ANALYSIS OF BEAMS
4.3 SHEAR IN BEAMS
Figure 4.4arepresents a simple beam with a uniformly distributed load over its entire length. Examination of an actual beam so loaded would
w=16 000= 20 800
, lb /ft or 800 plf (pounds per lineal foot)
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Figure 4.3 Determination of beam loads and display of the loaded beams for a framing system. (a) Plan. (b) Loading diagram for the beam. (c) Loading diagram for the girder.
probably not reveal any effects of the loading on the beam. However, there are three distinct major tendencies for the beam to fail. Figures 4.4b–dillustrate the three phenomena.
First, there is a tendency for the beam to fail by dropping between the supports (Figure 4.4b). This is called vertical shear. Second, the beam may fail by bending(Figure 4.4c). Third, there is a tendency in wood beams for the fibers of the beam to slide past each other in a horizontal direction (Figure 4.4d), an action described as horizontal shear. Natu- rally, a beam properly designed does not fail in any of the ways just mentioned, but these tendencies to fail are always present and must be considered in structural design.
Vertical Shear
Vertical shear is the tendency for one part of a beam to move vertically with respect to an adjacent part. The magnitude of the shear force at any section in the length of a beam is equal to the algebraic sum of the verti- cal forces on either side of the section. Vertical shear is usually repre- sented by the letter V. In computing its values in the examples and problems, consider the forces to the left of the section, but keep in mind that the same resulting force magnitude will be obtained with the forces on the right. To find the magnitude of the vertical shear at any section in the length of a beam, simply add up the forces to the right or the left of the section. It follows from this procedure that the maximum value of the shear for simple beams is equal to the greater reaction.
Example 1. Figure 4.5aillustrates a simple beam with concentrated loads of 600 lb and 1000 lb. The problem is to find the value of the ver- tical shear at various points along the length of the beam. Although the weight of the beam constitutes a uniformly distributed load, it is ne- glected in this example.
Solution: The reactions are computed as previously described, and are found to be R1= 1000 lb and R2= 600 lb.
Consider next the value of the vertical shear Vat an infinitely short distance to the right of R1. Applying the rule that the shear is equal to the reaction minus the loads to the left of the section, we write
V=R1– 0, or V= 1000 lb
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Figure 4.4 Characteristic forms of failure for a simple beam. (a) Beam with uniformly distributed load. (b) Vertical shear. (c) Bending. (d) Horizontal shear.
The zero represents the value of the loads to the left of the section, which of course, is zero. Now take a section 1 ft to the right of R1; again
V(x= 1)=R1– 0, or V(x= 1)= 1000 lb
The subscript (x= 1) indicates the position of the section at which the shear is taken, the distance of the section from R1. At this section, the shear is still 1000 lb and has the same magnitude up to the 600-lb load.
The next section to consider is a very short distance to the right of the 600-lb load. At this section,
V(x= 2+)= 1000 – 600 = 400 lb
Because there are no loads intervening, the shear continues to be the same magnitude up to the 1000 -lb load. At a section a short distance to the right of the 1000-lb load,
V(x= 6+)= 1000 – (600 + 1000) = – 600 lb This magnitude continues up to the right-hand reaction R2.
Figure 4.5 Examples 1 and 2.
Example 2. The beam shown in Figure 4.5b supports a concentrated load of 12,000 lb located 6 ft from R2and a uniformly distributed load of 800 pounds per linear foot (lb/ft) over its entire length. Compute the value of vertical shear at various sections along the span.
Solution: By use of the equations of equilibrium, the reactions are deter- mined to be R1= 10,900 lb and R2= 13,900 lb. Note that the total distrib- uted load is 800 ×16 = 12,800 lb. Now consider the vertical shear force at the following sections at a distance measured from the left support.
V(x = 0) = 10,900 – 0 = 10,900 lb V(x= 1) = 10,900 – (800 ×1) = 10,100 lb V(x= 5) = 10,900 – (800 ×5) = 6900 lb V(x= 10 –) = 10,900 – (800 ×10) = 2900 lb
V(x= 10 +) = 10,900 – {(800 ×10) + 12,000)} = –9100 lb V(x= 16) = 10,900 – {(800 ×16) + 12,000)} = –13,900 Shear Diagrams
In the two preceding examples, the value of the shear at several sections along the length of the beams was computed. In order to visualize the re- sults, it is common practice to plot these values on a diagram, called the shear diagram, which is constructed as explained below.
To make such a diagram, first draw the beam to scale and locate the loads. This has been done in Figures 4.6aandbby repeating the load di- agrams of Figures 4.5aandb, respectively. Beneath the beam draw a horizontal baseline representing zero shear. Above and below this line, plot at any convenient scale the values of the shear at the various sec- tions; the positive, or plus, values are placed above the line and the neg- ative, or minus, values below. In Figure 4.6a, for instance, the value of the shear at R1is +1000 lb. The shear continues to have the same value up to the load of 600 lb, at which point it drops to 400 lb. The same value continues up to the next load, 1000 lb, where it drops to –600 lb and continues to the right-hand reaction. Obviously, to draw a shear di- agram, it is necessary to compute the values at significant points only.
Having made the diagram, we may readily find the value of the shear at any section of the beam by scaling the vertical distance in the dia- gram. The shear diagram for the beam in Figure 4.6bis made in the same manner.
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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.