# Simple Beam - Load Increasing Uniformly to One End

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 One End formulas

$$R_1 \;=\ V_1 \;=\; W \;/\;3$$

$$R_2 \;=\ V_2 \;=\; 2\;W \;/\;3$$

$$V_x \;=\ (W\;/\;3) - (W\;x^2\;/\;L^2)$$

$$M_{max} \; (at \; x = L\;/\; \sqrt{3}\; ) \;=\; 2\;W\;L \;/\; 9\; \sqrt{3}$$

$$M_x \;=\; (W \;x\;/\;3\;L^2) \; ( L^2 - x^2 )$$

$$\Delta_{max} \; ( \;at \; x = L\; \sqrt{1 - ( 8/15 )^{\frac{1}{2} } } \;) \;=\; 0.01304 \; ( W \;L^3\;/\; \lambda \;I )$$

$$\Delta_x \;=\; (W \;x\;/\; 180\; \lambda \;I \;L^2 ) \; ( 3\;x^4 - 10\;L^2\;x^2 + 7\;L^4 )$$

### S B Load Increasing Unif to One End - Solve for R1

$$\large{ R_1 = \frac{ \frac{w\;L}{2} }{3} }$$

 load per unit length, w span length, L

### S B Load Increasing Unif to One End - Solve for R2

$$\large{ R_2 = \frac{2\; \frac{w\;L}{2} }{3} }$$

 load per unit length, w span length, L

### S B Load Increasing Unif to One End - Solve for Vx

$$\large{ V_x = \frac{ \frac{w\;L}{2} }{3} - \frac{ \frac{w\;L}{2} \;x^2}{L^2} }$$

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

### S B Load Increasing Unif to One End - Solve for Mmax

$$\large{ M_{max} = \frac{ 2\; \frac{w\;L}{2} \;L }{ 9\; \sqrt{3} } }$$

 load per unit length, w span length, L

### S B Load Increasing Unif to One End - Solve for Mx

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

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

### S B Load Increasing Unif to One End - Solve for Δmax

$$\large{ \Delta_{max} = 0.01304 \; \frac{ \frac{w\;L}{2} \;L^3}{ \lambda \;I} }$$

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

### S B Load Increasing Unif to One End - Solve for Δx

$$\large{ \Delta_x = \frac{ \frac{w\;L}{2} \;x}{ 180\; \lambda \;I \;L^2 } \; \left( 3\;x^4 - 10\;L^2\;x^2 + 7\;L^4 \right) }$$

 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