A bronze bar 3 m long with a cross sectional area of 320 mm' is placed between two rigid walls as shown in Fig. P-265. At a temperature of -20°C, the gap A = 25 mm. Find the temperature at which the compressive stress in the bar will be 35 MPa. Use a = 18.0 x 10* m/(m-°C) and E = 80 GPa. L-3m Figure P-265

Materials Science And Engineering Properties
1st Edition
ISBN:9781111988609
Author:Charles Gilmore
Publisher:Charles Gilmore
Chapter15: Experimental Methods
Section: Chapter Questions
Problem 25CQ
Question
ممكن حل هذا الاسئلة
(AE
AE
1200 la
P- aAE(T;- T.)
1200 - (6.5 x 10)(0.25) (29 x 10)(T;- 40)
T;= 65.46°F
Problem 265
A bronze bar 3 m long with a cross sectional area of 320 mm? is placed between two
rigid walls as shown in Fig. P-265. At a temperature of -20°C, the gap A = 25 mm. Find
the temperature at which the compressive stress in the bar will be 35 MPa. Use a =
18.0 x 10 m/(m.°C) and E= 80 GPa.
L= 3 m
Figure P-265
Solution 265
ol
+ 2.5
E
aL(AT) =
(18 x 10)(3000)(AT)-
35(3000)
+ 2.5
L-3 m
S0000
AT - 70.6°C
A- 2.05 mm
T= 70.6 – 20
T- 50.6°C
Problem 266
Calculate the increase in stress for each segment of the compound bar shown in Fig. P-
266 if the temperature increases by 100°F. Assume that the supports are unyielding
and that the bar is suitably braced against buckling.
Aluminum
A = 2.0 in
E- 10 x 10 psi
a - 12.8 x 10"F a-6.5 x 10F
Steel
A- 15 n?
E- 29 x 10 psi
10 in
15 in
Figure P-266
Solution 266
or - al AT
STa) - (6.5 x 10)(15)(100)
Orun = 0.00975
öra) = (12.8 x 10)(10)(100)
OTra) = 0.0128 in
10 in
15 in
PL
AE)
PL
+
- 0.00975 + 0.0128
AE
st
where P- P - Pa
P(15)
P(10)
- 0.02255
1.5(29 x 10°) 2(10×10°)
P- 26 691.84 psi
P
A
26691.84
= 17 794.56 psi
1.5
26691.84
- 13 345.92 psi
2.0
Transcribed Image Text:(AE AE 1200 la P- aAE(T;- T.) 1200 - (6.5 x 10)(0.25) (29 x 10)(T;- 40) T;= 65.46°F Problem 265 A bronze bar 3 m long with a cross sectional area of 320 mm? is placed between two rigid walls as shown in Fig. P-265. At a temperature of -20°C, the gap A = 25 mm. Find the temperature at which the compressive stress in the bar will be 35 MPa. Use a = 18.0 x 10 m/(m.°C) and E= 80 GPa. L= 3 m Figure P-265 Solution 265 ol + 2.5 E aL(AT) = (18 x 10)(3000)(AT)- 35(3000) + 2.5 L-3 m S0000 AT - 70.6°C A- 2.05 mm T= 70.6 – 20 T- 50.6°C Problem 266 Calculate the increase in stress for each segment of the compound bar shown in Fig. P- 266 if the temperature increases by 100°F. Assume that the supports are unyielding and that the bar is suitably braced against buckling. Aluminum A = 2.0 in E- 10 x 10 psi a - 12.8 x 10"F a-6.5 x 10F Steel A- 15 n? E- 29 x 10 psi 10 in 15 in Figure P-266 Solution 266 or - al AT STa) - (6.5 x 10)(15)(100) Orun = 0.00975 öra) = (12.8 x 10)(10)(100) OTra) = 0.0128 in 10 in 15 in PL AE) PL + - 0.00975 + 0.0128 AE st where P- P - Pa P(15) P(10) - 0.02255 1.5(29 x 10°) 2(10×10°) P- 26 691.84 psi P A 26691.84 = 17 794.56 psi 1.5 26691.84 - 13 345.92 psi 2.0
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