Page 28 - Marks Calculation for Machine Design

P1: Shibu
January 4, 2005
Brown˙C01
Brown.cls
10
U.S. Customary 12:26 STRENGTH OF MACHINES SI/Metric
Step 2. Solve for the force (P) in Eq. (1.7) to Step 2. Solve for the force (P) in Eq. (1.7) to
give give
δAE δAE
P = P =
L L
2
2
9
2
2
6
(0.0025 ft)(9in )(11 × 10 lb/in ) (0.0007 m)(0.0064 m )(71 × 10 N/m )
= =
(3ft) (1m)
247,500 ft · lb 318,000 N · m
= = 82,500 lb = = 318,000 N
3ft 1m
= 82.5 kips = 318 kN
Prismatic bar
FIGURE 1.10 Thermal strain.

Thermal Strain. If the temperature of the prismatic bar shown in Fig. 1.10 increases, then
an axial strain (ε T ) will be developed and given by Eq. (1.9),

ε T = α(
T ) (1.9)

and the bar will lengthen by an amount (δ T ) given by Eq. (1.10).

δ T = ε T L = α(
T )L (1.10)
where α = coefficient of thermal expansion

T = change in temperature
L = length of bar
For a temperature decrease, the thermal strain (ε T ) will be negative as given by Eq. (1.9),
and consequently the bar will shorten by an amount (δ T ) as given by Eq. (1.10).

Thermal Stress. If during a temperature change the bar is not constrained, no thermal
stress will develop. However, if the bar is constrained from lengthening or shortening, a
thermal stress (σ T ) will develop as given by Eq. (1.11).

σ T = Eε T = Eα(
T ) (1.11)

Notice that Eq. (1.11) represents Hooke’s law, Eq. (1.3), where the thermal strain (ε T )
given by Eq. (1.9) has been substituted for the axial strain (ε). Also notice that the cross-
sectional area (A) of the bar does not appear in Eqs. (1.9) to (1.11).