[역학] 재료역학(Mechanics of Materials) 해법 - James M.Gere 6th edition 입니다.
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작성일 20-10-30 07:24
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Download : 재료역학.zip
P 200 lb
max 4450 psi —
4
863.9mm2
Problem 1.2-4 A circular aluminum tube of length L 400
A
ft2
1
SECTION 1.2 Normal Stress and Strain 3
and inside diameters are 60 mm and 50 mm, respectively. A
Strain gage
P
smax
P P
in.2 ≤
설명
Solution 1.2-3 Long steel rod in tension
Strain gage
strains in the longitudinal direction.
d
다.
역학, 재료, 공대, 솔루션
2d1
shortening of the bar?
A
(b) If the compressive stress in the bar is intended to be 40
gL (490 lbft3)(110 ft)¢
A
144
P P
strain gage is placed on the outside of the bar to measure normal
4
레포트 > 공학,기술계열
Solution 1.2-4 Aluminum tube in compression
(Volume)
[d2
L 400 mm
max 374 psi 4074 psi 4448 psi
40 MPa
L 110 ft
e 550 106
mm is loaded in compression by forces P (see figure). The outside
eL (550 106)(400 mm)
gL
재료역학(Mechanics of Materials) 솔루션 - James M.Gere 6th edition 입니다.
200 lb
P
Download : 재료역학.zip( 77 )
[이용대상]
d 1⁄4 in.
AL
(a) If the measured strain is 550 106, what is the
34.6 kN —
(b) COMPRESSIVE LOAD P
재료역학(Mechanics of Materials) 해결책 - James M.Gere 6th edition 입니다.
A
4
d1 50 mm
챕터1부터 12까지 까지 입니다. 챕터1부터 12까지 까지 입니다. [이용대상]
2]
순서
P 200 lb
P = 200 lb
챕터1부터 12까지 까지 입니다.
L
W Weight of rod
374.3 psi
L = 400 mm
재료역학(Mechanics of Materials) 해법 - James M.Gere 6th edition 입니다. Weight density: 490 lb/ft3
d2 60 mm
MPa, what should be the load P?
WP
P A (40 MPa)(863.9 mm2)
0.220 mm —
[ (60 mm)2 (50 mm)2 ]
(a) SHORTENING OF THE BAR
Rounding, we get
(0.25 in.)2 4074 psi
[역학] 재료역학(Mechanics of Materials) 해법 - James M.Gere 6th edition 입니다.
