# Solution to Problem 425 | Relationship Between Load, Shear, and Moment

**Problem 425**

Beam loaded as shown in Fig. P-425.

# Relationship Between Load, Shear, and Moment

The vertical shear at C in the figure shown in previous section (also shown to the right) is taken as

$V_C = (\Sigma F_v)_L = R_1 - wx$

where R_{1} = R_{2} = wL/2

$V_c = \dfrac{wL}{2} - wx$

The moment at C is

$M_C = (\Sigma M_C) = \dfrac{wL}{2}x - wx \left( \dfrac{x}{2} \right)$

$M_C = \dfrac{wLx}{2} - \dfrac{wx^2}{2}$

If we differentiate M with respect to x:

$\dfrac{dM}{dx} = \dfrac{wL}{2} \cdot \dfrac{dx}{dx} - \dfrac{w}{2} \left( 2x \cdot \dfrac{dx}{dx} \right)$

$\dfrac{dM}{dx} = \dfrac{wL}{2} - wx = \text{shear}$

thus,

# Solution to Problem 422 | Shear and Moment Equations

**Problem 422**

Write the shear and moment equations for the semicircular arch as shown in Fig. P-422 if (a) the load P is vertical as shown, and (b) the load is applied horizontally to the left at the top of the arch.

# Solution to Problem 421 | Shear and Moment Equations

**Problem 421**

Write the shear and moment equations as functions of the angle θ for the built-in arch shown in Fig. P-421.

# Solution to Problem 420 | Shear and Moment Diagrams

**Problem 420**

A total distributed load of 30 kips supported by a uniformly distributed reaction as shown in Fig. P-420.

# Solution to Problem 419 | Shear and Moment Diagrams

**Problem 419**

Beam loaded as shown in Fig. P-419.

# Solution to Problem 418 | Shear and Moment Diagrams

**Problem 418**

Cantilever beam loaded as shown in Fig. P-418.

# Solution to Problem 417 | Shear and Moment Diagrams

**Problem 417**

Beam carrying the triangular loading shown in Fig. P-417.

# Solution to Problem 416 | Shear and Moment Diagrams

**Problem 416**

Beam carrying uniformly varying load shown in Fig. P-416.