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Many structures can be approximated as a straight beam or as a collection of straight beams. For this reason, the analysis of stresses and deflections in a beam is an important and useful topic.

This section covers shear force and bending moment in beams, shear and moment diagrams, stresses in beams, and a table of common beam deflection formulas.

Contents

Shear Force and Bending Moment

To find the shear force and bending moment over the length of a beam, first solve for the external reactions at the boundary conditions. For example, the cantilever beam below has an applied force shown in red, and the reactions are shown in blue at the fixed boundary condition:

Cantilever Beam

After the external reactions have been solved for, take section cuts along the length of the beam and solve for the reactions at each section cut. An example section cut is shown in the figure below:

Shear and Moment in a Beam

When the beam is cut at the section, either side of the beam can be considered when solving for the reactions. The side that is selected does not affect the results, so choose whichever side is easiest. In the figure above, the side of the beam to the right of the section cut was selected. The reactions at the section cut are shown with blue arrows.

Sign Convention

The signs of the shear and moment are important. The sign is determined after a section cut is taken and the reactions are solved for the portion of the beam to one side of the cut. The shear force at the section cut is considered positive if it causes clockwise rotation of the selected beam section, and it is considered negative if it causes counter-clockwise rotation. The bending moment at the section cut is considered positive if it compresses the top of the beam and elongates the bottom of the beam (i.e. if it makes the beam "smile").

Based on this sign convention, the shear force at the section cut in the figure above is positive since it causes clockwise rotation of the selected section. The moment is negative since it compresses the bottom of the beam and elongates the top (i.e. it makes the beam "frown").

Sign Convention


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Shear and Moment Diagrams

The shear and bending moment throughout a beam are commonly expressed with diagrams. A shear diagram shows the shear along the length of the beam, and a moment diagram shows the bending moment along the length of the beam. These diagrams are typically shown stacked on top of one another, and the combination of these two diagrams is a shear-moment diagram. Shear-moment diagrams for some common end conditions and loading configurations are shown within the beam deflection tables at the end of this page. An example of a shear-moment diagram is shown in the following figure:

Generic Shear-Moment Diagram

General rules for drawing shear-moment diagrams are given in the table below:

Shear Diagram Moment Diagram
  • Point loads cause a vertical jump in the shear diagram. The direction of the jump is the same as the sign of the point load.
  • Uniform distributed loads result in a straight, sloped line on the shear diagram. The slope of the line is equal to the value of the distributed load.
  • The shear diagram is horizontal for distances along the beam with no applied load.
  • The shear at any point along the beam is equal to the slope of the moment at that same point: $$ V = { dM \over dx } $$
  • The moment diagram is a straight, sloped line for distances along the beam with no applied load. The slope of the line is equal to the value of the shear.
  • Uniform distributed loads result in a parabolic curve on the moment diagram.
  • The maximum/minimum values of moment occur where the shear line crosses zero.
  • The moment at any point along the beam is equal to the area under the shear diagram up to that point: $$ M = \int{V dx} $$

Bending Stresses in Beams

The bending moment, \(M\), along the length of the beam can be determined from the moment diagram. The bending moment at any location along the beam can then be used to calculate the bending stress over the beam's cross section at that location. The bending moment varies over the height of the cross section according to the flexure formula below:

$$ \sigma_{b} = - { My \over I_c } $$

where \(M\) is the bending moment at the location of interest along the beam's length, \(I_c\) is the centroidal moment of inertia of the beam's cross section, and \(y\) is the distance from the beam's neutral axis to the point of interest along the height of the cross section. The negative sign indicates that a positive moment will result in a compressive stress above the neutral axis.

The bending stress is zero at the beam's neutral axis, which is coincident with the centroid of the beam's cross section. The bending stress increases linearly away from the neutral axis until the maximum values at the extreme fibers at the top and bottom of the beam.

Bending Stress in Beam

The maximum bending stress is given by:

$$ \sigma_{b.max} = { Mc \over I_c } $$

where \(c\) is the centroidal distance of the cross section (the distance from the centroid to the extreme fiber).

If the beam is asymmetric about the neutral axis such that the distances from the neutral axis to the top and to the bottom of the beam are not equal, the maximum stress will occur at the farthest location from the neutral axis. In the figure below, the tensile stress at the top of the beam is larger than the compressive stress at the bottom.

Bending Stress Distribution

The section modulus of a cross section combines the centroidal moment of inertia, \(I_c\), and the centroidal distance, \(c\):

$$ S = { I_c \over c } $$

The benefit of the section modulus is that it characterizes the bending resistance of a cross section in a single term. The section modulus can be substituted into the flexure formula to calculate the maximum bending stress in a cross section:

$$ \sigma_{b.max} = { M \over S } $$

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Shear Stresses in Beams

The shear force, \(V\), along the length of the beam can be determined from the shear diagram. The shear force at any location along the beam can then be used to calculate the shear stress over the beam's cross section at that location. The average shear stress over the cross section is given by:

$$ \tau_{avg} = { V \over A } $$

The shear stress varies over the height of the cross section, as shown in the figure below:

Shear Stress Distribution

The shear stress is zero at the free surfaces (the top and bottom of the beam), and it is maximum at the centroid. The equation for shear stress at any point located a distance \(y_1\) from the centroid of the cross section is given by:

$$ \tau = { VQ \over I_c b } $$

where \(V\) is the shear force acting at the location of the cross section, \(I_c\) is the centroidal moment of inertia of the cross section, and \(b\) is the width of the cross section. These terms are all constants. The \(Q\) term is the first moment of the area bounded by the point of interest and the extreme fiber of the cross section:

$$ Q = \int_{y1}^{c} {y ~ dA} $$

Shear stresses for several common cross sections are discussed in the sections below.

Shear Stresses in Rectangular Sections

The distribution of shear stress along the height of a rectangular cross section is shown in the figure below:

Shear Stress in Rectangular Section

The first moment of area at any given point \(y_1\) along the height of the cross section is calculated by:

$$ Q = { b \over 2 } \left( { h^2 \over 4 } - y_1^2 \right) $$

The maximum value of \(Q\) occurs at the neutral axis of the beam (where \(y_1 = 0\)):

$$ Q_{max} = { b h^2 \over 8 } $$

The shear stress at any given point \(y_1\) along the height of the cross section is calculated by:

$$ \tau = { VQ \over I_c b } = {V \over 2 I_c} \left( {h^2 \over 4} - y_1^2 \right) $$

where \(I_c = { b h^3 \over 12 } \) is the centroidal moment of inertia of the cross section. The maximum shear stress occurs at the neutral axis of the beam and is calculated by:

$$ \tau_{max} = { 3V \over 2A } $$

where \( A = bh \) is the area of the cross section.

Note that the maximum shear stress in the cross section is 50% higher than the average stress \( V/A \).

Shear Stresses in Circular Sections

A circular cross section is shown in the figure below:

Shear Stress in Circular Section

The equations for shear stress in a beam were derived using the assumption that the shear stress along the width of the beam is constant. This assumption is valid at the centroid of a circular cross section, although it is not valid anywhere else. Therefore, while the distribution of shear stress along the height of the cross section cannot be readily determined, the maximum shear stress in the section (occuring at the centroid) can still be calculated. The maximum value of first moment, \(Q\), occurring at the centroid, is given by:

$$ Q_{max} = { 2 r^3 \over 3 } $$

The maximum shear stress is then calculated by:

$$ \tau_{max} = { V Q_{max} \over I_c b } = { 4V \over 3A } $$

where \(b = 2r\) is the diameter (width) of the cross section, \( I_c = { \pi r^4 \over 4 } \) is the centroidal moment of inertia, and \( A = \pi r^2 \) is the area of the cross section.

Shear Stresses in Circular Tube Sections

A circular tube cross section is shown in the figure below:

Shear Stress in Circular Tube

The maximum value of first moment, \(Q\), occurring at the centroid, is given by:

$$ Q_{max} = {2 \over 3} \left( r_o^3 - r_i^3 \right) $$

The maximum shear stress is then calculated by:

$$ \tau_{max} = { V Q_{max} \over I_c b } = { 4V \over 3A } \left( { r_o^2 + r_o r_i + r_i^2 \over r_o^2 + r_i^2 } \right) $$

where \(b = 2 (r_o - r_i)\) is the effective width of the cross section, \( I_c = {\pi \over 4} \left( r_o^4 - r_i^4 \right) \) is the centroidal moment of inertia, and \( A = \pi (r_o^2 - r_i^2) \) is the area of the cross section.

Shear Stresses in I-Beams

The distribution of shear stress along the web of an I-Beam is shown in the figure below:

Shear Stress in I-Beam

The equations for shear stress in a beam were derived using the assumption that the shear stress along the width of the beam is constant. This assumption is valid over the web of an I-Beam, but it is invalid for the flanges (specifically where the web intersects the flanges). However, the web of an I-Beam takes the vast majority of the shear force (approximately 90% - 98%, according to Gere), and so it can be conservatively assumed that the web carries all of the shear force.

The first moment of the area of the web of an I-Beam is given by:

$$ Q = {b \over 8} (h^2 - h_w^2) + {t_w \over 8} (h_w^2 - 4 y_1^2) $$

The shear stress along the web of the I-Beam is given by:

$$ \tau = {VQ \over I_c t_w} = {V \over 8 I_c t_w} \left[ b (h^2 - h_w^2) + t_w (h_w^2 - 4 y_1^2) \right] $$

where \(t_w\) is the web thickness and \(I_c\) is the centroidal moment of inertia of the I-Beam:

$$ I_c = {bh^3 \over 12} - {(b - t_w) h_w^3 \over 12} = {1 \over 12} (bh^3 - b h_w^3 + t_w h_w^3) $$

The maximum value of shear stress occurs at the neutral axis ( \( y_1 = 0 \) ), and the minimum value of shear stress in the web occurs at the outer fibers of the web where it intersects the flanges ( \( y_1 = \pm h_w / 2 \) ):

$$ \tau_{max} = {V \over 8 I_c t_w} (bh^2 - b h_w^2 + t_w h_w^2) $$ $$ \tau_{min} = {Vb \over 8 I_c t_w} (h^2 - h_w^2) $$

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Beam Deflection Tables

The tables below give equations for the deflection, slope, shear, and moment along straight beams for different end conditions and loadings. You can find comprehensive tables in references such as Gere, Lindeburg, and Shigley. However, the tables below cover most of the common cases.

Cantilever Beams

Cantilever, End Load Cantilever, End Load

Deflection:

$$ \delta = - { Fx^2 \over 6EI } \left( 3L - x \right) $$
$$ \delta_{max} = {F L^3 \over 3EI} $$ @ x = L

Slope:

$$ \theta = - { Fx \over 2EI } \left( 2L - x \right) $$
$$ \theta_{max} = {F L^2 \over 2EI} $$ @ x = L

Shear:

$$ V = +F $$

Moment:

$$ M = -F (L - x) $$
$$ M_{max} = -FL $$ @ x = 0
Cantilever, Intermediate Load Cantilever, Intermediate Load

Deflection:

$$ \delta = - { Fx^2 \over 6EI } \left( 3a - x \right) $$ \(( 0 \le x \le a )\)
$$ \delta = - { Fa^2 \over 6EI } \left( 3x - a \right) $$ \(( a \le x \le L )\)
$$ \delta_{max} = {F a^2 \over 6EI} \left( 3L - a \right) $$ @ x = L

Slope:

$$ \theta = - { Fx \over 2EI } \left( 2a - x \right) $$ \(( 0 \le x \le a )\)
$$ \theta = - { Fa^2 \over 2EI } $$ \(( a \le x \le L )\)

Shear:

$$ V = +F $$ \(( 0 \le x \le a )\)
$$ V = 0 $$ \(( a \le x \le L )\)

Moment:

$$ M = -F (a - x) $$ \(( 0 \le x \le a )\)
$$ M = 0 $$ \(( a \le x \le L )\)
Cantilever, Uniform Distributed Load Cantilever, Uniform Distributed Load

Deflection:

$$ \delta = - { w x^2 \over 24EI } \left( 6L^2 - 4Lx + x^2 \right) $$
$$ \delta_{max} = {w L^4 \over 8EI} $$ @ x = L

Slope:

$$ \theta = - {wx \over 6EI} \left( 3L^2 - 3Lx + x^2 \right) $$
$$ \theta_{max} = {w L^3 \over 6EI} $$ @ x = L

Shear:

$$ V = +w (L - x) $$
$$ V_{max} = +wL $$ @ x = 0

Moment:

$$ M = -{w (L-x)^2 \over 2 } $$
$$ M_{max} = -{w L^2 \over 2 } $$ @ x = 0
Cantilever, Triangular Distributed Load Cantilever, Triangular Distributed Load

Deflection:

$$ \delta = -{w_1 x^2 \over 120 LEI} \left( 10L^3 - 10 L^2 x + 5Lx^2 - x^3 \right) $$
$$ \delta_{max} = {w_1 L^4 \over 30EI} $$ @ x = L

Slope:

$$ \theta = - {w_1 L \over 24LEI} \left( 4L^3 - 6 L^2 x + 4Lx^2 - x^3 \right) $$
$$ \theta_{max} = {w_1 L^3 \over 24EI} $$ @ x = L

Shear:

$$ V_{max} = +{w_1 L \over 2} $$ @ x = 0

Moment:

$$ M_{max} = -{w_1 L^2 \over 6 } $$ @ x = 0
Cantilever, End Moment Simply Supported, Center Moment

Deflection:

$$ \delta = - {M_0 x^2 \over 2EI} $$
$$ \delta_{max} = {M_0 L^2 \over 2EI} $$ @ x = L

Slope:

$$ \theta = - {M_0 x \over EI} $$
$$ \theta_{max} = {M_0 L \over EI} $$ @ x = L

Shear:

$$ V = 0 $$

Moment:

$$ M = - M_0 $$

Simply Supported Beams

Simply Supported, Intermediate Load Simply Supported, Intermediate Load

Deflection:

$$ \delta = -{Fbx \over 6LEI} \left( L^2 - b^2 - x^2 \right) $$ \(( 0 \le x \le a )\)

For \(a \ge b \):

$$ \delta_{max} = {Fb (L^2 - b^2)^{3/2} \over 9 \sqrt{3} LEI} $$ @ \( x = \sqrt{ {L^2 - b^2} \over 3 } \)

Slope:

$$ \theta = -{Fb \over 6LEI} \left( L^2 - b^2 - 3x^2 \right) $$ \(( 0 \le x \le a )\)
$$ \theta_1 = -{Fab (L+b) \over 6LEI} $$ @ x = 0
$$ \theta_2 = {Fab (L+a) \over 6LEI} $$ @ x = L

Shear:

$$ V_1 = +{Fb \over L} $$ \(( 0 \le x \le a )\)
$$ V_2 = -{Fa \over L} $$ \(( a \le x \le L )\)

Moment:

$$ M_{max} = +{Fab \over L} $$ @ x = a
Simply Supported, Center Load Simply Supported, Center Load

Deflection:

$$ \delta = -{Fx \over 48EI} \left( 3L^2 - 4x^2 \right) $$ \(( 0 \le x \le L/2 )\)
$$ \delta_{max} = {F L^3 \over 48EI} $$ @ x = L/2

Slope:

$$ \theta = -{F \over 16EI} \left( L^2 - 4x^2 \right) $$ \(( 0 \le x \le L/2 )\)
$$ \theta_1 = -{FL^2 \over 16EI} $$ @ x = 0
$$ \theta_2 = +{FL^2 \over 16EI} $$ @ x = L

Shear:

$$ V_1 = +{F \over 2} $$ \(( 0 \le x \le L/2 )\)
$$ V_2 = -{F \over 2} $$ \(( L/2 \le x \le L )\)

Moment:

$$ M_{max} = {FL \over 4} $$ @ x = L/2
Simply Supported, 2 Loads at Equal Distances from Supports Simply Supported, 2 Loads at Equal Distances from Supports

Deflection:

$$ \delta = - { Fx \over 6EI } \left( 3aL - 3a^2 - x^2 \right) $$ \(( 0 \le x \le a )\)
$$ \delta = - { Fa \over 6EI } \left( 3Lx - 3x^2 - a^2 \right) $$ \(( a \le x \le L - a )\)
$$ \delta_{max} = {Fa \over 24EI} \left( 3L^2 - 4a^2 \right) $$ @ x = L/2

Slope:

$$ \theta = - { F \over 2EI } \left( aL - a^2 - x^2 \right) $$ \(( 0 \le x \le a )\)
$$ \theta = - { Fa \over 2EI } \left( L - 2x \right) $$ \(( a \le x \le L - a )\)
$$ \theta_1 = - { Fa (L - a) \over 2EI } $$ @ x = 0
$$ \theta_2 = + { Fa (L - a) \over 2EI } $$ @ x = L

Shear:

$$ V_1 = +F $$ \(( 0 \le x \le a )\)
$$ V_2 = -F $$ \(( L-a \le x \le L )\)

Moment:

$$ M_{max} = Fa $$ \(( a \le x \le L - a )\)
Simply Supported, Uniform Distributed Load Simply Supported, Uniform Distributed Load

Deflection:

$$ \delta = - { wx \over 24EI } \left( L^3 - 2Lx^2 + x^3 \right) $$
$$ \delta_{max} = {5 w L^4 \over 384EI} $$ @ x = L/2

Slope:

$$ \theta = - { w \over 24EI } \left( L^3 - 6Lx^2 + 4x^3 \right) $$
$$ \theta_1 = -{ wL^3 \over 24EI } $$ @ x = 0
$$ \theta_2 = +{ wL^3 \over 24EI } $$ @ x = L

Shear:

$$ V = w \left( {L \over 2} - x \right) $$
$$ V_1 = + {wL \over 2} $$ @ x = 0
$$ V_2 = - {wL \over 2} $$ @ x = L

Moment:

$$ M_{max} = {w L^2 \over 8} $$ @ x = L/2
Simply Supported, Moment at Each Support Simply Supported, Moment at Each Support

Deflection:

$$ \delta = - {M_0 x \over 2EI} \left( L - x \right) $$
$$ \delta_{max} = {M_0 L^2 \over 8EI} $$ @ x = L/2

Slope:

$$ \theta = - {M_0 \over 2EI} \left( L - 2x \right) $$
$$ \theta_1 = - {M_0 L \over 2EI} $$ @ x = 0
$$ \theta_2 = + {M_0 L \over 2EI} $$ @ x = L

Shear:

$$ V = 0 $$

Moment:

$$ M = M_0 $$
Simply Supported, Moment at One Support Simply Supported, Moment at One Support

Deflection:

$$ \delta = - {M_0 x \over 6LEI} \left( 2L^2 - 3Lx + x^2 \right) $$
$$ \delta_{max} = {M_0 L^2 \over 9 \sqrt{3} EI} $$ @ \( x = L \left(1 - {\sqrt{3} \over 3 } \right) \)

Slope:

$$ \theta = - {M_0 \over 6LEI} \left( 2L^2 - 6Lx + 3x^2 \right) $$
$$ \theta_1 = - {M_0 L \over 3EI} $$ @ x = 0
$$ \theta_2 = + {M_0 L \over 6EI} $$ @ x = L

Shear:

$$ V = - {M_0 \over L} $$

Moment:

$$ M_{max} = M_0 $$ @ x = 0
Simply Supported, Center Moment Simply Supported, Center Moment

Deflection:

$$ \delta = - {M_0 x \over 24LEI} \left( L^2 - 4x^2 \right) $$ \(( 0 \le x \le L/2 )\)

Slope:

$$ \theta = - {M_0 \over 24LEI} \left( L^2 - 12x^2 \right) $$ \(( 0 \le x \le L/2 )\)
$$ \theta_1 = - {M_0 L \over 24EI} $$ @ x = 0
$$ \theta_2 = - {M_0 L \over 24EI} $$ @ x = L

Shear:

$$ V = + {M_0 \over L} $$

Moment:

$$ M = {M_0 x \over L} $$ \(( 0 \le x \le L/2 )\)
$$ M_{max} = {M_0 \over 2} $$ @ x = L/2

Fixed-Fixed Beams

Fixed-Fixed, Center Load Fixed-Fixed, Center Load

Deflection:

$$ \delta = -{Fx^2 \over 48EI} \left( 3L - 4x \right) $$ \(( 0 \le x \le L/2 )\)
$$ \delta_{max} = {F L^3 \over 192EI} $$ @ x = L/2

Shear:

$$ V_1 = +{F \over 2} $$ \(( 0 \le x \le L/2 )\)
$$ V_2 = -{F \over 2} $$ \(( L/2 \le x \le L )\)

Moment:

$$ M = { F \over 8 } \left( 4x - L \right) $$ \(( 0 \le x \le L/2 )\)
$$ M_1 = M_3 = -{FL \over 8} $$ @ x = 0 and x = L
$$ M_2 = +{FL \over 8} $$ @ x = L/2
Fixed-Fixed, Uniform Distributed Load Fixed-Fixed, Uniform Distributed Load

Deflection:

$$ \delta = - { w x^2 \over 24EI } \left( L - x \right)^2 $$
$$ \delta_{max} = {w L^4 \over 384EI} $$ @ x = L/2

Shear:

$$ V = w \left( {L \over 2} - x \right) $$
$$ V_1 = + {wL \over 2} $$ @ x = 0
$$ V_2 = - {wL \over 2} $$ @ x = L

Moment:

$$ M = {w \over 12} \left( 6Lx - 6x^2 - L^2 \right) $$
$$ M_1 = M_3 = - {w L^2 \over 12} $$ @ x = 0 and x = L
$$ M_2 = - {w L^2 \over 24} $$ @ x = L/2


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References

  1. Budynas-Nisbett, "Shigley's Mechanical Engineering Design," 8th Ed.
  2. Gere, James M., "Mechanics of Materials," 6th Ed.
  3. Lindeburg, Michael R., "Mechanical Engineering Reference Manual for the PE Exam," 13th Ed.