Step 1

We are given a cantilever beam which is fixed at point A and carries a uniform varying load and a uniformly distributed load.

We are asked to draw shear and moment diagrams of the beam.

 
Step 2

Make a section (1 – 1) at an arbitrary distance $x$ from the free end. Consider the segment of the beam left of the section.

Using similarity of triangles, calculate the load due to uniformly varying load at section (1 – 1):

\[\begin{array}{c} \frac{w}{x} = \frac{{2\;{\rm{kip}}/{\rm{ft}}}}{{15\;{\rm{ft}}}}\\ w = \left( {\frac{{2\;{\rm{kip}}/{\rm{ft}}}}{{15\;{\rm{ft}}}}} \right)x \end{array}\]

Let $V$ is the shear force acting downward at section (1 – 1) and $M$ is the moment in a counter-clockwise direction at section (1 – 1).

Apply equilibrium equation of forces in the vertical direction:

\[\begin{array}{c} \sum {{F_y}} = 0\\ 1\;{\rm{kip}}/{\rm{ft}} \times x - \frac{1}{2} \times w \times x - V = 0\\ V = 1\;{\rm{kip}}/{\rm{ft}} \times x - \frac{1}{2} \times w \times x \end{array}\]

Substitute the value of $w$ in the above equation:

\[V = 1\;{\rm{kip}}/{\rm{ft}} \times x - \frac{1}{2} \times \left( {\left( {\frac{{2\;{\rm{kip}}/{\rm{ft}}}}{{15\;{\rm{ft}}}}} \right)x} \right) \times x\]   ......(1)

Substitute $x = 0\;{\rm{ft}}$ in equation (1):

\[\begin{array}{c} {V_{x = 0}} = 1\;{\rm{kip}}/{\rm{ft}} \times \left( {0\;{\rm{ft}}} \right) - \frac{1}{2} \times \left( {\left( {\frac{{2\;{\rm{kip}}/{\rm{ft}}}}{{15\;{\rm{ft}}}}} \right)\left( {0\;{\rm{ft}}} \right)} \right) \times \left( {0\;{\rm{ft}}} \right)\\ = 0\;{\rm{kip}} \end{array}\]

Substitute $x = 7.5\;{\rm{ft}}$ in equation (1):

\[\begin{array}{c} {V_{x = 7.5}} = 1\;{\rm{kip}}/{\rm{ft}} \times \left( {7.5\;{\rm{ft}}} \right) - \frac{1}{2} \times \left( {\left( {\frac{{2\;{\rm{kip}}/{\rm{ft}}}}{{15\;{\rm{ft}}}}} \right)\left( {7.5\;{\rm{ft}}} \right)} \right) \times \left( {7.5\;{\rm{ft}}} \right)\\ = 3.75\;{\rm{kip}} \end{array}\]

Substitute $x = 15\;{\rm{ft}}$ in equation (1):

\[\begin{array}{c} {V_{x = 15}} = 1\;{\rm{kip}}/{\rm{ft}} \times \left( {15\;{\rm{ft}}} \right) - \frac{1}{2} \times \left( {\left( {\frac{{2\;{\rm{kip}}/{\rm{ft}}}}{{15\;{\rm{ft}}}}} \right)\left( {15\;{\rm{ft}}} \right)} \right) \times \left( {15\;{\rm{ft}}} \right)\\ = 0\;{\rm{kip}} \end{array}\]
 
Step 3

Apply equilibrium equation for the moment about the section:

\[\begin{array}{c} \sum M = 0\\ M + \left( {\frac{1}{2} \times w \times x} \right) \times \frac{x}{3} - \left( {1\;{\rm{kip}}/{\rm{ft}} \times x} \right) \times \frac{x}{2} = 0\\ M = \left( {1\;{\rm{kip}}/{\rm{ft}} \times x} \right) \times \frac{x}{2} - \left( {\frac{1}{2} \times w \times x} \right) \times \frac{x}{3}\\ M = 1\;{\rm{kip}}/{\rm{ft}} \times \frac{{{x^2}}}{2} - \frac{{{x^2}}}{6} \times w \end{array}\]

Substitute the value of $w$ in the above equation:

\[M = 1\;{\rm{kip}}/{\rm{ft}} \times \frac{{{x^2}}}{2} - \frac{{{x^2}}}{6} \times \left( {\frac{{2\;{\rm{kip}}/{\rm{ft}}}}{{15\;{\rm{ft}}}}} \right)x\]   ......(2)

Substitute $x = 0\;{\rm{ft}}$ in equation (2):

\[\begin{array}{c} {M_{x = 0}} = 1\;{\rm{kip}}/{\rm{ft}} \times \frac{{{{\left( {0\;{\rm{ft}}} \right)}^2}}}{2} - \frac{{{{\left( {0\;{\rm{ft}}} \right)}^2}}}{6} \times \left( {\frac{{2\;{\rm{kip}}/{\rm{ft}}}}{{15\;{\rm{ft}}}}} \right)\left( {0\;{\rm{ft}}} \right)\\ = 0\;{\rm{kip}} \cdot {\rm{ft}} \end{array}\]

Substitute $x = 7.5\;{\rm{ft}}$ in equation (2):

\[\begin{array}{c} {M_{x = 7.5}} = 1\;{\rm{kip}}/{\rm{ft}} \times \frac{{{{\left( {7.5\;{\rm{ft}}} \right)}^2}}}{2} - \frac{{{{\left( {7.5\;{\rm{ft}}} \right)}^2}}}{6} \times \left( {\frac{{2\;{\rm{kip}}/{\rm{ft}}}}{{15\;{\rm{ft}}}}} \right)\left( {7.5\;{\rm{ft}}} \right)\\ = 18.75\;{\rm{kip}} \cdot {\rm{ft}} \end{array}\]

Substitute $x = 15\;{\rm{ft}}$ in equation (2):

\[\begin{array}{c} {M_{x = 15}} = 1\;{\rm{kip}}/{\rm{ft}} \times \frac{{{{\left( {15\;{\rm{ft}}} \right)}^2}}}{2} - \frac{{{{\left( {15\;{\rm{ft}}} \right)}^2}}}{6} \times \left( {\frac{{2\;{\rm{kip}}/{\rm{ft}}}}{{15\;{\rm{ft}}}}} \right)\left( {15\;{\rm{ft}}} \right)\\ = 37.5\;{\rm{kip}} \cdot {\rm{ft}} \end{array}\]

Using the above-calculated values of shear force and moment, draw the shear force and moment diagram of the beam.