Beams - Fixed at Both Ends - Continuous and Point Loads
- 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
Beam Fixed at Both Ends - Single Point Load
M A = - F a b 2 / L 2 (1a)
where
M A = moment at the fixed end A (Nm, lb f ft)
F = load (N, lb f )
M B = - F a 2 b / L 2 (1b)
where
M B = moment at the fixed end B (Nm, lb f ft)
M F = 2 F a 2 b 2 / L 3 (1c)
where
M F = moment at the point load (Nm, lb f ft)
Deflection
δ F = F a 3 b 3 / (3 L 3 E I) (1d)
where
δ F = deflection at point load (m, ft)
E = Modulus of Elasticity (Pa (N/m 2 ), N/mm 2 , psi)
I = Area Moment of Inertia (m 4 , mm 4 , in 4 )
Support Reactions
R A = F (3 a + b) b 2 / L 3 (1f)
where
R A = support force at fixed end A (N, lb f )
R B = F (a + 3 b) a 2 / L 3 (1g)
where
R B = support force at fixed end B (N, lb f )
Beam Fixed at Both Ends - Uniform Continuous Distributed Load
M A = M B
= - q L 2 / 12 (2a)
where
M = moments at the fixed ends (Nm, lb f ft)
q = uniform load (N/m, lb f /ft)
M 1 = q L 2 / 24 (2b)
where
M 1 = moment at the center (Nm, lb f ft)
Deflection
δ max = q L 4 / (384 E I) (2c)
where
δ max = max deflection at center (m, ft)
E = Modulus of Elasticity (Pa (N/m 2 ), N/mm 2 , psi)
I = Area Moment of Inertia (m 4 , mm 4 , in 4 )
Support Reactions
R A = R B
= q L / 2 (2d)
where
R = support forces at the fixed ends (N, lb f )
Beam Fixed at Both Ends - Uniform Declining Distributed Load
M A = - q L 2 / 20 (3a)
where
M A = moments at the fixed end A (Nm, lb f ft)
q = uniform declining load (N/m, lb f /ft)
M B = - q L 2 / 30 (3b)
where
M B = moments at the fixed end B (Nm, lb f ft)
M 1 = q L 2 / 46.6 (3c)
where
M 1 = moment at x = 0.475 L (Nm, lb f ft)
Deflection
δ max = q L 4 / (764 E I) (3d)
where
δ max = max deflection at x = 0.475 L (m, ft)
E = Modulus of Elasticity (Pa (N/m 2 ), N/mm 2 , psi)
I = Area Moment of Inertia (m 4 , mm 4 , in 4 )
δ 1/2 = q L 4 / (768 E I) (3e)
where
δ 1/2 = deflection at x = 0.5 L (m, ft)
Support Reactions
R A = 7 q L / 20 (3f)
where
R A = support force at the fixed end A (N, lb f )
R B = 3 q L / 20 (3g)
where
R B = support force at the fixed end B (N, lb f )
Beam Fixed at Both Ends - Partly Uniform Continuous Distributed Load
M A = - (q a 2 / 6) (3 - 4 a / l + 1.5 (a / L) 2 ) (4a)
where
M A = moment at the fixed end A (Nm, lb f ft)
q = partly uniform load (N/m, lb f /ft)
M B = - (q a 2 / 3) (a / L - 0.75 (a / L) 2 ) (4b)
where
M B = moment at the fixed end B (Nm, lb f ft)
Support Reactions
R A = q a (L - 0.5 a) / L - (M A - M B ) / L (4c)
where
R A = support force at the fixed end A (N, lb f )
R B = q a 2 / (2 L) + (M A - M B ) / L (4d)
where
R B = support force at the fixed end B (N, lb f )
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Related Topics
• Beams and Columns
Deflection and stress, moment of inertia, section modulus and technical information of beams and columns.
• Mechanics
Forces, acceleration, displacement, vectors, motion, momentum, energy of objects and more.
• Statics
Loads - forces and torque, beams and columns.
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