Beam Forces and Moments
This page provides the sections on beam forces and moments from the "Stress Analysis Manual," Air Force Flight Dynamics Laboratory, October 1986.
Other related chapters from the Air Force "Stress Analysis Manual" can be seen to the right.
 Simple Beam Bending
 Shear Web Beam Bending
 Partial Tension Beam Bending
 Beam Forces & Moments
 Beam Columns
 Beam Torsion
Nomenclature
A  =  area of moment diagram 
a  =  linear dimension 
a  =  distance from the left end of a span to the centroid of its moment diagram 
b  =  linear dimension 
b  =  distance from the right end of a span to the centroid of its moment diagram 
C  =  centroid of moment diagram 
c  =  linear dimension 
d  =  linear dimension 
E  =  modulus of elasticity 
I  =  moment of inertia 
L  =  length 
M  =  applied bending moment 
P  =  applied concentrated load 
R  =  reaction 
W  =  concentrated transverse load 
w  =  distributed transverse load 
1.3.4 Introduction to Reaction Forces and Moments on Beams Under Transverse Loading
Figure 130 shows a beam under transverse loading. Two equations of equilibrium may be applied to find the reaction loads applied to such a beam by the supports. These consist of a summation of forces in the vertical direction and a summation of moments. If a beam has two reaction loads supplied by the supports, as in the case of a cantilever beam or a beam simply supported at two points, the reaction loads may be found by the equilibrium equations and the beam is statically determinate. However, if a beam has more than two reaction loads, as in the case of a beam fixed at one end and either pinned or fixed at the other end, it is statically indeterminate and beam deflection equations must be applied in addition to the equations of statics to determine the reaction loads.
Section 1.3.4.1 presents a method for determining reaction loads on beams fixed at one end and pinned at another point, and Section 1.3.4.3 treats reaction loads for beams fixed at both ends. Beams on three or more supports are treated in Section 1.3.4.5.
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1.3.4.1 Reaction Forces and Moments on Beams with One Fixed End and One Pinned Support
Figure 131(a) shows a uniform beam with one fixed and one pinned support. The following procedure may be used to determine the support reactions on such a beam if its stresses are in the elastic range.
 Split the beam at the pinned support as in Figure 131(b) and find M_{A} from the equations of statics.
 Consider the right section of the beam as a single beam simply supported at both ends as in Figure 131(b). Find the moment diagram for this beam as in Figure 131(c). A is the area of this moment diagram and C is the centroid of this area.

Find M_{B} by the equation
$$ M_B = { 3 A \bar{a} \over L^2 }  { M_A \over 2 } $$The evaluation of the first term of this equation may be facilitated by the use of Table 110.(138)
 Evaluate R_{A} and R_{B} by applying the equations of statics to Figure 131(d).
Once the support reactions have been determined, the moment and shear diagrams may be constructed for the beam. If the pinned support is at the end of the beam, M_{A} may be set equal to zero.
1.3.4.2 Sample Problem  Reactions on Beam with One Fixed and One Pinned Support
Given: The beam shown in Figure 132.
Find: The reaction moments and forces on the beam.
Solution: Figure 133(a) may be obtained by redrawing the beam as in Figure 131(b). The moment diagram may then be drawn for the right portion; and A, a, and M_{A} may be determined as in Figure 133(b).
From Equation (138),
Now that M_{B} is known, R_{A} and R_{B} may be found by applying the equations of statics to Figure 133(c). Doing this gives R_{A} = 781 lb and R_{B} = 219 lb.
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1.3.4.3 Reaction Forces and Moments on Beams with Both Ends Fixed
Figure 134(a) shows a uniform beam with both ends fixed. The following procedure may be used to determine the support reactions on such a beam if its stresses are in the elastic range.
 Consider the beam to be simply supported as in Figure 134(b).
 Find the moment diagram for this simply supported beam as in Figure 134(c). A is the area of the moment diagram and C is the centroid of this area.

Find M_{A} and M_{B} by the equations
$$ M_A = { 2 A \over L^2 } ~( 2 \bar{b}  \bar{a} ) $$(139)$$ M_B = { 2 A \over L^2 } ~( 2 \bar{a}  \bar{b} ) $$The evaluation of the terms in these equations may be facilitated by the use of Table 110.(140)
 Evaluate R_{A} and R_{B} by applying the equations of statics to Figure 134(d).
Once the end reactions have been determined, the moment and shear diagrams may be constructed for the beam.
The above procedure may be avoided by using Table 19 which gives equations for the reaction moments for beams fixed at both ends under various loadings. The sign convention for this table are as shown in Figure 134(d).
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1.3.4.4 Reaction Forces and Moments on Continuous Beams
A continuous beam is one with three or more supports. Such a beam is statically indeterminate and deflection equations must be applied to find the support reactions. The threemoment equation is such an equation.
1.3.4.5 Application of the Three Moment Equation to Solving for the Reactions on Continuous Beams
Figure 135(a) shows a uniform beam that is simply supported at three colinear points, A, B, and C. In order to obtain the reactions, the beam is broken into two simply supported sections with no end moments, as shown in Figure 135(b). The moment diagrams are then found for these sections and the area A and centroid C of these diagrams are found as shown in Figure 135(c). The quantities found may now be substituted into the three moment equation:
If M_{A} and M_{C} are known, this equation may be solved for the moment at B, M_{B}. Knowing this moment, the support reactions at A, B, and C may be found by applying the equations of statics.
The terms to the right of Equation (141) may be found for various simple loadings by use of Table 110.
If a beam has a number of concentrated loads as shown in Figure 136, Equation (142) becomes
where P_{1} denotes any one of several concentrated loads which may act on the left span at a distance a_{1} from support A. Similarly, P_{2} denotes any load in the right span at a distance from support C.
If a beam is simply supported at more than thre.e points, the threemoment equation may be written for each intermediate support. The equations may then be solved simultaneously to obtain the moments at each support. This procedure is illustrated by the sample problem in Section 1.3.4.6.
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1.3.4.6 Sample Problem  Reactions on Continuous Beams by the Three Moment Equation
Given: The continuous beam shown in Figure 137.
Find: The support reactions.
Solution: The threemoment equation may be written for spans 1 and 2. Since only concentrated loads are present, the special case given by Equation (142) may be used. Thus,
Inserting numerical values gives
Simplifying gives 5M_{2} + 2M_{3} = 3875.
The more general form of the threemoment equation given by Equation (141) may now be written for spans 2 and 3 with the aid of cases one and three of Table 110.
Simplifying gives 3M_{2} + 14M_{3} = 15,000.
The two equations in M_{2} and M_{3} that were just obtained may be solved simultaneously to find that M_{2} = 376 and M_{3} = 990.
The equations of statics may now be applied as illustrated in Figure 138 to find the reaction forces.
The beam may now be drawn as in Figure 139.