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

## Support loads, stress and deflections

### 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)

MB = - F a2 b / L2                              (1b)

where

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

MF = 2 F a2 b2 / 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

#### 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

#### 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

#### 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)

## 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 - force and torque, beams and columns

## Tag Search

• en: beams shafts stresses deflection load calculator modulus elasticity moment inertia

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## Citation

• Engineering ToolBox, (2004). Beams - Fixed at Both Ends - Continuous and Point Loads . [online] Available at: https://www.engineeringtoolbox.com/beams-fixed-both-ends-support-loads-deflection-d_809.html [Accessed Day Mo. Year].

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