ANALYSIS OF BEAMS
4.4 BENDING MOMENTS IN BEAMS
There are two important facts to note concerning the vertical shear.
The first is the maximum value. The diagrams in each case confirm the earlier observation that the maximum shear is at the reaction having the greater value, and its magnitude is equal to that of the greater reaction. In Figure 4.6a, the maximum shear is 1000 lb, and in Figure 4.6b, it is 13,900 lb. We disregard the positive or negative signs in reading the maximum values of the shear, for the diagrams are merely conventional methods of representing the absolute numerical values.
Another important fact to note is the point at which the shear changes from a plus to a minus quantity. We call this the point at which the shear passes through zero. In Figure 4.6a, it is under the 1000-lb load, 6 ft from R1. In Figure 4.6b, it is under the 12,000-lb load, 10 ft from R1. A major con- cern for noting this point is that it indicates the location of the maximum value of bending moment in the beam, as discussed in the next section.
Problems 4.3.A–F
For the beams shown in Figures 4.7a–f, draw the shear diagrams and note all critical values for shear. Note particularly the maximum value for shear and the point at which the shear passes through zero.
141
Figure4.7Problems 4.3.A–F.
or 2000 lb, tends to cause a clockwise rotation about this point. Because the force is 2000 lb and the lever arm is 6 ft, the moment of the force is 2000
×6 = 12,000 ft-lb. This same value may be found by considering the forces to the right of the section X-X: R2, which is 6000 lb, and the load 8000 lb, with lever arms of 10 and 6 ft, respectively. The moment of the reaction is 6000×10 = 60,000 ft-lb, and its direction is counterclockwise with respect to the section X-X. The moment of force 8000 lb is 8000 ×6 = 48,000 ft-lb, and its direction is clockwise. Then 60,000 ft-lb – 48,000 ft-lb = 12,000 ft-lb, the resultant moment tending to cause counterclockwise rota- tion about the section X-X. This is the same magnitude as the moment of the forces on the left, which tends to cause a clockwise rotation.
Thus, it makes no difference whether use is made of the forces to the right of the section or the left, the magnitude of the moment is the same.
It is called the bending moment(or the internal bending moment) because it is the moment of the forces that causes bending stresses in the beam. Its magnitude varies throughout the length of the beam. For instance, at 4 ft fromR1, it is only 2000 ×4, or 8000 ft-lb. The bending moment is the al- gebraic sum of the moments of the forces on either side of the section.
For simplicity, take the forces on the left; then the bending moment at any section of a beam is equal to the moments of the reactions minus the moments of the loads to the left of the section. Because the bending mo- ment is the result of multiplying forces by distances, the denominations are foot-pounds or kip-feet.
Bending Moment Diagrams
The construction of bending moment diagrams follows the procedure used for shear diagrams. The beam span is drawn to scale showing the
Figure 4.8 Development of bending at a selected cross section.
locations of the loads. Below this, and usually below the shear diagram, a horizontal baseline is drawn representing zero bending moment. Then the bending moments are computed at various sections along the beam span, and the values are plotted vertically to any convenient scale. In sim- ple beams, all bending moments are positive and therefore are plotted above the baseline. In overhanging or continuous beams, there are also negative moments, and these are plotted below the baseline.
Example 3. The load diagram in Figure 4.9 shows a simple beam with two concentrated loads. Draw the shear and bending moment diagrams.
Solution: R1andR2are computed first, and are found to be 16,000 lb and 14,000 lb, respectively. These values are recorded on the load diagram.
The shear diagram is drawn as described in Section 4.3. Note that, in this instance, it is only necessary to compute the shear at one section (between the concentrated loads) because there is no distributed load, and we know that the shear at the supports is equal in magnitude to the reactions.
Because the value of the bending moment at any section of the beam is equal to the moments of the reactions minus the moments of the loads to the left of the section, the moment at R1must be zero, for there are no forces to the left. Other values in the length of the beam are computed as follows. The subscripts (x= 1, etc.) show the distance from R1at which the bending moment is computed.
M(x= 1)2= (16,000 ×1) = 16,000 ft-lb M(x= 2)2= (16,000 ×2) = 32,000 ft-lb
M(x= 5)2= (16,000 ×5) – (12,000 ×3) = 44,000 ft-lb M(x= 8)2= (16,000 ×8) – (12,000 ×6) = 56,000 ft-lb
M(x= 10)= (16,000 ×10) – {(12,000 ×8) + (18,000 ×2)} = 28,000 ft-lb M(x= 12)= (16,000 ×12) – {(12,000 ×10) + (18,000 ×4)} = 0
The result of plotting these values is shown in the bending moment di- agram of Figure 4.9. More moments were computed than were necessary.
We know that the bending moments at the supports of simple beams are zero, and in this instance, only the bending moments directly under the loads were needed.
BENDING MOMENTS IN BEAMS 143
spect is that the bending moment has a maximum magnitude wherever the shear passes through zero. In Figure 4.9, the shear passes through zero under the 18,000-lb load, that is, at x= 8 ft. Note that the bending moment has its greatest value at this same point, 56,000 ft-lb.
Example 4. Draw the shear and bending moment diagrams for the beam shown in Figure 4.10, which carries a uniformly distributed load of 400 lb/ft and a concentrated load of 21,000 lb located 4 ft from R1.
Solution: Computing the reactions, we find R1= 17,800 lb and R2= 8800 lb. By use of the process described in Section 4.3, the critical shear values are determined and the shear diagram is drawn as shown in the figure.
Figure 4.9 Example 3.
Although the only value of bending moment that must be computed is that where the shear passes through zero, some additional values are de- termined in order to plot the true form of the moment diagram. Thus, M(x= 2)2= (17,800 ×2) – (400 ×2×1) = 34,800 ft-lb
M(x= 4)2= (17,800 ×4) – (400 ×4×2) = 68,000 ft-lb
M(x= 8)2= (17,800 ×8) – {(400 ×8×4) + (21,000 ×4)} = 45,600 ft-lb M(x= 12)= (17,800 ×12) – {(400 ×12×6) + (21,000 ×8)} = 16,800 ft-lb From the two preceding examples (Figures 4.9 and 4.10), it will be ob- served that the shear diagram for the parts of the beam on which no loads occur is represented by horizontal lines. For the parts of the beam on which a uniformly distributed load occurs, the shear diagram consists of straight inclined lines. The bending moment diagram is represented by straight inclined lines when only concentrated loads occur, and by a curved line if the load is distributed.
BENDING MOMENTS IN BEAMS 145
Figure 4.10 Example 4.
sary in designing beams to find the maximum bending moment, we must know the point at which it occurs. This, of course, is the point where the shear passes through zero, and its location is readily determined by the procedure illustrated in the following example.
Example 5. The load diagram in Figure 4.11 shows a beam with a con- centrated load of 7000 lb, applied 4 ft from the left reaction, and a uni- formly distributed load of 800 lb/ft extending over the full span.
Compute the maximum bending moment on the beam.
Solution: The values of the reactions are found to be R1= 10,600 lb and R2= 7600 lb and are recorded on the load diagram.
The shear diagram is constructed, and it is observed that the shear passes through zero at some point between the concentrated load of 7000 lb and the right reaction. Call this distance xft from R2. The value of the shear at this section is zero; therefore, an expression for the shear for this point, using the reaction and loads, is equal to zero. This equation con- tains the distance x:
V(atx)= −7600 800+ x=0, x=7600= . ft 800 9 5
Figure 4.11 Example 5.
The zero shear point is thus at 9.5 ft from the right support and (as shown in the diagram) at 4.5 ft from the left support. This location can also be determined by writing an equation for the summation of shear from the left of the point, which should produce the answer of 4.5 ft.
Following the convention of summing up the moments from the left of the section, the maximum moment is determined as
Problems 4.4.A–F
Draw the shear and bending moment diagrams for the beams in Figure 4.7, indicating all critical values for shear and moment and all significant dimensions. (Note:These are the beams for Problem 4.3, for which the shear diagrams were constructed.)