Chapter 4
Shear and Moment in Beams
4-1 Introduction
In civil engineering, the term beam refers to a slender horizontal bar that supports transverse loadings
or forces that are perpendicular to the bar. There are internal forces that act inside the beam when
subjected to applied forces. Those internal resultant loads are called the shearing force and bending
moment. As discussed in the previous chapters, axial and torsional loads resulted in some internal
forces that are found inside the bar or over portions of the bar. The effect of shear force and bending
moment vary continuously from section to section of the beam.
Due to internal forces acting in the beam, two kinds of stresses are developed in the beam’s
section, namely, shearing stress and flexural stress. In order to properly design a beam, it is necessary to
determine the maximum shear and moment which are essential in the computation of stresses. These
stresses are part of the criteria that need to be satisfied in the actual design of the structural element.
This chapter is concerned with the variation of the shear and moment under different combination of
loadings and support system.
4-2 Types of Loadings
Beams are subjected to various loadings caused by dead loads, live loads, earthquake loads, wind loads,
and other types of loads. Loadings applied to the beam consist of concentrated loads ( Figure 4-2a ),
uniformly distributed load ( Figure 4-2b ), uniformly varying load ( Figure 4-2c ), or a couple moment
( Figure 4-2d ). These loadings are shown in Figure 4-2.
Figure 4-2a Figure 4-2b
Figure 4-2c Figure 4-2d
Figure 4-2
4-3 Shear and Moment
Shear is defined as the sum of the external loads that are found to the left or to the right of a section
acting perpendicular to the horizontal axis of the beam. Shear is denoted by V and expressed
mathematically as:
V = ( Σ Loads )Left
or V = ( Σ Loads )Right
where the subscripts left and right denote that the summation of forces must include only the loads that
act on the segment either to the left or to the right of the section.
Bending moment is defined as the sum of the moments taken about the centroidal axis of the
beam’s section of all loads acting either to the left or to the right of the section. This can be expressed
as:
M = ( Σ Moment )Left = ( Σ Moment )Right
where the subscript left denotes that the bending moment is calculated for the loads to the left of the
section and right refers to the loads found to the right of the section.
Consider the beam shown in Figure 4-3 which is subjected to a uniform load “w” and is held in
equilibrium by reactions R1 and R2 at supports A and B, respectively. Assume that the beam is cut at
point C and the left portion of the beam is considered. The part of the beam removed is replaced by
vertical shearing force V and moment M to hold the left portion in equilibrium under the action of R 1
and wx. The couple moment M is called the resisting moment or simply moment and the force V is
termed resisting shear or simply shear.
Figure 4-3
4-4 Statically Determinate and Statically Indeterminate Beams
Beams are called statically determinate when the support reactions may be determined using the
equations of static equilibrium. On the other hand, if the number of reaction elements exceeded the
equations of static equilibrium, the beam is said to be indeterminate. In order to compute the reactions
at the supports of the beam, the equilibrium equations must be supported by other equations based
upon the elastic deformation of the beam.
The degree of indeterminacy of a beam is the difference between the number of reactions and
the number of equilibrium equations. Shown in Figure 4-4a and Figure 4-4b are examples of statically
determinate and statically indeterminate beams.
Figure 4-4a Statically Determinate Beams
Figure 4-4b Statically Indeterminate Beams
4-5 Sign Conventions for Load, Shear and Moment
In this chapter, loads acting upward are considered positive, and negative if directed downward. For
shearing force, a positive shear force tends to move the left segment upward relative to the right
segment, and vice versa as shown in Figure 4-5a.
Figure 4-5a
Bending moment is considered positive if it causes the beam segment to bend concave upward
and negative if otherwise. The concavity of the beam segments is shown in Figure 4-5b.
Figure 4-5b
Illustrative Problems:
1. Write the shear and moment equations for the beams loaded as shown. In each problem, let “x” be
the distance measured from the left end of the beam. Also, construct the shear and moment diagrams
indicating values at every change of loading positions and at points of zero shear.
(a)
Solution: Calculate the support reactions;
Σ MB = 0
– 30 kN ( 1 m ) + 50 kN ( 3 m ) – R D ( 5 m ) = 0
RD = 24 kN
Σ FV = 0
RB + RD = 30 kN + 50 kN
RB = 56 kN
Consider segment AB; Consider segment BC;
VAB = – 30 kN VBC = – 30 kN + 56 kN
MAB = – 30x kN-m VBC = 26 kN
MBC = – 30x + 56 ( x -1 )
MBC = 26x – 56 kN-m
Consider segment CD;
VCD = – 30 kN + 56 kN – 50 kN MCD = – 30x + 56 ( x -1 ) – 50 ( x -4 )
VCD = – 24 kN MCD = – 24x + 144 kN-m
To determine the ordinate of the shear diagram, downward forces are subtracted and upward forces are
added. For the moment diagram, substitute the values of lower and upper limits to horizontal distances.
For example, consider segment AB with the distance varying from zero to 1. Substitute the values of
zero and 1 in MAB = – 30x kN-m to obtain zero and – 30 kN-m, respectively as reflected on the moment
diagram.
(b)
Solution: Calculate the support reactions;
Σ MC = 0
– 80 kN ( 8 m ) – 10 kN/ m ( 10 m ) ( 5 m ) + R A ( 10 m ) = 0
RA = 114 kN
Σ FV = 0
RA + RC = 80 kN + 10 kN/ m ( 10 m )
RC = 66 kN
Consider segment AB; Consider segment BC;
VAB = 114 kN – 10x VBC = 114 kN – 80 kN – 10x
MAB = 114x – 10x ( x/2 ) VBC = 34 kN – 10x
2
MAB = 114x – 5x MBC = 114x – 10x ( x/2 ) – 80 ( x - 2 )
After creating the shear and moment equations, substitute the values of “x” and draw the shear and
moment diagrams.
From the diagram, the maximum moment occurs when the value of shear is zero. It is also noted that
the ordinate of moment diagram depends on the area of the shear diagram. Positive areas of the shear
diagram are added and negative areas are subtracted in order to obtain the ordinate of the moment
diagram.
2. Draw the shear and moment diagram for the beam loaded as shown without writing the shear and
moment equations. Give the numerical values at change of loading positions at point of zero shear.
Use the semi-graphical method.
Solution: Compute the support reactions;
Σ M2 = 0
– ½ ( 60 kN/ m ) ( 3 m ) ( 2 m ) – 20 kN/ m ( 4 m ) ( 3 m ) + R 1 ( 5 m ) = 0
R1 = 84 kN
Σ M1 = 0
½ ( 60 kN/ m ) ( 3 m ) ( 3 m ) + 20 kN/ m ( 4 m ) ( 2 m ) – R 2 ( 5 m ) = 0
R2 = 86 kN
To draw the shear diagram;
VA = R1 = 84 kN
VB = 84 kN – 20 kN/ m ( 1 m ) = 64 kN
VC = 64 kN – ½ ( 20 + 80 ) ( 3 m ) = – 86 kN
VD = – 86 kN + 86 kN = 0
Compute for distance “x”;
E
x
Σ FV = 0
84 kN – 20 kN/ m ( 1 + x ) – ½ ( y ) ( x ) = 0
where y = 20x
84 kN – 20 kN/ m ( 1 + x ) – ½ ( 20x ) ( x ) = 0
X = 1.72 m
To construct the moment diagram;
MA = 0
MB = MA + ½ ( 84 + 64 ) ( 1 ) = 74 kN-m
Σ ME = M E
ME = 84 kN ( 2.72 m ) – 20 kN/ m ( 2.72 ) ( 2.72/2 ) – ½ ( 34.4 ) ( 1.72 ) ( 1.72/3 )
ME = 137.50 kN-m
Σ MC = M C
MC = ( 86 kN ) ( 1 m ) = 86 kN-m
Exercises:
1. Write the shear and moment equations to solve the following problems. Then, construct the shear
and moment diagrams indicating values at all change of loading positions and at points of zero shear.
Neglect the weight of the beam.
( a ) Mmax = 5250 lb-ft 1800 lb
(b)
500 lb 300 lb
( c ) Mmax = 44.45 kN-m
20 kN/ m
(d)
( e ) Mmax = 67.22 kN-m
40 kN/ m
(f)
700 lb/ ft
(g) 90 lb/ ft
( h ) Mmax = 28 kN-m
( i ) Assume wo = 14 lb/ ft and L = 10 ft Mmax = 116.67 lb-ft
(j)
(k)
w = 3 kip/ ft
( l ) Mmax = 80 kN-m
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2. Draw the shear and moment diagrams for the beams loaded as shown without writing the shear and
moment equations. Give the numerical values at change of loading positions at point of zero shear.
Use the semi-graphical method.
( a ) Mmax = 36 kN-m
4 kN/ m
(b)
750 lb
( c ) Mmax = 16 kN-m
8 kN/ m
(d)
130 lb
( e ) Mmax = 105 kN-m
30 kN 15 kN/ m 20 kN
(f)
40 kN 25 kN/ m
( g ) Mmax = 72 kN-m
30 kN
(h)
350 lb/ ft
( i ) Mmax = 9880 lb-ft
1200 lb 410 lb/ ft
(j)
3400 lb 360 lb/ ft
( k ) Mmax = 151 kN-m
90 kN/ m
(l)
70 kN 70 kN
3. Construct the load and moment diagrams that correspond to the given shear diagram. Specify values
at all change of loading positions and at all points of zero shear.
(a)
(b)
(c)
(d)
(e)
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4-6 Moving Loads
A container van, ten-wheeler truck, or other moving vehicles rolling across a beam or girder comprises a
system of concentrated loads at constant distances from one another. For beams carrying series of
concentrated loads, the maximum bending moment occurs under one of the loads. The problem here is
to compute the bending moment under each load and identify the position of the loads to come up with
the maximum values of the functions.
Beams and girders such as those used in bridges are subject to moving loads. For a single
moving load, the maximum moment occurs when the load is located at the midspan. For two moving
loads, the maximum shear occurs at the reaction where the larger load is positioned over the support as
shown in Figure 4-6.
Figure 4-6
In general, the bending moment under a particular load becomes a maximum when the center
of the beam is midway between that load and the resultant of all loads then on the span. For the
calculation of the maximum shear, the largest value of shearing force occurs at, and is equal to the
maximum reaction. Maximum reaction is the reaction to which the resultant load is nearest.
Illustrative Problems:
1. A truck with axle loads of 40 kN and 60 kN rolls over a 10-m span beam. Determine the maximum
bending moment and maximum shearing force developed in the beam due to moving loads.
Solution: Locate the position of the resultant force R;
The value of resultant force R = 40 kN + 60 kN = 100 kN
MR = Σ Mcomp
R ( x ) = 40 kN ( 5 m )
x=2m
Calculate the maximum moment under the 40-kN load;
Σ MR2 = 0
R1 ( 10 m ) = 100 kN ( 3.5 m )
R1 = 35 kN
Pass a section through the 40-kN load and consider to the left of the section;
Σ M40 = M40
M40 = 35 kN ( 3.5 m )
M40 = 122.50 kN-m
Calculate the maximum moment under the 60-kN load;
Σ MR1 = 0
R2 ( 10 m ) = 100 kN ( 4 m )
R2 = 40 kN
Pass a section through the 60-kN load and consider to the right of the section;
Σ M60 = M60
M60 = 40 kN ( 4 m )
M60 = 160 kN-m
Thus, Mmax = 160 kN-m Ans
The maximum shearing force occurs when the 60-kN load is placed over the right support.
Σ MR1 = 0
R2 ( 10 m ) = 100 kN ( 8 m )
R2 = 80 kN
Thus, Vmax = 80 kN Ans
2. The three moving loads roll over a 44-ft span beam. Determine the maximum values of shearing
force and bending moment developed in the beam due to the moving loads.
Solution: Locate the position of the resultant force R;
The value of resultant force R = 4 kip + 8 kip + 6 kip = 18 kip
MR = Σ Mcomp
R ( x ) = 8 kip ( 9’ ) + 6 kip ( 27’ )
x = 13’
Case 1: Calculate the maximum moment under the 4-kip load;
Σ MR2 = 0
R1 ( 44’ ) = 18 kip ( 15.5’ )
R1 = 6.34 kip
Pass a section through the 4-kip load and consider to the left of the section;
Σ M4 = M 4
M4 = 6.34 kip ( 15.5’ )
M4 = 98.27 kip-ft
Case 2: Calculate the maximum moment under the 8-kip load;
Σ MR2 = 0
R1 ( 44’ ) = 18 kip ( 20’ )
R1 = 8.18 kip
Pass a section through the 8-kip load and consider to the left of the section;
Σ M8 = M 8
M8 = 8.18 kip ( 20’ ) – 4 kip ( 9’ )
M8 = 127.60 kip-ft
Case 3: Calculate the maximum moment under the 6-kip load;
Σ MR1 = 0
R2 ( 44’ ) = 18 kip ( 15’ )
R2 = 6.14 kip
Pass a section through the 8-kip load and consider to the right of the section;
Σ M6 = M 6
M6 = 6.14 kip ( 15’ )
M6 = 92.10 kip-ft
Thus, Mmax = 127.60 kip-ft Ans
The maximum shearing force occurs when the 4-kip load is placed over the left support.
Σ MR2 = 0
R1 ( 44’ ) = 18 kip ( 35’ )
R1 = 14.32 kip
Thus, Vmax = 14.32 kip Ans
Exercises:
1. A truck with axle loads of 35 kN and 50 kN on a wheel base of 6 meters rolls across a 12-m span.
Calculate the maximum shearing force and bending moment developed in the span.
Answer: Mmax = 160.82 kN-m
2. Repeat problem 1 using axle loads of 20 kN and 45 kN on a wheel base of 4 meters crossing an 8-m
span.
3. A tractor with axle loads of 5 kN and 8 kN rolls across a 7-m span beam. It has a wheel base of 3
meters. Determine the maximum shear and moment developed in the beam.
Answer: Mmax = 15.89 kN-m
5 kN R 8 kN
4. Three equal wheel loads of 25 kN each, separated by 2 m between each load, roll over a 10-m span
beam. Compute the maximum shearing force and bending moment developed in the beam.
5. A trailer truck crossing a 12-m span has axle loads of 10 kN, 18 kN, and 26 kN separated respectively
by distances of 3 m and 5 m. Calculate the maximum shear and moment developed in the span.
Answer: Vmax = 39.8 kN
10 kN 18 kN 26 kN
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