Example 3.1.1

Given: A pipe of initial radius, R = 0.2 m, carries a pressurized liquid that causes the circumferential strain in the pipe to be 0.1% (the circumference increases by 0.1%).

Req'd: What is the change in radius of the pipe, DR, due to the pressure?

Sol'n: Strain, e, is the change in length divided by the original length. The circumferential strain is 0.1% (0.001). Take L as the circumference of the pipe, and L+D as the circuference under pressure.

e = [(L+D)-L]/L = [2p(R+DR) - 2pR] / [2pR] = DR/R = 1/1000 = 0.001

The radius increases by: DR = R/1000 = 0.2m/1000 = DR = 0.0002 m = 0.2 mm

Example 3.1.2

Given: A lamp weighing W = 10 lb hangs from the ceiling by a steel wire of diameter, D = 0.050 inches.

Req'd: What is the stress in the wire?

Sol'n: Stress, s, is the force divided by the area. By statics, the force in the wire is: P = W.  

Thus:    s = P/A = P/(pD²/4) = (10 lb)/[p(0.05in)²/4]

s = 5093 lb/in² = 5100 psi = 5.1 ksi.

Example 3.1.3: Factor of Safety

Given: The wire holding up the lamp in the above problem is considered to have a Failure (Yield) Strength of 60 ksi. The Factor of Safety for the design of the wire is to be 1.5 (just in case the user pulls down on the lamp, or for some other reason).

Req'd: What is the maximum force (Allowable Force) that the wire can support without exceeding the Allowable (Design) Stress?

Sol'n: Factor of Safety is: F.S. = [Failure Strength]/[Allowable Stress]

s_allow = S_u/F.S. = (60 ksi)/1.5 = 40 ksi
P_allow = s_allowA = 78.5 lb = 78 lb

Note: Don't round up allowables: 79 lb exceeds allowable.