of the beam.

Sign convention:

The positive direction is considered when the distributed load acts upward on the beam; the internal shear force causes a clockwise rotation of the beam segment on which it acts. The internal moment causes compression of the top fibers of the beam such that that it causes the beam to sag downwards.

Loadings that are opposite to those are considered negative.

(a)

The free body diagram of the given beam is shown as in Figure (4).

Here, moment about point C is , reaction at C along x-direction is , and reaction at C along y-direction is .

Apply equilibrium condition for forces along the x-direction.

$C=0 $

For Figure (4), apply equilibrium condition for forces along y-direction.

Apply equilibrium condition for moments about point C.

(b)

The free body diagram of the segment is shown as in Figure (5).

The internal shear of the segment is obtained as follows.

Apply the force law of equilibrium along x-direction.

Here, internal shear force is V.

The bending moment for the segment is calculated as follows:

Apply the moment law of equilibrium in the Figure (5).

Here, bending moment is M.

The free body diagram of the segment is shown as in Figure (6).

The shear force equation for the segment is obtained as follows:

Apply the force law of equilibrium along y-direction.

The bending moment over the segment is calculated as follows:

Apply the moment law of equilibrium in the Figure (6).

(c)

The free body diagram of the segment is shown as in Figure (5).

The internal shear of the segment is obtained as follows.

Apply the force law of equilibrium along x-direction.

Here, internal shear force is V.

The bending moment for the segment is calculated as follows:

Apply the moment law of equilibrium in the Figure (5).

Here, bending moment is M.

(d)

The free body diagram of the segment is shown as in Figure (6).

The shear force equation for the segment is obtained as follows:

Apply the force law of equilibrium along y-direction.

At

At

The bending moment over the segment is calculated as follows:

Apply the moment law of equilibrium in the Figure (6).

At

At

(e)

The shear force diagram for the entire beam is shown in Figure (7)

(f)

The bending moment diagram for the entire beam is shown in Figure.

Ans: Part a

The reactions at support C along x-direction is 0 kip along y-direction is 30 kip and the moment about point C is .

Part b

The internal shear over the segment is 0 kip and the internal bending moment over the segment is .

The internal shear over the segment is and the internal bending moment over the segment is .

Part c

The internal shear and bending moment at 0 ft are 0 kip, and the internal shear and bending moment at 8 ft are 0 kip, .

Part d

The internal shear and bending moment at 8 ft are 0 kip, and the internal shear and bending moment at 14 ft are ,.

Part e

The shear force diagram drawn using the expression is

Part f

The bending moment diagram drawn using the expression is