# Beam Fixed at Both Ends - 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.

### beam fixed at both ends - Uniformly Distributed Load formulas

$$R \;=\; V \;=\; w\; L\;/\;2$$

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

$$M_{max} \; (at \;ends ) \;=\; w\; L^2\;/\;12$$

$$M_1 \; (at\; center ) \;=\; w\; L^2\;/\;24$$

$$M_x \;=\; (w\;/\;12) \; ( 6\;L\;x - L^2 - 6\;x^2 )$$

$$M_{max} \; (at \;center ) \;=\; w\; L^4\;/\;384\; \lambda\; I$$

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

$$x \; (points \;of \;contraflexure ) \;=\; (\sqrt{3} - 3 )\; L$$

### B F at B E - Uniformly Distributed Load - Solve for R

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

 load per unit length, w span length, L

### B F at B E - Uniformly Distributed Load - Solve for Vx

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

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

### B F at B E - Uniformly Distributed Load - Solve for MmaxE

$$\large{ M_{max} = \frac{w\; L^2}{12} }$$

 load per unit length, w span length, L

### B F at B E - Uniformly Distributed Load - Solve for M1

$$\large{ M_1 = \frac{w\; L^2}{24} }$$

 load per unit length, w span length, L

### B F at B E - Uniformly Distributed Load - Solve for Mx

$$\large{ M_x = \frac{w}{12} \; \left( 6\;L\;x - L^2 - 6\;x^2 \right) }$$

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

### B F at B E - Uniformly Distributed Load - Solve for MmaxC

$$\large{ M_{max} = \frac{w\; L^4}{384\; \lambda\; I} }$$

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

### B F at B E - Uniformly Distributed Load - Solve for Δx

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

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

### B F at B E - Uniformly Distributed Load - Solve for x

$$\large{ x = \left(\sqrt{3} - 3 \right) L }$$

 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