Beams - Supported at Both Ends - Continuous and Point Loads

The stress in a bending beam can be expressed as

σ = y M / I                                     (1)            

where

σ = stress (Pa (N/m²), N/mm², psi)

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

M = bending moment (Nm, lb in)

I = moment of Inertia (m⁴, mm⁴, in⁴)

• Beams - Supported at Both Ends - Continuous and Point Loads
• Beams - Fixed at One End and Supported at the Other - Continuous and Point Loads
• Beams - Fixed at Both Ends - Continuous and Point Loads

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

Beam Supported at Both Ends - Uniform Continuous Distributed Load

The moment in a beam with uniform load supported at both ends in position x can be expressed as

M_x = q x (L - x) / 2                                          (2)

where

M_x = moment in position x (Nm, lb in)

x = distance from end (m, mm, in)

The maximum moment is at the center of the beam at distance L/2 and can be expressed as

M_max = q L² / 8                                          (2a)

where

M_max = maximum moment (Nm, lb in)

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

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

Maximum Stress

 

Equation 1 and 2a can be combined to express maximum stress in a beam with uniform load supported at both ends at distance L/2 as

σ_max = y_max q L² / (8 I)                                     (2b)

where

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

y_max = distance to extreme point from neutral axis (m, mm, in)

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

• Ulimate tensile strength for some common materials

Maximum deflection:

δ_max = 5 q L⁴ / (384 E I)                                     (2c)

where

δ_max = maximum deflection (m, mm, in)

E = Modulus of Elasticity (Pa (N/m²), N/mm², psi)

Deflection in position x:

δ_x = q x (L³ - 2 L x² + x³)  / (24 E I)                                     (2d)

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

Forces acting on the ends:

R₁ = R₂

     = q L / 2                                      (2e)

where 

R = reaction force (N, lb)   

Example - Beam with Uniform Load, Metric Units

A UB 305 x 127 x 42 beam with length 5000 mm carries a uniform load of 6 N/mm. The moment of inertia for the beam is 8196 cm⁴ (81960000 mm⁴) and the modulus of elasticity for the steel used in the beam is 200 GPa (200000 N/mm²). The height of the beam is 300 mm (the distance of the extreme point to the neutral axis is 150 mm). 

The maximum stress in the beam can be calculated

σ_max = (150 mm) (6 N/mm) (5000 mm)² / (8 (81960000 mm⁴))

  = 34.3 N/mm²

  = 34.3 10⁶ N/m² (Pa)

  = 34.3 MPa

The maximum deflection in the beam can be calculated

δ_max = 5 (6 N/mm) (5000 mm)⁴ / ((200000 N/mm²) (81960000 mm⁴) 384)

   = 2.98 mm

Uniform Load Beam Calculator - Metric Units

q - Uniform load (N/mm)

L - Length of Beam (mm)

I - Moment of Inertia (mm⁴)

E - Modulus of Elasticity (N/mm²)

y - Distance of extreme point off 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)

Uniform Load Beam Calculator - Imperial Units

q - Load (lb/in)

L - Length of Beam (in)

I - Moment of Inertia (in⁴)

E - Modulus of Elasticity (psi)

y - Distance of extreme point off neutral axis(in)

Example - Beam with Uniform Load, Imperial 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_max = y_max 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_max = 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

Maximum moment in a beam with center load supported at both ends:

M_max = F L / 4                                          (3a)

Maximum Stress

Maximum stress in a beam with single center load supported at both ends:

σ_max = y_max F L / (4 I)                                    (3b)

where

F = load (N, lb)

Maximum deflection can be expressed as

δ_max = F L³ / (48 E I)                                  (3c)

Forces acting on the ends:

R₁ = R₂

     = F / 2                                      (3d)

Single Center Load Beam Calculator - Metric Units

F - Load (N)

L - Length of Beam (mm)

I - Moment of Inertia (mm⁴)

E - Modulus of Elasticity (N/mm²)

y - Distance of extreme point off neutral axis (mm)

Single Center Load Beam Calculator - Imperial Units

F - Load (lb)

L - Length of Beam (in)

I - Moment of Inertia (in⁴)

E - Modulus of Elasticity (psi)

y - Distance of extreme point off 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_max = y_max 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_max = F L³ / E I 48

    = (10000 lb) (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

Beam Supported at Both Ends - Eccentric Load

Maximum moment in a beam with single eccentric load at point of load:

M_max = F a b / L                                          (4a)

Maximum Stress

Maximum stress in a beam with single center load supported at both ends:

σ_max = y_max F a b / (L I)                                    (4b)

Maximum deflection at point of load can be expressed as

δ_F = F a² b² / (3 E I L)                                  (4c)

Forces acting on the ends:

R₁ = F b / L                                 (4d)

R₂ = F a / L                                 (4e)

Beam Supported at Both Ends - Two Eccentric Loads

Maximum moment (between loads) in a beam with two eccentric loads:

M_max = F a                                          (5a)

Maximum Stress

Maximum stress in a beam with two eccentric loads supported at both ends:

σ_max = y_max F a / I                                    (5b)

Maximum deflection at point of load can be expressed as

δ_F = F a (3L² - 4 a²) / (24 E I)                                  (5c)

Forces acting on the ends:

R₁ = R₂

     = F                                  (5d)

Insert beams to your Sketchup model with the Engineering ToolBox Sketchup Extension

Beam Supported at Both Ends - Three Point Loads

Maximum moment (between loads) in a beam with three point loads:

M_max = F L / 2                                          (6a)

Maximum Stress

Maximum stress in a beam with three point loads supported at both ends:

σ_max = y_max F L / (2 I)                                    (6b)

Maximum deflection at the center of the beam can be expressed as

δ_F = F L³ / (20.22 E I)                               (6c)

Forces acting on the ends:

R₁ = R₂

     = 1.5 F                                  (6d)

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