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2. Axial Force, Shear Force, Torque and Bending Moment Diagrams

2. Axial Force, Shear Force, Torque and Bending Moment Diagrams

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<strong>2.</strong> <strong>Axial</strong> <strong>Force</strong>, <strong>Shear</strong> <strong>Force</strong>, <strong>Torque</strong> <strong>and</strong> <strong>Bending</strong> <strong>Moment</strong><br />

<strong>Diagrams</strong><br />

In this section, we learn how to summarize the internal actions (shear force <strong>and</strong> bending<br />

moment) that occur throughout an axial member, shaft, or beam. Later, we learn how to<br />

convert these internal forces to stresses. (Note: <strong>Axial</strong> force <strong>and</strong> torsion bending diagrams are<br />

easy to derive. Will show an example to illustrate.)<br />

.: mention that axial force <strong>and</strong> torsion bending diagrams are easy to derive. Will showcouple<br />

of examples to ill<br />

Outline<br />

- Purpose of <strong>Axial</strong> <strong>Force</strong>, <strong>Shear</strong> <strong>Force</strong> <strong>and</strong> <strong>Bending</strong> <strong>Moment</strong> <strong>Diagrams</strong><br />

- Sign Convention<br />

- Basic Method Examples<br />

- Mathematical Relationship Between External Loading, Internal <strong>Shear</strong>, <strong>and</strong> Internal <strong>Bending</strong><br />

- Integration/Graphical Method Examples<br />

Purpose of <strong>Axial</strong> <strong>Force</strong>, <strong>Shear</strong> <strong>Force</strong> <strong>and</strong> <strong>Bending</strong> <strong>Moment</strong> <strong>Diagrams</strong><br />

We learned that we can get the internal forces at any point in a structure by: (1) taking an<br />

imaginary cut at the point; (2) drawing the free-body diagram; <strong>and</strong> (3) using the equations of<br />

equilibrium to calculate the internal forces. The challenge with a beam is that the internal<br />

forces can vary a great deal along the length of the beam. Therefore, we draw what are called<br />

shear force <strong>and</strong> bending moment diagrams, which show the internal forces along the<br />

length of the beam. This helps us visualize where the maximum stresses will occur.<br />

We will learn two methods for drawing shear force <strong>and</strong> bending moment diagrams: 1) the<br />

Basic Method; <strong>and</strong> 2) the Integration (also called Graphical) Method. In each case,<br />

mathematical functions describing the shear force <strong>and</strong> bending moment throughout the beam<br />

are derived <strong>and</strong> plotted along the length of the beam. In the Basic Method, we derive these<br />

functions from first principles. In the Integration/Graphical Method, the relationships<br />

between the applied load, shear force <strong>and</strong> bending moment are used to rapidly sketch out the<br />

diagrams.<br />

2-1


The approach used depends on the problem <strong>and</strong> what the designer requires from the<br />

diagrams. For example, many problems require only the maximum values of the shear <strong>and</strong><br />

moment, <strong>and</strong> the locations at which these values occur. The Graphical Method is most useful<br />

for these situations. In other cases, a beam may be subjected to a loading that is a fairly<br />

complicated function. For these situations, the Basic Method may be needed.<br />

You should know how to use both methods <strong>and</strong> recognize when to apply them!<br />

Sign Convention (Section 6.1 in Mechanics of Materials)<br />

We define the sign convention for internal shear force <strong>and</strong> bending moment:<br />

<strong>Shear</strong> <strong>Force</strong> <strong>Bending</strong> <strong>Moment</strong><br />

Positive internal shear force tends to rotate the free-body diagram clockwise.<br />

Positive internal bending moment causes the beam to sag. Also known (informally) as the<br />

smile rule. <strong>Bending</strong> moment is drawn on the compression side of the member.<br />

Summary of Steps for Basic Method:<br />

1) Determine the support reactions for the beam.<br />

2) Specify an origin for a co-ordinate x along the length of the beam.<br />

3) Section the beam with an imaginary cut at a distance x, <strong>and</strong> draw the free-body diagram.<br />

4) Determine shear <strong>and</strong> bending moment as a function of x using equilibrium equations.<br />

5) Repeat steps 3 <strong>and</strong> 4 for all regions between any two discontinuities of loading.<br />

6) Draw, to scale, the functions on a sketch of the beam.<br />

2-2


Basic Method Example<br />

Consider beam ABC from the example in Section 1.<br />

1.0 m<br />

A<br />

D<br />

0.5 m<br />

E<br />

θ<br />

B<br />

Pin<br />

F<br />

60 N/m<br />

0.5 m 0.5 m 0.5 m<br />

C<br />

Determine the axial force, shear force, <strong>and</strong> bending moment diagrams for the beam ABC.<br />

2-3


2-4


Mathematical Relationship Between External Loading, Internal <strong>Shear</strong>, <strong>and</strong> Internal <strong>Bending</strong><br />

The previous example shows that:<br />

1) axial force (<strong>and</strong> shear force) diagrams change abruptly at the location of a concentrated<br />

axial force (or applied force);<br />

2) for a region of a beam without external applied forces, the shear force has a constant<br />

value, <strong>and</strong> the bending moment is a function of x;<br />

[see section AB of the beam in the previous example]<br />

3) for a region of the beam subjected to a uniformly distributed load (UDL), the shear force is<br />

a function of x, <strong>and</strong> the bending moment is a function of x 2 .<br />

[see section BC of the beam in the previous example].<br />

In fact, we can prove that (see Section 6.2 of the textbook):<br />

dV ( x)<br />

= −w(<br />

x)<br />

(2-1)<br />

dx<br />

i.e. the slope of the shear force diagram at x is equal to the negative of the value of the<br />

loading function at x; <strong>and</strong>,<br />

dM ( x)<br />

= V ( x)<br />

(2-2)<br />

dx<br />

i.e. the slope of the moment diagram at x is equal to the value of the shear function at x.<br />

dV ( x)<br />

From Eq. 2-1 we have: = −w(<br />

x)<br />

through integration:<br />

dx<br />

(note what each side of the equation represents)<br />

From Eq. 2-2 we have through integration:<br />

(note what each side of the equation represents)<br />

2-5<br />

(2-3).<br />

(2-4).


Graphically:<br />

2-6


The following table, taken from your textbook, illustrates a number of common loading<br />

cases. It shows how shear <strong>and</strong> moment diagrams can be constructed on the basis of knowing<br />

the variation of the slope from the load <strong>and</strong> shear diagrams. Make sure that the relationships<br />

make sense, but you should not memorize the table!! You should always work from the<br />

basic relationships, Eqs. <strong>2.</strong>1 to <strong>2.</strong>4.<br />

(show overhead of table)<br />

2-7


Graphical Method Example 1<br />

Consider the cantilever beam subjected to the loading w(x). Draw the internal forces on a<br />

typical element of the beam, <strong>and</strong> hence derive the differential equations relating bending<br />

moments, shear forces, <strong>and</strong> applied loading, i.e. Eqs. 2-1, 2-<strong>2.</strong> Use these differential<br />

equations to draw shear force <strong>and</strong> bending moment diagrams for the following cases: (a) w(x)<br />

= 20 kN/m; (b) w(x) = 20x kN/m<br />

L<br />

w(x)<br />

2-8<br />

(add soln from Campbell notes)


2-9


Example<br />

Draw the shear force <strong>and</strong> bending moment diagrams.<br />

1 m<br />

35 kN/m<br />

2 m<br />

2-10<br />

3 m<br />

40 kN/m


2-11


Finally, we end this section with a discussion of torque diagrams. These are usually simpler<br />

than shear <strong>and</strong> bending moment diagrams, <strong>and</strong> can be illustrated with the following example.<br />

Example<br />

Draw the torque diagram for the cantilever shaft shown. Determine the maximum torque in<br />

the shaft.<br />

1 m 1 m 1 m<br />

5 kNm 7 kNm<br />

12 kNm<br />

2-12<br />

fixed<br />

end


2-13

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