STRUCTURES: PURPOSE AND FUNCTION
2.12 FORCES ON A BEAM
Problem 2.11.C
If the pier in Figure 2.11bweighs 5000 lb, find the magnitude required for force Pto cause overturning.
1. The algebraic sum of the horizontal forces is zero.
2. The algebraic sum of the vertical forces is zero.
3. The algebraic sum of the moments of all the forces about any point in the plane is zero.
These summations can be made for any coplanar system of forces.
However, any additional qualifications of the forces may result in sim- plification of the algebraic conditions. For example, when the forces are concurrent(all meeting at a single point), they have no moments with re- spect to each other and the condition for equilibrium of moments can be eliminated, leaving only the two force equations. This was the case for the system shown in Figure 2.26. An even simpler qualification is that of colinearforces, all acting on a single line of action, such as the system shown in Figure 2.30a. Such a system, if in equilibrium, consists of two equal forces of opposite sense.
Beams are generally operated on by parallel, coplanar forces. This eliminates one of the force summations from the condition for general coplanar systems, since all the forces are in a single direction. There are thus only two equations of equilibrium necessary for the parallel system, and consequently only two available for solution of the system. Elimi- nating one force equation from the general set leaves
1. The sum of the vertical forces equals zero.
2. The sum of the moments about any point equals zero.
FORCES ON A BEAM 103
Figure 2.30 Moment effects on a beam.
1. The sum of the moments about point Ais zero.
2. The sum of the moments about point Bis zero
Where:pointAis a different point in the plane than point B.
Consider the simple beam in Figure 2.30b. Four vertical forces act on this beam and are in equilibrium. The two downward forces, or loads, are 4 kips and 8 kips. Opposing these are the support reaction forces at the ends of the beam, 4.4 kips and 7.6 kips. If these parallel forces are indeed in equilibrium, they should satisfy the equilibrium equations for a paral- lel system. Thus,
ΣFv= 0 = +4.4 – 4 – 8 + 7.6 = (+12) + (–12) and the forces are in balance.
ΣMA= 0 = +(4.4 ×20) – (4 ×14) – (8 ×4) = (+88) + (–88) and the sum of the moments about point Ais indeed zero.
To further demonstrate the equilibrium of the force values, moments may be taken about any other point in the plane. For example, for point B, which is the location of the 4-kip load,
ΣMB= +(4.4 ×6) + (8 ×10) – (7.6 ×14) = +(106.4) – (106.4) which verifies the balance of moments about point B.
Another type of problem involves the finding of some unknown forces in a parallel system. Remember that the two conditions of equi- librium for the parallel system provide two algebraic equations, which potentially may be used to find two unknown forces in the system. Con- sider the beam shown in Figure 2.31, with a single support and a load of 800 lb at one end. The problem is to determine the required value for a load at the other end of the beam that will maintain equilibrium and the value for the single support reaction. A summation of vertical forces will produce an equation with two unknowns. Indeed, the two unknown
forces could be solved using two equations in two unknowns. However, a simpler procedure frequently used is to write equations involving only one unknown in a single equation at a time, if possible. For example, an equation for the sum of moments about either the right end or the support will produce such an equation. Thus, for moments about the support, calling the unknown load x,
ΣM= 0 = –(800 ×6) + (x×3); thus, x= 1600 lb
Then, from a summation of vertical forces, calling the reaction force R, ΣF= 0 = –800 +R –1600; thus, R= 2400 lb
This form of solution is frequently used to find reactions for ordinary beams with two supports, which is discussed next.
Problem 2.12.A
Write the two equations for moments for the four forces in Figure 2.30b, taking points CandDas the centers of moments, to verify the equilib- rium of the system.
Determination of Reactions for Beams
As noted earlier, reactionsare the forces at the supports of beams that hold the loads in equilibrium. A single-span beam is shown in Figure 2.32, with two supports, one at each end of the beam. As these supports are not shown to have resistance to rotation (called fixed supports), they are assumed to be resistant only to the necessary vertical forces, and de- scribed as simple supports. This common beam, with a single span and
FORCES ON A BEAM 105
Figure 2.31 Beam with a single support.
two simple supports, is referred to as a simple beam. The computations that follow will demonstrate the common procedure for finding the val- ues for the magnitudes of the two support reactions for a simple beam.
Note that the two reactions in Figure 2.32 are designated R1andR2, for the left and right reactions, respectively. This is a common practice that is followed throughout the work in this book.
Example 9. Compute the reactions for the beam in Figure 2.32.
Solution: Taking the right reaction as the center of moments,
Taking the left reaction as the center of moments,
To see whether a mistake has been made, the three forces (load and two reactions) may be checked for equilibrium of the vertical forces; thus,
ΣF= 0 = +450 –1800 +1350, and the net force is indeed zero.
Example 10. Compute the reactions for the simple beam in Figure 2.33 with three concentrated loads.
Σ = = +M 0 1800 9× − R ×12 R =16 200= 12 1350
2 2
( ) ( ); ,
thus, lb
Σ = = +M 0 R ×12 − 1800 3× R = 5400= 12 450
1 1
( ) ( ); thus, lb
Figure 2.32 Example 9.
Solution: Regardless of the type or number of loads, the procedure is the same. Thus, considering the right reaction as the center of moments,
ΣM= 0 = +(R1×15) –(400 ×12) –(1000 ×10) –(600 ×4) Thus,
Using the same procedure with the left reaction as the center of moments,
And, for a check, the summation of vertical forces is ΣF= +1146.7 – 400 – 1000 – 600 + 853.3 = 0
For any beam with two simple supports, the procedure is the same.
Care must be taken, however, to note carefully the sign of the moments:
that is, plus for clockwise moments and minus for counterclockwise mo- ments about the selected center of moments. The following example has its supports drawn in from the ends of the beam, producing cantilevered oroverhangingends.
Example 11. Compute the reactions for the beam in Figure 2.34 with overhanging ends.
R2 400 3 1000 5 600 11 15
12 800 15 853 3
=( × +) ( × +) ( × )= , =
. lb R1 4800 10 000 2400
15
17 200
15 1146 7
= + , + = , =
. lb
FORCES ON A BEAM 107
Figure 2.33 Example 10.
Solution: Using the same procedure as in the preceding two examples, first take moments about the right reaction; thus,
ΣM= 0 = –(200 ×22) + (R1×18) – (1000 ×10) – (800 ×4) + (600 ×2) from which
Then, with a summation of moments about the left reaction,
ΣM= 0 = –(200 ×4) + (1000 ×8) + (800 ×14) – (R2×18) + (600 ×20) Thus,
A summation of vertical forces can be used to verify the answers.
Example 12. The simple beam shown in Figure 2.35ahas a single con- centrated load and a uniformly distributed load over a portion of the span. Compute the reactions.
R2 30 400
18 1688 9
= , =
. lb R1 16 400
18 911 1
= , =
. lb
Figure 2.34 Example 11.
Solution: For a simplification in finding the reactions, it is common to consider the uniformly distributed load to be replaced by its resultant in the form of a single concentrated load at the center of the distributed load.
The total of the uniform load is 200 ×8 = 1600 lb, and the beam is thus considered to be as shown in Figure 2.35b. With the modified beam, a summation of moments about the right reaction is
A summation of moments about the left reaction will determine a value of 1940 lb for R2, and a summation of vertical forces may be used to ver- ify the answers.
This shortcut, consisting of replacing the distributed load by its resul- tant, is acceptable for finding the reactions, but the real nature of the dis- tributed load must be considered for other investigations of the beam, as will be demonstrated in some of the later chapters.
Problems 2.12.B–G
Compute the reactions for the beams shown in Figures 2.36b–g.
Σ = = +M 0 R ×20 − 2200 14× − 1600×4 R =37 200= 20 1860
1 1
( ) ( ) ( ), ,
lb
FORCES ON A BEAM 109
Figure 2.35 Example 12.
Figure 2.36 Problems 2.12.B–G.
111