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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

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

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- YOUR SUB ANSWERS MUST BE IN 6 DECIMAL PLACES.

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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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