The block shown is made of a magnesium alloy for which E = 45 GPa and v = 0.35. Knowing that ox = -180 MPa, determine (a) the magnitude of oy for which the change in the height of the block will be zero, (b) the corresponding change in the area of the face ABCD, (c) the corresponding change in the volume of the block. (10 points) Ty 25 mm 40 mm D G E 100 mm

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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STRAIN AND DEFORMATION TOPIC

- GET THE GIVEN, WHAT IS ASKED, AND THE ANSWER.

- MAKE A PROPER FREE BODY DIAGRAM. (IT DOES HAVE THE GREATER POINTS IN THIS ACTIVITY)

- YOUR SUB ANSWERS MUST BE IN 6 DECIMAL PLACES.

- YOUR FINAL ANSWER MUST BE IN 4 DECIMAL PLACES.

- TELL THE DIRECTION AND WHETHER IT IS TENSILE OR COMPRESSION.

- WRITE YOUR SOLUTION PROPERLY (EASY TO UNDERSTAND).

- PLEASE BASE YOUR SOLUTIONS ON THE SAMPLE PROBLEM I'VE GIVEN.

1
Ey
-νσχ + σ, - νσ,]
1
Ez =
voy + 02]
1
3
4. Generalized Hooke's Law
7
TRIAXIAL LOADING E-
Ox
- EL
Ea
+ Consi der
the effects of Ox
A- = ?3:3
E
E
the effects of Oy
* Consider
Oy
Ey:0, - vớ, - vO]
Ey-EVOX + Oy -V2]
+ Const der the effects of o,
E
1
2
4. Generalized Hooke's Law
6
7
Sample Problem 8
1.2 ksi
The polyethylene sheet is
subjected to
loading shown. Determine
the resulting elongations of 2
sides AB
D
the biaxial
y
4 ft
and AC. The
6 ft
properties of polyethylene
are E
В
300ksi and v = 0.4.
%|
1.2 ksi
1
2
3
4. Generalized Hooke's Law
6.
7
Sample Problem 8
[Ex
-v] [Ox]
-v||Oy
[0z
1
-v
Given:
1
Ey
E
[Ez
1
= -
E = 300ksi
v = 0.4
Required: 6na . Sne
Solution:
-v
AC
1.2 ksi
Ex: EAB
LAB
4 ft
Sne
Ey: Enc:
6 ft
LAC
1.2 ksi
1
3
4. Generalized Hooke's Law
7
1.2 ksi
Sample Problem 8
* Consider SAD
L.
4 ft
Ox
6 ft
[2•04(12) - 0 4(0)]
SAB
В
300
EAB : 0 005 067
+ Consider SAc
EAB -
LAB
O 00 5067:
72
1.2 ksi
1SAR : 0:364n
Enc = Ey: Ě [vort Oy-vo] •
[-04(2) +12 -04(0]
3D0
SAC
E: 0 001 333 EAc:
LAC
[Ex]
1
-v
-v] [0x
1
Ey
E.
O-001333=
48
1
-v||Oy
= -
SA : 0 064in
E
-v
2 ksi
2 ksi
2 ksi
2 ksi
2 ksi
Transcribed Image Text:1 Ey -νσχ + σ, - νσ,] 1 Ez = voy + 02] 1 3 4. Generalized Hooke's Law 7 TRIAXIAL LOADING E- Ox - EL Ea + Consi der the effects of Ox A- = ?3:3 E E the effects of Oy * Consider Oy Ey:0, - vớ, - vO] Ey-EVOX + Oy -V2] + Const der the effects of o, E 1 2 4. Generalized Hooke's Law 6 7 Sample Problem 8 1.2 ksi The polyethylene sheet is subjected to loading shown. Determine the resulting elongations of 2 sides AB D the biaxial y 4 ft and AC. The 6 ft properties of polyethylene are E В 300ksi and v = 0.4. %| 1.2 ksi 1 2 3 4. Generalized Hooke's Law 6. 7 Sample Problem 8 [Ex -v] [Ox] -v||Oy [0z 1 -v Given: 1 Ey E [Ez 1 = - E = 300ksi v = 0.4 Required: 6na . Sne Solution: -v AC 1.2 ksi Ex: EAB LAB 4 ft Sne Ey: Enc: 6 ft LAC 1.2 ksi 1 3 4. Generalized Hooke's Law 7 1.2 ksi Sample Problem 8 * Consider SAD L. 4 ft Ox 6 ft [2•04(12) - 0 4(0)] SAB В 300 EAB : 0 005 067 + Consider SAc EAB - LAB O 00 5067: 72 1.2 ksi 1SAR : 0:364n Enc = Ey: Ě [vort Oy-vo] • [-04(2) +12 -04(0] 3D0 SAC E: 0 001 333 EAc: LAC [Ex] 1 -v -v] [0x 1 Ey E. O-001333= 48 1 -v||Oy = - SA : 0 064in E -v 2 ksi 2 ksi 2 ksi 2 ksi 2 ksi
The block shown is made of a magnesium alloy for which E = 45 GPa and v = 0.35. Knowing that ox = -180 MPa, determine (a)
the magnitude of oy for which the change in the height of the block will be zero, (b) the corresponding change in the area of the
face ABCD, (c) the corresponding change in the volume of the block. (10 points)
Ty
25 mm
40 mm
D
G
E
x
100 mm
Transcribed Image Text:The block shown is made of a magnesium alloy for which E = 45 GPa and v = 0.35. Knowing that ox = -180 MPa, determine (a) the magnitude of oy for which the change in the height of the block will be zero, (b) the corresponding change in the area of the face ABCD, (c) the corresponding change in the volume of the block. (10 points) Ty 25 mm 40 mm D G E x 100 mm
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