Learning Objectives:

• Define different types of loads, understand their symbols and able to convert them into equivalent point loads.

1. Point loads-concentrated forces exerted at point or location



2. Distributed loads-a force applied along a length or over an area. The distribution can be uniform or non-uniform.

   1. Uniformly distributed Load (UDL):

   

   2. Triangular Load:

   

Solved Example: 23-2-01

U.D.L. stands for?

A. Uniformly diluted length

B. Uniformly developed loads

C. Uniaxial distributed load

D. Uniformly distributed load

UDL are uniformly spread over a portion or whole area.

Correct Answer: D

Solved Example: 23-2-02

What are the units of U.D.L.?

A. $KN/m$

B. $KN-m$

C. $KN-m^2$

D. $KN$

Uniformly distributed loads (UDL) distribute over span the units for this kind of loads will be load per meter length i.e KN/m. It is denoted by “w”. They are generally represented as rate of load that is Kilo Newton per meter length (KN/m).

Correct Answer: A

Solved Example: 23-2-03

Weight of a beam is an example of:

A. Concentrated load

B. Uniformly distributed load

C. Linearly varying load

D. Varying load

Correct Answer: B

Learning Objectives:

• Be able to identify and draw appropriate free body diagram for given concurrent force systems.

A free body diagram is a simplified repersentation of given problem that consists of a memeber with loading conditions. In the free body diagram (FBD) we replace the following things

• Supports are replaced by support reactions.
• Surfaces are replaced by normal reactions.
• Uniformly distributed load (UDL) and triangular loads (also called Uniformly Varying Loads- UVL) by their equivalent point loads.
• Springs will be replaced by spring force. F = kx
• Wires, ropes and threads are replaced by the tensions in them.

Mets501, CC BY-SA 3.0, via Wikimedia Commons

Solved Example: 23-3-01

Two people pull on opposite ends of a rope, each with a force F. The tension in the rope is:

A. F/2

B. F

C. 2F

D. F$^2$

Correct Answer: B

Solved Example: 23-3-02

The limiting force P, in terms of Q, required for impending motion of Block N to just move it in the upward direction is given as P = $\alpha$Q. The value of coefficient '$\alpha$' (round off to one decimal place) is:

A. 0.5

B. 0.9

C. 2.0

D. 0.6

Free Body Diagram of Block N

\[\Sigma F_Y = 0\] \[N_2 \sin 60^\circ - 0.2 N_2 \sin 30^\circ - Q = 0\] \[Q = 0.766 N_2\]

Free Body Diagram of Wedge M

\[\Sigma F_x = 0\] \[0.2 N_2 \cos 30^\circ + N_2 \cos 60^\circ - P = 0\] \[P = 0.67 N_2 = 0.67 \times \left(\dfrac{Q}{0.766}\right)\]

Correct Answer: B