Question

A)

Draw the shear diagram for the beam.

Begin by placing vertical lines. Place the appropriate function between the vertical lines, ensuring the endpoints have the correct values. NOTE - You should not draw an “extra” vertical line at the location of applied moment.

Part A Draw the shear diagram for the beam Begin by placing vertical lines. Place the appropriate function between the vertical lines, ensuring the endpoints have the correct values. NOTE- You should not draw an "extra" vertical line at the location of applied moment. add vertical line on delete add segment ▼ resethelp 2 kip/ft 30 kip ft 5 ft 5 ft 5 ft ki ft 10 15 -6 -10

B)

Draw the moment diagram for the beam.

Begin by placing vertical lines. Then click on "add segment" button to add functions between the lines. NOTE - The curve you choose from the drop-down is only a pictorial representation of a real quadratic/cubic curve. The equation of this curve is not mathematically equivalent to the correct answer. Consequently, slopes at discontinuities and intercepts with the x-axis (if any) are not accurate.

Answer 1

Concepts and reason

Equilibrium of a rigid body:

An object is said to be in equilibrium when the sum of external forces and couples are zero.

For a rigid body to be in equilibrium in three dimensions, the sum of external forces acting along , and directions have to zero.

For a rigid body to be in equilibrium in three dimensions, the sum of external couples about any point should be zero.

Normal force:

It is the force component acting perpendicular to the contact surface or the cross sectional area of a body.

Shear force:

When force acting on a body pushes two parts of the body in opposite direction then the force is referred as shear force. Shear force on an object acts normal to its cross section.

Bending moment:

It is the degree of bending caused on the beam when an arbitrary load acts on it. Usually moment is said to prevail on a beam or structure that is acted upon by a perpendicular force triggering rotation.

Uniformly distributed load:

It refers to the load that is laid over a particular span of the beam at constant rate. The load that has constant intensity over its span on the beam is referred as uniformly distributed load.

Sign convention of force:

The forces acting towards right side are considered as positive and the forces acting towards left side are considered as negative.

Fundamentals

Consider the inclined force acting at the end of the rod AB as shown in Figure (1).

The horizontal component of force is expressed as follows:

Here, magnitude of the force is and the angle made by the force with the horizontal axis is.

The vertical component of force is expressed as follows:

Write the expression for calculating moment about point .

Here, magnitude of the force is and its perpendicular distance is .

Write the equilibrium condition for forces along the -axis.

Here, the sum of forces along x-direction is .

Write the equilibrium condition for forces along the -axis.

Here, the sum of forces along y-direction is .

Write the relation for equilibrium of moments.

Here, the sum of moments about any point is .

Consider the uniformly distributed load of constant intensity as shown in Figure (2).

Write the expression for concentrated load equivalent to the uniformly distributed load.

Here, magnitude of the uniformly distributed load is and its span is .

Draw the free body diagram of the beam as shown in the Figure (3).

$2 kip ft 30 kip ft RBx C \RBy RA 5 ft 5 ft 5 ft Figure 3 $

Apply equilibrium of moments about point .

Apply equilibrium condition for forces along the direction.

Substitute for .

Apply equilibrium condition for forces along the x direction.

Calculate shear force at point .

Calculate shear force at point .

Calculate shear force at point .

Calculate shear force at point .

Calculate the bending moment at point .

Calculate the bending moment at point .

Substitute for .

Calculate the bending moment at point .

Substitute for and for

Ans:

The shear force for the beam is