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:

@ x = L

Slope:

@ x = L

Shear:

V = +F

Moment:

M = −F (L − x)
Mmax = −FL @ x = 0
Cantilever, Intermediate Load Cantilever, Intermediate Load

Deflection:

( 0 ≤ x ≤ a )
( a ≤ x ≤ L )
@ x = L

Slope:

( 0 ≤ x ≤ a )
( a ≤ x ≤ L )

Shear:

V = +F ( 0 ≤ x ≤ a )
V = 0 ( a ≤ x ≤ L )

Moment:

M = −F (a − x) ( 0 ≤ x ≤ a )
M = 0 ( a ≤ x ≤ L )
Cantilever, Uniform Distributed Load Cantilever, Uniform Distributed Load

Deflection:

@ x = L

Slope:

@ x = L

Shear:

V = +w (L − x)
Vmax = +wL @ x = 0

Moment:

M = −w (L − x)2 / 2
Mmax = −wL2 / 2 @ x = 0
Cantilever, Triangular Distributed Load Cantilever, Triangular Distributed Load

Deflection:

@ x = L

Slope:

@ x = L

Shear:

Vmax = +w1L / 2 @ x = 0

Moment:

Mmax = −w1L2 / 6 @ x = 0
Cantilever, End Moment Simply Supported, Center Moment

Deflection:

@ x = L

Slope:

@ x = L

Shear:

V = 0

Moment:

M = −M0

Simply Supported Beams

Simply Supported, Intermediate Load Simply Supported, Intermediate Load

Deflection:

( 0 ≤ x ≤ a )

For a ≥ b:

@

Slope:

( 0 ≤ x ≤ a )
@ x = 0
@ x = L

Shear:

V1 = +Fb / L ( 0 ≤ x ≤ a )
V2 = −Fa / L ( a ≤ x ≤ L )

Moment:

Mmax = +Fab / L @ x = a
Simply Supported, Center Load Simply Supported, Center Load

Deflection:

( 0 ≤ x ≤ L/2 )
@ x = L/2

Slope:

( 0 ≤ x ≤ L/2 )
@ x = 0
@ x = L

Shear:

V1 = +F / 2 ( 0 ≤ x ≤ L/2 )
V2 = −F / 2 ( L/2 ≤ x ≤ L )

Moment:

Mmax = FL / 4 @ x = L/2
Simply Supported, 2 Loads at Equal Distances from Supports Simply Supported, 2 Loads at Equal Distances from Supports

Deflection:

( 0 ≤ x ≤ a )
( a ≤ x ≤ L − a )
@ x = L/2

Slope:

( 0 ≤ x ≤ a )
( a ≤ x ≤ L − a )
@ x = 0
@ x = L

Shear:

V1 = +F ( 0 ≤ x ≤ a )
V2 = −F ( L − a ≤ x ≤ L )

Moment:

Mmax = Fa ( a ≤ x ≤ L − a )
Simply Supported, Uniform Distributed Load Simply Supported, Uniform Distributed Load

Deflection:

@ x = L/2

Slope:

@ x = 0
@ x = L

Shear:

V = w (L/2 − x)
V1 = +wL / 2 @ x = 0
V2 = −wL / 2 @ x = L

Moment:

Mmax = wL2 / 8 @ x = L/2
Simply Supported, Moment at Each Support Simply Supported, Moment at Each Support

Deflection:

@ x = L/2

Slope:

@ x = 0
@ x = L

Shear:

V = 0

Moment:

M = M0
Simply Supported, Moment at One Support Simply Supported, Moment at One Support

Deflection:

x = L (1 − √3/3)

Slope:

@ x = 0
@ x = L

Shear:

V = −M0 / L

Moment:

Mmax = M0 @ x = 0
Simply Supported, Center Moment Simply Supported, Center Moment

Deflection:

( 0 ≤ x ≤ L/2 )

Slope:

( 0 ≤ x ≤ L/2 )
@ x = 0
@ x = L

Shear:

V = +M0 / L

Moment:

M = M0x / L ( 0 ≤ x ≤ L/2 )
Mmax = M0 / 2 @ x = L/2

Fixed-Fixed Beams

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

Deflection:

( 0 ≤ x ≤ L/2 )
@ x = L/2

Shear:

V1 = +F / 2 ( 0 ≤ x ≤ L/2 )
V2 = −F / 2 ( L/2 ≤ x ≤ L )

Moment:

M = F (4x − L) / 8 ( 0 ≤ x ≤ L/2 )
M1 = M3 = −FL / 8 @ x = 0 and x = L
M2 = +FL / 8 @ x = L/2
Fixed-Fixed, Uniform Distributed Load Fixed-Fixed, Uniform Distributed Load

Deflection:

@ x = L/2

Shear:

V = w (L/2 − x)
V1 = +wL / 2 @ x = 0
V2 = −wL / 2 @ x = L

Moment:

M = w (6Lx − 6x2 − L2) / 12
M1 = M3 = −wL2 / 12 @ x = 0 and x = L
M2 = −wL2 / 24 @ x = L/2

Check out our beam calculator based on the methodology described here.

  • Calculates stresses and deflections in straight beams
  • Builds shear and moment diagrams
  • Can specify any configuration of constraints, concentrated forces, and distributed forces

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.