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Beams - Fixed at Both Ends - Continuous and Point Loads

Stress, deflections and supporting loads.

Beam Fixed at Both Ends - Single Point Load

Beam fixed at both ends - single point load Bending Moment

MA = - F a b2/ L2(1a)

where

MA = moment at the fixed end A (Nm, lbf ft)

F = load (N, lbf )

MB = - F a2b / L2(1b)

where

MB = moment at the fixed end B (Nm, lbf ft)

MF = 2 F a2b2/ L3 (1c)

where

MF = moment at the point load (Nm, lbf ft)

Deflection

δF = F a3 b3 / (3 L3 E I)                                  (1d)

where

δF = deflection at point load (m, ft)

E = Modulus of Elasticity (Pa (N/m2), N/mm2, psi)

I = Area Moment of Inertia (m4, mm4, in4 )

Support Reactions

RA = F (3 a + b) b2/ L3 (1f)

where

RA = support force at fixed end A (N, lbf )

RB = F (a + 3 b) a2/ L3 (1g)

where

RB = support force at fixed end B  (N, lbf )

Beam Fixed at Both Ends - Uniform Continuous Distributed Load

Beam fixed at both ends - uniform load Bending Moment

MA = MB

= - q L2/ 12                                   (2a)

where

M = moments at the fixed ends  (Nm, lbf ft)

q = uniform load (N/m, lbf /ft)

M1 = q L2/ 24                         (2b)

where

M1 = moment at the center (Nm, lbf ft)

Deflection

δmax = q L4 / (384 E I)                                  (2c)

where

δmax = max deflection at center (m, ft)

E = Modulus of Elasticity (Pa (N/m2), N/mm2, psi)

I = Area Moment of Inertia (m4, mm4, in4 )

Support Reactions

RA = RB

=  q L / 2                               (2d)

where

R = support forces at the fixed ends  (N, lbf )

Beam Fixed at Both Ends - Uniform Declining Distributed Load

Beam fixed at both ends - uniform declining load Bending Moment

MA = - q L2/ 20                                  (3a)

where

MA = moments at the fixed end A  (Nm, lbf ft)

q = uniform declining load (N/m, lbf /ft)

MB = - q L2/ 30                                 (3b)

where

MB = moments at the fixed end B  (Nm, lbf ft)

M1 = q L2/ 46.6                         (3c)

where

M1 = moment at x = 0.475 L (Nm, lbf ft)

Deflection

δmax = q L4 / (764 E I)                                  (3d)

where

δmax = max deflection at x = 0.475 L (m, ft)

E = Modulus of Elasticity (Pa (N/m2), N/mm2, psi)

I = Area Moment of Inertia (m4, mm4, in4 )

δ1/2 = q L4 / (768 E I)                                  (3e)

where

δ1/2 = deflection at x = 0.5 L (m, ft)

Support Reactions

RA = 7 q L / 20                               (3f)

where

RA = support force at the fixed end A  (N, lbf )

RB =  3 q L / 20                               (3g)

where

RB = support force at the fixed end B  (N, lbf )

Beam Fixed at Both Ends - Partly Uniform Continuous Distributed Load

Beam fixed at both ends - partly uniform load Bending Moment

MA = - (q a2/ 6) (3 - 4 a / l + 1.5 (a / L)2)                                  (4a)

where

MA = moment at the fixed end A  (Nm, lbf ft)

q = partly uniform load (N/m, lbf /ft)

MB = - (q a2/ 3) (a / L - 0.75 (a / L)2)                                 (4b)

where

MB = moment at the fixed end B  (Nm, lbf ft)

Support Reactions

RA = q a (L - 0.5 a) / L - (MA - MB ) / L                              (4c)

where

RA = support force at the fixed end A  (N, lbf )

RB =  q a2/ (2 L) + (MA - MB ) / L                              (4d)

where

RB = support force at the fixed end B  (N, lbf )

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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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