# Beams - Supported at Both Ends - Continuous and Point Loads

The bending stress in a bending beam can be expressed as

* σ = y M / I (1) *

* where *

* σ = bending stress (Pa (N/m ^{2}), N/mm^{2}, psi) *

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

* M = bending moment (Nm, lb in) *

* I = moment of Inertia (m ^{4}, mm^{4}, in^{4} ) *

- 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

Distance y in a typical steel beam profile.

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 L^{2}/ 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 L^{2}/ (8 I) (2b) *

* where *

* σ _{ max } = maximum stress (Pa (N/m^{2}), N/mm^{2}, psi) *

* y _{ max } = distance to extreme point from neutral axis (m, mm, in) *

*1 N/m*^{2}= 1x10^{-6}N/mm^{2}= 1 Pa = 1.4504x10^{-4}psi*1 psi (lb/in*^{2}) = 144 psf (lb_{f}/ft^{2}) = 6,894.8 Pa (N/m^{2}) = 6.895x10^{-3}N/mm^{2}

Maximum ** deflection ** :

* δ _{ max } = 5 q L^{4} / (384 E I) (2c) *

* where *

* δ _{ max } = maximum deflection (m, mm, in) *

* E = Modulus of Elasticity (Pa (N/m ^{2}), N/mm^{2}, psi) *

Deflection in position x:

* δ _{ x } = q x (L^{3} - 2 L x^{2}+ x^{3} ) / (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:

* R _{1} = R_{2}*

* = 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 cm ^{4} (81960000 mm^{4} ) * and the modulus of elasticity for the steel used in the beam is

*200 GPa (200000 N/mm*. The height of the beam is

^{2})*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 mm^{4} )) *

* = 34.3 N/mm ^{2}*

* = 34.3 10 ^{6} N/m^{2}(Pa) *

* = 34.3 MPa *

The maximum deflection in the beam can be calculated

* δ _{ max } = 5 (6 N/mm) (5000 mm)^{4} / ((200000 N/mm^{2}) (81960000 mm^{4} ) 384) *

* = 2.98 mm *

#### Uniform Load Beam Calculator - Metric Units

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*1 mm*^{4}= 10^{-4}cm^{4}= 10^{-12}m^{4}*1 cm*^{4}= 10^{-8}m = 10^{4}mm*1 in*^{4}= 4.16x10^{5}mm^{4}= 41.6 cm^{4}*1 N/mm*^{2}= 10^{6}N/m^{2}(Pa)

#### Uniform Load Beam Calculator - Imperial Units

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#### 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 in ^{4} *, modulus of elasticity

*29000000 psi*, with uniform load

*100 lb/in*can be calculated as

* σ _{ max } = y _{ max } q L^{2}/ (8 I) *

* = (6.25 in) (100 lb/in) (100 in) ^{2}/ (8 (285 in^{4} )) *

* = 2741 (lb/in ^{2}, psi) *

The maximum deflection can be calculated as

δ_{ max }= 5 q L^{4}/ (E I 384)

= 5 (100 lb/in) (100 in)^{4}/ ((29000000 lb/in^{2}) (285 in^{4}) 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 L^{3} / (48 E I) (3c) *

Forces acting on the ends:

* R _{1} = R_{2}*

* = F / 2 (3d) *

#### Single Center Load Beam Calculator - Metric Units

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#### Single Center Load Beam Calculator - Imperial Units

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#### 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 in ^{4} *, 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 in ^{4} )) *

* = 5482 (lb/in ^{2}, psi) *

The maximum deflection can be calculated as

* δ _{ max } = F L^{3} / E I 48 *

* = (10000 lb) (100 in) ^{3} / ((29000000 lb/in^{2}) (285 in^{4} ) 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 a^{2}b^{2}/ (3 E I L) (4c) *

Forces acting on the ends:

* R _{1} = F b / L (4d) *

* R _{2}= 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 (3L^{2}- 4 a^{2}) / (24 E I) (5c) *

Forces acting on the ends:

* R _{1} = R_{2}*

* = 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 L^{3} / (20.22 E I) (6c) *

Forces acting on the ends:

* R _{1} = R_{2}*

* = 1.5 F (6d) *

## Related Topics

### • Beams and Columns

Deflection and stress in beams and columns, moment of inertia, section modulus and technical information.

### • Mechanics

The relationships between forces, acceleration, displacement, vectors, motion, momentum, energy of objects and more.

### • Statics

Forces acting on bodies at rest under equilibrium conditions - loads, forces and torque, beams and columns.

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