# Simple Beam - Load Increasing Uniformly to Center

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 - Load Increasing Uniformly to Center formulas

$$R = V_{max} \;=\; W\;/\;2$$

$$V_x \; [ \; x < (L\;/\;2) \;] \;=\; (W\;/\;2\;L^2) \; ( L^2 - 4\;x^2 )$$

$$M_{max} \; (at \;center) \;=\; W \;L\;/\;6$$

$$M_x \; [\; x < (L\;/\;2) \;] \;=\; W\;x \; [\; (1/2) - (2\;x^2\;/\;3\;L^2) \;]$$

$$\Delta_{max} \; (at \;center) \;=\; W \;L^3 \;/\; 60 \; \lambda \;I$$

$$\Delta_x \; [\; x < (L\;/\;2) \;] \;=\; (W\; x \;/\;480\; \lambda \;I \;L^2) \; ( 5\;L^2 - 4\;x^2 )^2$$

### S B - Load Increasing Unif to Center - Solve for R

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

 load per unit length, w span length, L

### S B - Load Increasing Unif to Center - Solve for Vx

$$\large{ V_x = \frac{ \frac{w\;L}{2} }{2\;L^2} \; \left( L^2 - 4\;x^2 \right) }$$

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

### S B - Load Increasing Unif to Center - Solve for Mmax

$$\large{ M_{max} = \frac{ \frac{w\;L}{2} \;L}{6} }$$

 load per unit length, w span length, L

### S B - Load Increasing Unif to Center - Solve for Mx

$$\large{ M_x = \frac{w\;L}{2} \; x \; \left( \frac{1}{2} - \frac {2\;x^2}{3\;L^2} \right) }$$

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

### S B - Load Increasing Unif to Center - Solve for Δmax

$$\large{ \Delta_{max} = \frac{ \frac{w\;L}{2} \;L^3} {60 \; \lambda \;I } }$$

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

### S B - Load Increasing Unif to Center - Solve for Δx

$$\large{ \Delta_x = \frac{ \frac{w\;L}{2} \; x}{480\; \lambda \;I \;L^2} \; \left( 5\;L^2 - 4\;x^2 \right)^2 }$$

 w (load per unit length, w) L (span length, L) x (dist from reaction, x) lambda (modulus of elasticity, λ) I (second moment of area, I)

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 $$in$$ $$mm$$
$$W$$ = total load or $$w\;L\;/\;2$$ $$lbf$$ $$N$$
$$w$$ = highest load per unit length of UIL $$lbf\;/\;in$$ $$N\;/\;m$$
$$L$$ = 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