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Tutorial 6 -- Shear and Moment Diagrams

Questions and Solutions

[ 11 Questions ]

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Question 1 (Example 6.7)

● Draw the shear and moment diagrams for the beam shown below.

Note: fixed support, a concentrated force (No distributed load No couple moment)

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load, No couple moment).

Procedure for analysis (R.S.F.P.)Determine Support Reactions;Separate x and determine shear and moment Functions;Plot Shear and Moment Diagrams.

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Step 1: Support Reactions. By drawing a FBD of the beam and applying the equations of equilibrium, we can determine the support reactions as shown in Figure (b).

Question 1(Solution)

The FDB of a beam segment of length

Step 2: Coordinates, Shear and Moment Functions.

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The FDB of a beam segment of length x is shown in the figure. The unknown shear and moment are indicated acting in positive and negative senses of the right-hand face of the segment.

V

M

x

V

MApplying the equations of equilibrium

+↑ Fy = 0; ... V = P

+ M = 0; M =Px PL = P(x L)

Question 1(Solution) (continued)

V+ M = 0; ... M =Px-PL = P(x-L)

Step 3: Shear and Moment Diagrams. The shear diagram represents a plot of the first equation and the moment diagram represents a plot of the second equation. At x=0, V=P, M=-PL and at x=L, V=P, M=0.

x

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Question 2 (Example 6.9)

● Draw the shear and moment diagrams for beam shown below.

Note: fixed support, a couple moment (No concentrated force, No distributed load).

Solution

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Step 1: Support Reactions, as shown in the figure.

Question 2 (Solution)(continued)

Step 2: Coordinates, Shear and Moment Functions. A FDB of a beam segment of length x is shown in the figure. The unknown moment is indicated acting in positive sense.

M

x

Applying the equations of equilibrium

+↑ Fy = 0; ... V = 0

+ M = 0; ... M =M0

Step 3: Shear and Moment Diagrams.

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p g

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Question 3●Draw the shear and moment diagrams for the cantilevered beam.

Note: fixed support, a couple moment at point B, and a concentrated force at point C.

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Procedure for analysis (R.S.F.P.)

Question 3(Solution)

Step 1: Support reactions at the fixed point as shown:

Step 2: Separate coordinates and shear and moment functions: Separate into two regions:

x

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For the second region:

Question 3 (Solution)(continued)

Step 3: Shear and moment diagrams:

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Step 3: Shear and moment diagrams:

Question 4

●The suspender bar supports the 3000N engine. Draw the shear and moment diagrams for the bar.

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Note: the bar has three concentrated forces.

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Question 4 (Solution)

Steps 1 and 2: Support reactions and separate coordinates: Separate into two regions:

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Step 3: Plot the shear and moment functions: Shear and moment diagrams: (Two values at the end points).

Question 4 (Solution)(Continued)

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Question 5

●Draw the shear and moment diagrams for the beam (a) in terms of the parameters shown; (b) set M0=500N m, L=8m.

Note: Simply supported beam, two couple moments.

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Procedure for analysis (R.S.F.P

Question 5 (Solution)

(a) Step 1: Support reactions: (The reactions are zero)

Step 2: Separate into three regions:

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Step 3: Shear and moment diagrams.

Question 5 (Solution)(continued)

||||||||||

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FBD

||||

Shear and moment diagrams

Question 5 (Solution)(continued)

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Question 6

●Draw the shear and moment diagrams for the beam.

Note: Simply supported beam, a distributed load acting on segment BC.

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Solution: Step1: Support Reactions:

Question 6(Solution)

Step 2: Separate into two regions:

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Question 6(Solution)(continued)

Step 3: Shear and moment diagrams.

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Question 7

●Draw the shear and moment diagrams for the beam.

Note: two couple moments at points A and B, and a distributed load acting on the beam AB.

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Procedure for analysis (R.S.F.P.)

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Step 1: Support reactions:

Question 7(Solution)

Step 2: Separate coordinates, Shear and moment functions:

FBD of the beam

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FBD of segment

Step 3: Shear and moment diagrams.

Question 7(Solution)(continued)

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Question 8

●Draw the shear and moment diagrams for the beam.

Note: a couple moment at point C, a concentrated force at point C and a distributed load on AB

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at point C, and a distributed load on AB.

Step 1: Support reactions as shown in the figure:

Question 8(Solution)

Step 2: Separate coordinates, shear and moment functions: Separate into two regions:

For

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Ans

Ans

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Question 8(Solution)(continued)

Take the FBD of the left part (recommended):

133.75 - 40*8 + 206.25 – V =0V = 20kN.

Or take the FBD of the right part (why?):

For

-133.75*x + 40*8*(x-4) - 206.25*x + M =0

M = 20*x -370 (kN.m).

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Ans

Ans

Step 3: Shear and moment diagrams

Question 8(Solution)(continued)

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Question 9 (Problem 6-21)

● Draw the shear and moment diagrams for the beam and determine the shear and moment in the beam as functions of x, where 1.2m<x<3m.

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Solution: Step 1: Support reactions and FBD.

YA

= YB

= 2.25kN.

Question 9 (Solution)(continued)

● Step 2: Separate coordinates into three segments:

Segment 1: For 0< x <1.2m;

Shear force: V=0, Bending moment: M =-300Nm.

Segment 2: For 1.2m< x < 3.0m;

28Segment 3: For 3.0m < x < 4.2m;

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● Step 3: Shear and moment diagrams.

Question 9 (Solution)(continued)Shear force; V=0N. Bending moment: M = -300Nm.

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Question 9 (Solution) (continued)

● Check whether the functions for shear force and bending moment are correct. Can you find the errors?

● The shear force function for 1.2m<x<3.0 should be

V= 5250 – 2500x (N)

● The bending moment function for 1.2m<x<3.0 should be

M =-1250x*x + 5250x – 4800 (Nm)

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● The maximum moment should be 712.5 Nm.

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Question 10 (Problem 6-25)

● Draw the shear and moment diagrams for the beam. The two segments are joined together at B.

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Question 10 (Solution)

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Question 11

● Determine the shear force and bending moment throughout the beam as functions of x (the origin of the coordinate is at the left end of the beam), then draw thecoordinate is at the left end of the beam), then draw the shear force and bending moment diagrams for the beam.

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