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

## Supporting loads, stress and deflections.

The stress in a bending beam can be expressed as

*σ = y M / I (1) *

*where *

*σ = 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

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*and the modulus of elasticity for the steel used in the beam is

^{4})*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

*q - Uniform load (N/mm)*

*L - Length of Beam (mm)*

*I - Moment of Inertia (mm ^{4})*

*E - Modulus of Elasticity (N/mm ^{2})*

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

*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

*q - Load (lb/in)*

*L - Length of Beam (in)*

*I - Moment of Inertia (in ^{4})*

*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 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

*F - Load (N)*

*L - Length of Beam (mm)*

*I - Moment of Inertia (mm ^{4})*

*E - Modulus of Elasticity (N/mm ^{2})*

*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 (in ^{4})*

*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 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)*