Beam Calculator

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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.

For information on beam deflection, see our reference on stresses and deflections in beams.

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



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.