Question

Part D - Draw the Shear Diagram
Draw the shear diagram for this beam.
Begin by placing the lines of discontinuity. Place the appropriate function between the lines of discontinuity, ensuring the endpoints have the correct values.

Draw the Shear Diagram Draw the shear diagram for this beam. Begin by placing the lines of discontinuity. Place the appropriate function between the lines of discontinuity, ensuring the endpoints have the correct values.

Answer 1

Concepts and reason

The concepts required to solve the given problem are static equilibrium and moment of forces.

Free Body Diagram (FBD): It is a method to analyze the external forces and reactions acting on a given body by sketching the direction of the forces on the body. The free body diagram is sketched after all the other adjoining bodies have been removed. Generally, the direction of a force is assumed if the direction is not predefined. Upon continuing the problem, if the value of such a force is obtained as negative, then it can be said that the force acts in the opposite direction to which it was assumed.

Static equilibrium: Static equilibrium for a system of forces exists when the algebraic sum of all the forces along a particular direction is zero and the moment of all the forces about any point is zero.

Moment of force: Moment is the turning effect of force and is usually defined with respect to a fixed reference point. The evaluation of moment is done by multiplying the magnitude of the force and the perpendicular distance of force from a reference point. Moment about a point not only evaluates the magnitude but also the direction in which the given arrangement tends to rotate.

Concept of the graphical representation of the shear force variation on the beam can be

used to indicate the variation of the shear force across the beam.

Initially, apply static equilibrium of the forces to check whether the given system is in static equilibrium or not. Then, draw free body diagrams by considering at point between AB and BC. Determine the shear force acting at each section and finally draw the shear force diagram.

Fundamentals

Equation for the static equilibrium are as follows,

$\begin{array}{l}\\\sum {{F_V}} = 0\\\\\sum {{F_H}} = 0\\\\\sum M = 0\\\end{array}$

Here, ${F_V}$ represent the forces in the vertical direction, ${F_H}$ represent the force in the horizontal direction, $M$ represents the moment of the forces.

Sign convention for forces: The forces that are acting in upward direction are taken as positive and forces acting in downward direction are taken as negative.

Sign convention for moment: Consider the moment in clockwise direction as positive and the moments in counterclockwise direction as negative.

Draw the free body diagram of the beam.

As it is obvious from the free body diagram that all the forces on the beam are in equilibrium, it can be confirmed by applying the equilibrium equation in the vertical direction,

$\begin{array}{c}\\\sum {{F_V}} = \left( {5{\rm{ kips}}} \right) - \left( {15{\rm{ kips}}} \right) + \left( {10{\rm{ kips}}} \right)\\\\ = 0\\\end{array}$

Now taking moment of all the forces about the left most point (point of action of force ) of the beam.

$\begin{array}{c}\\\sum {{M_A}} = \left( {15{\rm{ kips}}} \right)\left( {14{\rm{ in}}} \right) - \left( {10{\rm{ kips}}} \right)\left( {21{\rm{ in}}} \right)\\\\ = 210 - 210\\\\ = 0\\\end{array}$

There is no net force in the vertical direction and also the moment about a point of all the forces is also zero. It can be inferred that the beam is in static equilibrium under the application of the given forces.

Consider a section between points A and B (very close to point B), at a distance of $x$ $\left( {0{\rm{ units}} \le x \le 14{\rm{ units}}} \right)$ from point A. Draw the free body diagram as,

Here, ${V_x}$ is the shear force at the considered section, ${N_x}$ is the horizontal reaction at that section and ${M_x}$ is the moment reaction at that section.

Apply vertical equilibrium condition to obtain the shear force at that section $\left( {{V_x}} \right)$ .

$\begin{array}{l}\\\sum {{F_y} = 0} \\\\5{\rm{ kips}} - {V_x} = 0\\\\{V_x} = 5{\rm{ kips}}\\\end{array}$

Calculate the shear force at point A:

Substitute $0{\rm{ unit}}$ for x.

${V_A} = 5{\rm{ kips}}$

Calculate the shear force at point B.

Substitute $14{\rm{ unit}}$ for x.

${V_B} = 5{\rm{ kips}}$

So, the shear force at point A, point B and at any point between A and B is $5{\rm{ kips}}$ .

Now consider a section to point between B and C (very close to point C), at a distance of $x'$ $\left( {{\rm{14 units}} \le x < 21{\rm{ units}}} \right)$ from point A. Draw the free body diagram as,

Here, ${V_{x'}}$ is the shear force at the considered section, ${N_{x'}}$ is the horizontal reaction at that section and ${M_{x'}}$ is the moment reaction at that section.

Apply vertical equilibrium condition to obtain the shear force at that section $\left( {{V_{x'}}} \right)$ .

$\begin{array}{l}\\\sum {{F_y} = 0} \\\\5{\rm{ kips}} - 15{\rm{ kips}} - {V_{x'}} = 0\\\\{V_{x'}} = - 10{\rm{ kips}}\\\end{array}$

Calculate the shear force at point B, by substituting $14{\rm{ unit}}$ for $x'$ .

${V_B} = - 10{\rm{ kips}}$

Calculate the shear force at point C, by substituting $21{\rm{ unit}}$ for $x'$ .

${V_C} = - 10{\rm{ kips}}$

So, the shear force at point B, point C and at any point between B and C is $- 10{\rm{ kips}}$ .

But in the previous step it was obtained that shear force at point B is $5{\rm{ kips}}$ . This shows that the shear force changes it sign and magnitude at point B. So, obtain the final shear force diagram as,

Ans:

Shear force diagram along with the free body diagram of the beam is,