Beams  Supported at Both Ends  Continuous and Point Loads
Supporting loads, stress and deflections.
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^{2}) . 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 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
 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
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
Single Center Load Beam Calculator  Imperial Units
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

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Deflection and stress in beams and columns, moment of inertia, section modulus and technical information. 
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The relationships between forces, acceleration, displacement, vectors, motion, momentum, energy of objects and more. 
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Forces acting on bodies at rest under equilibrium conditions  loads, forces and torque, beams and columns.
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