For the block loaded triaxially as described in Prob. 2, find the uniformly distributed load that must be added in the x direction to produce no deformation in the y direction.
For the block loaded triaxially as described in Prob. 2, find the uniformly distributed load that must be added in the x direction to produce no deformation in the y direction.
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Question
For the block loaded triaxially as described in Prob. 2, find the uniformly
distributed load that must be added in the x direction to produce no deformation
in the y direction.
Transcribed Image Text:2 - A rectangular steel block is
subjected to a triaxial loading of three
uniformly distributed forces. If v = 0.30
and E = 200 GPa, determine the single
uniformly distributed load in the x
direction that would produce the same
deformation in the z direction as the
original loading. Fx = 10kN, Fy = 15 kN,
F.
F2 = 20 kN, L = 600 mm, W = 400 mm, &
2- axis
D = 800 mm.
X-axis
![* Solution : • Dimensions of the block:
D = 800 mm
L= 600 mm ;
• Force acting in x, y and z direction:
Fx = 10KN ; Fy = 15 kN , Fz = 20 kN
• Poisson's Ratio :
• Modulus of Elasticity : E = 200 GPa = 200 X 103 N/mm2
W= 400 mm
V= 0. 30
Stress in the x, y and z-direction :
10 x 10 N
(800mm) x (400 mm)
Fx
0.03125 Mpa (tension )
+ 6x
%3D
%3D
DXW
15x 10 N
(400 mm) x (600 mm)
• 5y =
Fy
0.06250 MPa (Tension )
%3D
%3D
wXL
- 20 X 10 N
(800mm) x (600 mm)
Fz
MPa =-0.04167 MPa
24
DXL
( Compressin )
• Stzain in the z- direction:
Ez
%3D
[( + %) xA - 3] ->
(200 x 1o N/mm2)
Ez
x
24
0. 30 x (0.031 25 + 0.06 250)
Ez =
-3.489583333 x 10 7
force is tequired in the x-direction
Ez is negati ve, thus tensile
to produce the same deformation in the ydirection as the original
forces.
• For equivalent single forre in the x-direction :
K = - Ez
Ex
.: V x €x = - Ez
= - Ez
6x =
- Ez XE
-(-3.489583333 x 167)× C200x103)
0.30
6x =
0. 2326388889 MPu](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F00e13490-4e98-49be-94d1-207ac69ee0be%2Fea8094c1-d072-4742-9b84-cdd9180901ac%2Fwkgt9xp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:* Solution : • Dimensions of the block:
D = 800 mm
L= 600 mm ;
• Force acting in x, y and z direction:
Fx = 10KN ; Fy = 15 kN , Fz = 20 kN
• Poisson's Ratio :
• Modulus of Elasticity : E = 200 GPa = 200 X 103 N/mm2
W= 400 mm
V= 0. 30
Stress in the x, y and z-direction :
10 x 10 N
(800mm) x (400 mm)
Fx
0.03125 Mpa (tension )
+ 6x
%3D
%3D
DXW
15x 10 N
(400 mm) x (600 mm)
• 5y =
Fy
0.06250 MPa (Tension )
%3D
%3D
wXL
- 20 X 10 N
(800mm) x (600 mm)
Fz
MPa =-0.04167 MPa
24
DXL
( Compressin )
• Stzain in the z- direction:
Ez
%3D
[( + %) xA - 3] ->
(200 x 1o N/mm2)
Ez
x
24
0. 30 x (0.031 25 + 0.06 250)
Ez =
-3.489583333 x 10 7
force is tequired in the x-direction
Ez is negati ve, thus tensile
to produce the same deformation in the ydirection as the original
forces.
• For equivalent single forre in the x-direction :
K = - Ez
Ex
.: V x €x = - Ez
= - Ez
6x =
- Ez XE
-(-3.489583333 x 167)× C200x103)
0.30
6x =
0. 2326388889 MPu
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