The amount of deflection, in inches, $\displaystyle {x}$ inches from the left end of a 2 in x 4 in x 10 ft western red cedar beam freely supported at the ends with a uniformly distributed load of $\displaystyle {w}$ pounds per square inch is given by δ $\displaystyle =\frac{{{w}{x}}}{{{24}{E}{I}}}{\left({l}^{{3}}-{2}{l}{x}^{{2}}+{x}^{{3}}\right)}$.

• δ is the amount of deflection in inches, which is a positive value.
• $\displaystyle {l}$ is the length of the board in inches
• $\displaystyle {E}$ is the modulus of elasticity in lbs/i$\displaystyle {n}^{{2}}$
• $\displaystyle {I}$ is the moment of inertia in i$\displaystyle {n}^{{4}}$
• $\displaystyle {w}$ is the force in lbs/i$\displaystyle {n}^{{2}}$

δ is the amount of deflection, which is a positive value.



At 30 inches from the end of the beam the deflection is 0.2 inches. If the modulus of elasticity is 1,051,000 lbs/i$\displaystyle {n}^{{2}}$ and the momment of inertia is 2.5 i$\displaystyle {n}^{{4}}$. What is the force of the uniformly distributed load?

Select an answer in lbs per square in in^2

Numbers may not be realistic.