This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Sec_2.5.qxd 9/25/08 156 3:00 PM Page 156 CHAPTER 2 Axially Loaded Members Problem 2.5-7 A circular steel rod AB (diameter d1 1.0 in., length L1
3.0 ft) has a bronze sleeve (outer diameter d2 1.25 in., length L2 1.0 ft)
shrunk onto it so that the two parts are securely bonded (see figure).
Calculate the total elongation of the steel bar due to a temperature rise
T 500°F. (Material properties are as follows: for steel, Es 30 106 psi and
6.5 10 6/°F; for bronze, Eb 15 106 psi and b 11 10 6/°F.)
s Solution 2.5-7 Steel rod with bronze sleeve
SUBSTITUTE NUMERICAL VALUES:
6.5 10 6/°F Es 30 106 psi d1 1.0 in. As p2
41 d2 1.25 in. Ab p
42 T 500°F 2 0.04493 in. s L1 36 in. L2 12 in. ELONGATION OF THE TWO OUTER PARTS OF THE BAR
1 s( (6.5 T)(L1 L2)
6 10 /°F)(500°F)(36 in. 12 in.) 0.07800 in.
ELONGATION OF THE MIDDLE PART OF THE BAR
The steel rod and bronze sleeve lengthen the same
amount, so they are in the same condition as the bolt
and sleeve of Example 2-8. Thus, we can calculate the
elongation from Eq. (2-21):
d2 b 11 10 6/°F Eb 15 106 psi 0.78540 in.2 d12)
L2 0.44179 in.2 12.0 in. TOTAL ELONGATION (as Es As + ab Eb Ab)( ¢ T)L2
Es As + Eb Ab Problem 2.5-8 A brass sleeve S is fitted over a steel bolt B (see
figure), and the nut is tightened until it is just snug. The bolt has a
diameter dB 25 mm, and the sleeve has inside and outside diameters
d1 26 mm and d2 36 mm, respectively.
Calculate the temperature rise T that is required to produce a compressive stress of 25 MPa in the sleeve. (Use material properties as
follows: for the sleeve, S 21 10 6/°C and ES 100 GPa; for
the bolt, B 10 10 6/°C and EB 200 GPa.)
(Suggestion: Use the results of Example 2-8.) 1 2 0.123 in. ; ...
View Full Document
This note was uploaded on 12/22/2011 for the course MEEG 310 taught by Professor Staff during the Fall '11 term at University of Delaware.
- Fall '11