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Jan 6, 2005 · Figures 1 through 32 provide a series of shear and moment diagrams with accompanying formulas for design of beams under various static loading conditions. Shear and moment diagrams and formulas are excerpted from the Western Woods Use Book, 4th edition, and are provided herein as a courtesy of Western Wood Products Association.
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- Simply Supported Beam – Uniformly Distributed Line Load (UDL) Formulas
- Simply Supported Beam – Uniformly Distributed Load
- Simply Supported Beam – Point Load at Midspan
- Simply Supported Beam – Point Load Not at Midspan
- Simply Supported Beam – 2 Point Loads – Equally Spaced
- Simply Supported Beam – 3 Point Loads – Equally Spaced
- Simply Supported Beam – 2 Point Loads – Unequally Spaced
- Simply Supported Beam – One Side Triangular Line Load
- Simply Supported Beam – Double Triangular Line Load
Bending moment
M(x)=1/2⋅q⋅x⋅(l−x)
Max bending moment
Mmax=1/8⋅q⋅l2
Shear forces at supports
Va=−Vb=1/2⋅q⋅l
Max bending moment
Mmax=q⋅bl2⋅(c+b2)⋅(a+b2)⋅(l−b2) if x=a+b⋅(c+b2)l
Shear forces at supports
Va=q⋅b⋅c+b2l Vb=−q⋅b⋅a+b2l
Reaction forces
Ra=q⋅b⋅c+b2l Rb=q⋅b⋅a+b2l
Bending moment
M(x)=1/2⋅Q⋅xif x < l/2 M(x)=1/2⋅Q⋅(l−x)if x > l/2
Max bending moment
Mmax=1/4⋅Q⋅l
Shear forces at supports
Va=1/2⋅Q Vb=−1/2⋅Q
Bending moment
M(x)=Q⋅b⋅xlif x < a M(x)=Q⋅a⋅l−xlif x > a
Max bending moment
Mmax=Q⋅a⋅bl
Shear forces at supports
Va=Q⋅bl Vb=−Q⋅al
Bending moment
M(x)=Q⋅xif x < a
Max bending moment
Mmax=Q⋅a
Shear forces at supports
Va=Q Vb=−Q
Bending moment
M(x)=3/2Q⋅xif x < l/4
Max bending moment
Mmax=1/2⋅Q⋅l
Shear forces at supports
Va=3/2⋅Q Vb=−3/2⋅Q
Bending moment M1
M=Ql⋅(l–a+b)⋅amax Moment if a>b
Max bending moment M2
Mmax=Ql⋅(l–b+a)⋅b
Shear forces at supports
Va=Ql⋅(l–a+b) Vb=−Ql⋅(l–b+a)
Bending moment
M(x)=16⋅q⋅l⋅x⋅(1−(xl)2)
Max bending moment
Mmax=0.064⋅q⋅l2 at x=13⋅l
Shear forces at supports
Va=16⋅q⋅l Vb=−13⋅q⋅l
Bending moment
M(x)=14⋅q⋅l⋅x⋅(1−43⋅(xl)2)for 0 < x < l/2
Max bending moment
Mmax=112⋅q⋅l2
Shear forces at supports
Va=14⋅q⋅l Vb=−14⋅q⋅l
A summary of the most important formulas, equations & diagrams. This Structural Design Cheatsheet covers the engineering formulas and equations I use regularly as a structural engineer. It’s a “short” summary of many of the blog posts we published on structuralbasics.com. While this document covers many important formulas, be aware that ...
Beam Overhanging Both Supports – Unequal Overhangs – Uniformly Distributed Load. Beam Fixed at Both Ends – Uniformly Distributed Load. Beam Fixed at Both Ends – Concentrated Load at Center. Beam Fixed at Both Ends – Concentrated Load at Any Point. Continuous Beam – Two Equal Spans – Uniform Load on One Span.
Establish the M and x axes and plot the values of the moment at the ends of the beam. Shear and Moment Diagrams. – Since dM/dx = V, the slope of the moment diagram at any point is equal to the intensity of the shear at the point.
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Example: Draw the shear and moment diagrams for the following frame: by Superposition. We have learned how to construct a moment diagram from either writing the moment as a function of x or from the slope relationship with the shear diagram.
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This page provides a quick reference formula sheet for the calculation of stresses and deflections in beams. • Beam Analysis (Full Reference) • Strength of Materials. • Beam Calculator. Shear Force and Bending Moment.
