# Solution to Problem 689 | Beam Deflection by Method of Superposition

**Problem 689**

The beam shown in Fig. P-689 has a rectangular cross section 4 inches wide by 8 inches deep. Compute the value of P that will limit the midspan deflection to 0.5 inch. Use E = 1.5 × 10^{6} psi.

**Solution 689**

$M = 2000(2)(1) + 2P$

$M = 4000 + 2P$

Moment of inertia of beam section

$I = \dfrac{bd^3}{12} = \dfrac{4(8^3)}{12}$

$I = \dfrac{512}{3} \, \text{ in}^4$

The midspan deflection is a combination of deflection due to uniform load and two end moments. Use Case No. 8 and Cases No. 8, 11, and 12 to solve for the midspan deflection.

Type of Loading |
Midspan Deflection |

$\delta = \dfrac{5w_oL^4}{384EI}$ | |

$\delta = \dfrac{ML^2}{16EI}$ | |

$\delta = \dfrac{ML^2}{16EI}$ |

$\delta_{midspan} = \dfrac{5w_oL^4}{384EI} - 2\left[ \dfrac{ML^2}{16EI} \right]$

$0.5 = \dfrac{5(2000)(10^4)(12^3)}{384(1.5 \times 10^6)(\frac{512}{3})} - 2\left[ \dfrac{(4000 + 2P)(10^2)(12^3)}{16(1.5 \times 10^6)(\frac{512}{3})} \right]$

$0.5 = \dfrac{225}{128} - \dfrac{27(4000 + 2P)}{320\,000}$

$160\,000 = 562\,500 - 27(4000 + 2P)$

$27(4000 + 2P) = 402\,500$

$2P = 10\,907.40$

$P = 5453.7 \, \text{ lb}$ *answer*