Stress and Deflections in Beams

Beams and shafts - deflection and stress calculator

The calculator below can be used to calculate maximum stress and deflection of beams with one or uniform loads.

Beam Supported at Both Ends, Uniform Load

beam stress deflection uniform load

Maximum Stress

 

beam flange

Maximum stress in a beam with uniform load supported at both ends can be calculated as

σ = y q L2 / (8 I)         (1)

where

σ = maximum stress (Pa (N/m2), N/mm2, psi)

y = Perpendicular distance from to neutral axis X (m, mm, in)

q = uniform load per length unit (N/m, N/mm, lb/in)

L = length of beam (m, mm, in)

I = moment of Inertia (m4, mm4, in4)

  • 1 N/m2 = 1x10-6 N/mm2 = 1 Pa = 1.4504x10-4 psi
  • 1 psi (lb/in2) = 144 psf (lbf/ft2) = 6,894.8 Pa (N/m2) = 6.895x10-3 N/mm2

Maximum deflection can be expressed as

δ = 5 q L4 / (E I 384)         (2)

where

δ = maximum deflection (m, mm, in)

E = modulus of elasticity (Pa (N/m2), N/mm2, psi)

Note! - deflection is often a limiting factor in beam design. In some applications beams must be stronger than required by the maximum loads to avoid unacceptable deflections.

Metric Units

q - Uniform load (N/mm)

L - Length of Beam (mm)

I - Moment of Inertia (mm4)

E - Modulus of Elasticity (N/mm2))

y - Perpendicular distance from to neutral axis (mm)

  • 1 mm4 = 10-4 cm4 = 10-12 m4
  • 1 cm4 = 10-8 m = 104 mm
  • 1 in4 = 4.16x105 mm4 = 41.6 cm4
  • 1 N/mm2 = 106 N/m2 (Pa)

Imperial Units

q - Load (lb/in)

L - Length of Beam (in)

I - Moment of Inertia (in4)

E - Modulus of Elasticity (psi)

y - Perpendicular distance from to neutral axis X (in)

Example - Beam with Uniform Load, English Units

The maximum stress in a "W 12 x 35" Steel Wide Flange beam, 100 inches long, moment of inertia 285 in4, modulus of elasticity 29000000 psi, with uniform load 100 lb/in can be calculated as

σ = y q L2 / (8 I)

    = (6.25 in) (100 lb/in) (100 in)2 / (8 (285 in4))

    = 2741 (lb/in2, psi)

The maximum deflection can be calculated as

δ = 5 q L4 / (E I 384)

    = 5 (100 lb/in) (100 in)4 / ((29000000 lb/in2) (285 in4) 384)

    = 0.016 in

Beam Supported at Both Ends, Load at Center

beam stress deflection single load

Maximum Stress

Maximum stress in a beam with uniform load supported at both ends can be calculated as

σ = y F L / (4 I)         (3)

where

σ = maximum stress (Pa (N/m2), N/mm2, psi)

y = Perpendicular distance from to neutral axis (m, mm, in)

F = load (N, lb)

L = length of beam (m, mm, in)

I = moment of Inertia (m4,mm4, in4)

Maximum deflection can be expressed as

δ = F L3 / (E I 48)         (4)

where

δ = maximum deflection (m, mm, in)

E = modulus of elasticity (Pa (N/m2), N/mm2, psi)

Metric Units

F - Load (N)

L - Length of Beam (mm)

I - Moment of Inertia (mm4)

E - Modulus of Elasticity (N/mm2)

y - Perpendicular distance from to neutral axis (mm)

Imperial Units

F - Load (lb)

L - Length of Beam (in)

I - Moment of Inertia (in4)

E - Modulus of Elasticity (psi)

y - Perpendicular distance from to neutral axis (in)

Example - Beam with a Single Center Load

The maximum stress in a "W 12 x 35" Steel Wide Flange beam, 100 inches long, moment of inertia 285 in4, modulus of elasticity 29000000 psi, with a center load 10000 lb can be calculated like

σ = y F L / (4 I)

    = (6.25 in) (10000 lb) (100 in) / (4 (285 in4))

    = 5482 (lb/in2, psi)

The maximum deflection can be calculated as

δ = F L3 / E I 48

    = (10000 lb/in) (100 in)3 / ((29000000 lb/in2) (285 in4) 48)

    = 0.025 in

Some Typical Vertical Deflection Limits

  • total deflection : span/250
  • live load deflection : span/360
  • cantilevers : span/180
  • domestic timber floor joists : span/330 (max 14 mm)
  • brittle elements : span/500
  • crane girders : span/600

Related Topics

  • Mechanics - Kinematics, forces, vectors, motion, momentum, energy and the dynamics of objects
  • Beams and Columns - Deflection and stress, moment of inertia, section modulus and technical information of beams and columns
  • Statics - Loads - force and torque

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