# rectangular load

## Solution to Problem 681 | Midspan Deflection

**Problem 681**

Show that the midspan value of EIδ is (w_{o}b/48)(L^{3} - 2Lb^{2} + b^{3}) for the beam in part (a) of Fig. P-681. Then use this result to find the midspan EIδ of the loading in part (b) by assuming the loading to exceed over two separate intervals that start from midspan and adding the results.

## Solution to Problem 675 | Midspan Deflection

**Problem 675**

Repeat Prob. 674 for the overhanging beam shown in Fig. P-675.

## Solution to Problem 666 | Deflections in Simply Supported Beams

**Problem 666**

Determine the value of EIδ at the right end of the overhanging beam shown in Fig. P-666.

## Solution to Problem 665 | Deflections in Simply Supported Beams

**Problem 665**

Replace the concentrated load in Prob. 664 by a uniformly distributed load of intensity w_{o} acting over the middle half of the beam. Find the maximum deflection.

## Resultant of Parallel Force System

**Coplanar Parallel Force System**

Parallel forces can be in the same or in opposite directions. The sign of the direction can be chosen arbitrarily, meaning, taking one direction as positive makes the opposite direction negative. The complete definition of the resultant is according to its magnitude, direction, and line of action.

## Solution to Problem 632 | Moment Diagrams by Parts

**Problem 632**

For the beam loaded as shown in Fig. P-632, compute the value of (Area_{AB}) barred(X)_{A}. From this result, is the tangent drawn to the elastic curve at B directed up or down to the right? (Hint: Refer to the deviation equations and rules of sign.)

## Solution to Problem 631 | Moment Diagrams by Parts

**Problem 631**

Determine the value of the couple M for the beam loaded as shown in Fig. P-631 so that the moment of area about A of the M diagram between A and B will be zero. What is the physical significance of this result?