# 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

$$\large{ R = V_{max} = \frac{W}{2} }$$

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

$$\large{ M_{max} \; \left(at \;center\right) = \frac{W \;L}{6} }$$

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

$$\large{ \Delta_{max} \; \left(at \;center\right) = \frac{W \;L^3} {60 \; \lambda \;I } }$$

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

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