MATHalino

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 R₁ = R₂ = 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,