[역학] 재료역학(Mechanics of Materials) 解法(솔루션) - James M.Gere 6th edition…
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작성일 20-10-16 02:19
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Download : 재료역학.zip
[d2
P P
SECTION 1.2 Normal Stress and Strain 3
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P 200 lb
4
200 lb
설명
Solution 1.2-3 Long steel rod in tension
다. [이용대상]
(a) If the measured strain is 550 106, what is the
(0.25 in.)2 4074 psi
smax
W Weight of rod
gL
L
Problem 1.2-4 A circular aluminum tube of length L 400
ft2
Solution 1.2-4 Aluminum tube in compression
P = 200 lb
A
[역학] 재료역학(Mechanics of Materials) 解法(솔루션) - James M.Gere 6th edition 입니다.
Download : 재료역학.zip( 16 )
0.220 mm —
P P
A
e 550 106
4
4
[ (60 mm)2 (50 mm)2 ]
eL (550 106)(400 mm)
MPa, what should be the load P?
A
1
d1 50 mm
shortening of the bar?
863.9mm2
374.3 psi
L 400 mm
Weight density: 490 lb/ft3
max 4450 psi —
2]
WP
역학, 재료, 공대, 솔루션
재료역학(Mechanics of Materials) 解法(해법) - James M.Gere 6th edition 입니다.
L 110 ft
in.2 ≤
strain gage is placed on the outside of the bar to measure normal
(b) COMPRESSIVE LOAD P
A
레포트 > 공학,기술계열
2d1
L = 400 mm
Strain gage
P
d
AL
mm is loaded in compression by forces P (see figure). The outside
d2 60 mm
재료역학(Mechanics of Materials) 解法(솔루션) - James M.Gere 6th edition 입니다. 챕터1부터 12까지 까지 입니다.
P A (40 MPa)(863.9 mm2)
P 200 lb
P
(Volume)
재료역학(Mechanics of Materials) 솔루션 - James M.Gere 6th edition 입니다.
(a) SHORTENING OF THE BAR
d 1⁄4 in.
순서
and inside diameters are 60 mm and 50 mm, respectively. A
34.6 kN —
40 MPa
Strain gage
Rounding, we get
gL (490 lbft3)(110 ft)¢
(b) If the compressive stress in the bar is intended to be 40
144
max 374 psi 4074 psi 4448 psi
[이용대상]
strains in the longitudinal direction.
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