Beams  Fixed at Both Ends  Continuous and Point Loads
Stress, deflections and supporting 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
Bending Moment
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
Bending Moment
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
Bending Moment
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
Bending Moment
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

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