# Simple Beam - Uniformly Distributed Load

on . Posted in Structural Engineering

### diagram Symbols

• Bending moment diagram (BMD)  -  Used to determine the bending moment at a given point of a structural element.  The diagram can help determine the type, size, and material of a member in a structure so that a given set of loads can be supported without structural failure.
• Free body diagram (FBD)  -  Used to visualize the applied forces, moments, and resulting reactions on a structure in a given condition.
• Shear force diagram (SFD)  -  Used to determine the shear force at a given point of a structural element.  The diagram can help determine the type, size, and material of a member in a structure so that a given set of loads can be supported without structural failure.
• Uniformly distributed load (UDL)  -  A load that is distributed evenly across the entire length of the support area.

### Simple Beam - Uniformly Distributed Load formulas

$$R \;=\; V_{max} \;=\; w \; L\;/\;2$$

$$V_x \;=\; w \; [\; (L\;/\;2) - x \;]$$

$$M_{max} \; \left(at \;center \right) \;=\; w \; L^2\;/\;8$$

$$M_x \;=\; (w \; x\;/\;2) \; ( L - x )$$

$$\Delta_{max} \; \left(at \;center \right) \;=\; 5 \;w \;L^4\;/\;384\; \lambda \;I$$

$$\Delta_x \;=\; (w\; x\;/\;24\; \lambda \;I ) \; ( L^3 - 2\;L\;x^2 + x^3 )$$

### S B Uniformly Distributed Load - Solve for R

$$\large{ R = V_{max} = \frac{w \; L}{2} }$$

 load per unit length, w span length, L

### S B Uniformly Distributed Load - Solve for Vx

$$\large{ V_x = w \; \left( \frac{L}{2} - x \right) }$$

 load per unit length, w span length, L distance from reaction, x

### S B Uniformly Distributed Load - Solve for Mmax

$$\large{ M_{max} \; \left(at \;center \right) = \frac{w \; L^2}{8} }$$

 load per unit length, w span length, L

### S B Uniformly Distributed Load - Solve for Mx

$$\large{ M_x = \frac{w \; x}{2} \; \left( L - x \right) }$$

 load per unit length, w distance from reaction, x span length, L

### S B Uniformly Distributed Load - Solve for Δmax

$$\large{ \Delta_{max} \; \left(at \;center \right) = \frac{5 \;w \;L^4}{384\; \lambda \;I} }$$

 load per unit length, w span length, L modulus of elasticity, λ second moment of area, I

### S B Uniformly Distributed Load - Solve for Δx

$$\large{ \Delta_x = \frac{w\; x}{24\; \lambda \;I} \; \left( L^3 - 2\;L\;x^2 + x^3 \right) }$$

 load per unit length, w distance from reaction, x modulus of elasticity, λ second moment of area, I span length, L

Symbol English Metric
$$R$$ = reaction load at bearing point $$lbf$$ $$N$$
$$V$$ = maximum shear force $$lbf$$ $$N$$
$$M$$ = maximum bending moment $$lbf-in$$ $$N-mm$$
$$\Delta$$ = deflection or deformation $$in$$ $$mm$$
$$w$$ = load per unit length $$lbf\;/\;in$$ $$N\;/\;m$$
$$L$$ = span length of the bending member $$in$$ $$mm$$
$$x$$ = horizontal distance from reaction to point on beam $$in$$ $$mm$$
$$\lambda$$   (Greek symbol lambda) = modulus of elasticity $$lbf\;/\;in^2$$ $$Pa$$
$$I$$ = second moment of area (moment of inertia) $$in^4$$ $$mm^4$$

Tags: Beam Support