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Beams Fixed End

Reference data and engineering information about beams fixed end for mechanics applications.

beamsfixedend

Overview

Engineering reference data for Beams Fixed End in mechanics.

Key Formulas

Newton's Second Law

F=maF = ma

Force = mass × acceleration.

Work

W=FdcosθW = Fd\cos\theta

Work = force × displacement × cos(angle).

Kinetic Energy

Ek=12mv2E_k = \frac{1}{2}mv^2

Energy of motion.

Potential Energy

Ep=mghE_p = mgh

Gravitational potential energy.

Variables

SymbolDescriptionUnit
FFForceN
mmMasskg
aaAccelerationm/s²
vvVelocitym/s

References

Beam Fixed at One End and Supported at the Other: Load Case Analysis

This section details the specific formulas and key parameters for analyzing a beam that is fixed at one end (A) and simply supported at the other end (B), under various loading conditions.

Single Point Load (F)

  • Fixed End Moment (MA): The reaction moment at the fixed end is negative (hogging). MA=Fab(L+b)2L2M_A = - \frac{F a b (L + b)}{2 L^2}
  • Deflection at Load Point (δF): Maximum deflection occurs at or near the point of load application. δF=Fa3b2(3L+b)12L3EI\delta_F = \frac{F a^3 b^2 (3 L + b)}{12 L^3 E I}
  • Support Reactions: The fixed support (A) carries more load than the simple support (B) for most load positions. RA=Fb(3L2b2)2L3R_A = \frac{F b (3 L^2 - b^2)}{2 L^3} RB=Fa2(b+2L)2L3R_B = \frac{F a^2 (b + 2 L)}{2 L^3}

Continuous Uniformly Distributed Load (q)

  • Fixed End Moment (MA): MA=qL28M_A = - \frac{q L^2}{8}
  • Maximum Bending Moment (M1): Occurs at a distance of x = 0.625 L from the fixed end. M1=9qL2128M_1 = \frac{9 q L^2}{128}
  • Maximum Deflection (δmax): Occurs at x ≈ 0.579 L from the fixed end. δmax=qL4185EI\delta_{max} = \frac{q L^4}{185 E I}
  • Mid-span Deflection (δ1/2): δ1/2=qL4192EI\delta_{1/2} = \frac{q L^4}{192 E I}
  • Support Reactions: The fixed support (A) carries 5/8 of the total load. RA=5qL8R_A = \frac{5 q L}{8} RB=3qL8R_B = \frac{3 q L}{8}

Continuous Declining (Triangular) Load (q)

  • Fixed End Moment (MA): MA=qL215M_A = - \frac{q L^2}{15}
  • Maximum Bending Moment (M1): Occurs at x = 0.553 L from the fixed end. M1=qL233.6M_1 = \frac{q L^2}{33.6}
  • Maximum Deflection (δmax): Occurs at x ≈ 0.553 L. δmax=qL4419EI\delta_{max} = \frac{q L^4}{419 E I}
  • Mid-span Deflection (δ1/2): δ1/2=qL4427EI\delta_{1/2} = \frac{q L^4}{427 E I}
  • Support Reactions: RA=2qL5R_A = \frac{2 q L}{5} RB=qL10R_B = \frac{q L}{10}

Applied Moment (MB) at the Supported End

  • Fixed End Moment (MA): A moment applied at the simple support induces a moment at the fixed end with the same magnitude but opposite sign. MA=MB2M_A = -\frac{M_B}{2}
  • Maximum Deflection (δmax): Occurs at x = 2/3 L from the fixed end. δmax=MBL227EI\delta_{max} = \frac{M_B L^2}{27 E I}
  • Support Reactions: The reactions are equal in magnitude and opposite in direction, creating a couple. RA=3MB2LR_A = \frac{3 M_B}{2 L} RB=3MB2LR_B = -\frac{3 M_B}{2 L}

Single Point Load

For a beam fixed at end A and supported at end B with a single point load F applied at distance a from the fixed end and b from the supported end (L = a + b):

  • Bending Moment at Fixed End (A):
MA=Fab(L+b)2L2M_A = -\frac{F a b (L + b)}{2 L^2}
  • Bending Moment at Load Point:
MF=RBbM_F = R_B b
  • Deflection at Load Point:
δF=Fa3b2(3L+b)12L3EI\delta_F = \frac{F a^3 b^2 (3 L + b)}{12 L^3 E I}
  • Support Reactions:
    • At fixed end A:
RA=Fb(3L2b2)2L3R_A = \frac{F b (3 L^2 - b^2)}{2 L^3}
  • At support B:
RB=Fa2(b+2L)2L3R_B = \frac{F a^2 (b + 2 L)}{2 L^3}

Continuous Uniform Load

For a beam fixed at end A and supported at end B with a continuous uniform load q over the entire length L:

  • Bending Moment at Fixed End (A):
MA=qL28M_A = -\frac{q L^2}{8}
  • Maximum Bending Moment:
M1=9qL2128atx=0.625LM_1 = \frac{9 q L^2}{128} \quad \text{at} \quad x = 0.625 L
  • Maximum Deflection:
δmax=qL4185EIatx=0.579L\delta_{\text{max}} = \frac{q L^4}{185 E I} \quad \text{at} \quad x = 0.579 L
  • Deflection at Midspan (x = L/2):
δ1/2=qL4192EI\delta_{1/2} = \frac{q L^4}{192 E I}
  • Support Reactions:
    • At fixed end A:
RA=5qL8R_A = \frac{5 q L}{8}
  • At support B:
RB=3qL8R_B = \frac{3 q L}{8}

Continuous Declining Load

For a beam fixed at end A and supported at end B with a continuous declining load q from the fixed end to zero at the supported end:

  • Bending Moment at Fixed End (A):
MA=qL215M_A = -\frac{q L^2}{15}
  • Maximum Bending Moment:
M1=qL233.6atx=0.553LM_1 = \frac{q L^2}{33.6} \quad \text{at} \quad x = 0.553 L
  • Maximum Deflection:
δmax=qL4419EIatx=0.553L\delta_{\text{max}} = \frac{q L^4}{419 E I} \quad \text{at} \quad x = 0.553 L
  • Deflection at Midspan (x = L/2):
δ1/2=qL4427EI\delta_{1/2} = \frac{q L^4}{427 E I}
  • Support Reactions:
    • At fixed end A:
RA=2qL5R_A = \frac{2 q L}{5}
  • At support B:
RB=qL10R_B = \frac{q L}{10}

Moment at Supported End

For a beam fixed at end A and supported at end B with a moment MBM_B applied at the supported end B:

  • Bending Moment at Fixed End (A):
MA=MB2M_A = -\frac{M_B}{2}
  • Maximum Deflection:
δmax=MBL227EIatx=23L\delta_{\text{max}} = \frac{M_B L^2}{27 E I} \quad \text{at} \quad x = \frac{2}{3} L
  • Support Reactions:
    • At fixed end A:
RA=3MB2LR_A = \frac{3 M_B}{2 L}
  • At support B:
RB=3MB2LR_B = -\frac{3 M_B}{2 L}

Note: Negative reaction indicates direction opposite to the assumed positive direction.

Comparative Summary: Beam Fixed at One End and Supported at the Other

This section provides a summary table for quick reference across the different load cases analyzed on this page.

4 rows
Summary of key results for a beam fixed at end A and supported at end B under various load cases. All formulas use consistent variable definitions.
loadCase
Single Point Load (F)
Continuous Uniform Load (q)
Continuous Declining Load (q₀)
Moment at Supported End (Mᵦ)

Source: engineeringtoolbox.com

Structural Behavior Notes

The extracted data reveals important patterns in the structural response of a propped cantilever (fixed at one end, simply supported at the other).

Maximum Moment Location: For uniform and triangular distributed loads, the maximum bending moment does not occur at the fixed support. It shifts toward the simple support. For a continuous uniform load, it occurs at x=0.625Lx = 0.625L, and for a triangular load, it occurs at x=0.553Lx = 0.553L.

Maximum Deflection Location: Similarly, the point of maximum deflection is not at mid-span for these statically indeterminate beams. For a uniform load, maximum deflection occurs at x=0.579Lx = 0.579L, and for a triangular load, it also occurs at x=0.553Lx = 0.553L.

Reaction Distribution: The fixed support carries a larger share of the load than the simple support. For a uniform load, the fixed end reaction RAR_A is 58qL\frac{5}{8}qL (62.5% of total load), while RBR_B is 38qL\frac{3}{8}qL (37.5%). This ratio is even more pronounced for a triangular load (RAR_A is 80% of the total load). An applied moment at the simple support creates an equal and opposite reaction pair that is inversely proportional to the beam length.