problem-1
DESCRIPTION
PROBLEM-1. State of stress at a p oin t is represented by the element shown. Determine the state of stress at the p oin t on another element orientated 30 clockwise from the position shown. PROBLEM-1. PROBLEM-1. PROBLEM-2. - PowerPoint PPT PresentationTRANSCRIPT
2005 Pearson Education South Asia Pte Ltd
TUTORIAL-7: STRESS TRANSFORMATION
1
PROBLEM-1
State of stress at a point is represented by
the element shown. Determine the state
of stress at the point on another element
orientated 30 clockwise from the position
shown.
2005 Pearson Education South Asia Pte Ltd
TUTORIAL-7: STRESS TRANSFORMATION
2
PROBLEM-1
2222 xy
yxyx'x sincos
MPa8.25
6)25(6025080
25080 o
o0-sincos
2222 xy
yxyx'y sincos
MPa15.4
6)25(6025080
25080 o
o0-sincos
2005 Pearson Education South Asia Pte Ltd
TUTORIAL-7: STRESS TRANSFORMATION
3
PROBLEM-1
222 xy
yx'y'x cossin
MPa8.68
6)25(6025080 o
o0-cossin
2005 Pearson Education South Asia Pte Ltd
TUTORIAL-7: STRESS TRANSFORMATION
4
PROBLEM-2State of plane stress at a point on a body is represented on the element shown. Represent this stress state in terms of1. the maximum in-plane shear stress and associated
average normal stress, and2. the in-plane principal stress and its orientation.
2005 Pearson Education South Asia Pte Ltd
TUTORIAL-7: STRESS TRANSFORMATION
5
PROBLEM-21. The maximum in-plane shear stress and associated average
normal stress.
2xy
2yxmax
planein 2
MPa4.81
6029020 2
2
xy
yxs
22tan
9167.0
6029020
s = 21.3o
2005 Pearson Education South Asia Pte Ltd
TUTORIAL-7: STRESS TRANSFORMATION
6
PROBLEM-2
MPa3529020
2yx
avg
2. The in-plane principal stress and its orientation.
2xy
2yxyx
2,1 22
MPa)4.8135(
6029020
29020 2
2
2005 Pearson Education South Asia Pte Ltd
TUTORIAL-7: STRESS TRANSFORMATION
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PROBLEM-2
MPa4.46MPa)4.8135(MPa4.116MPa)4.8135(
2
1
2yx
xyp
2tan
Principal stress orientation.
091.129020
60
p = – 23.7o
2005 Pearson Education South Asia Pte Ltd
TUTORIAL-7: STRESS TRANSFORMATION
8
PROBLEM-3Steel pipe has inner diameter of 60 mm and outer diameter of 80 mm. It is subjected to a torsional moment of 8 kN·m and a bending moment of 3.5 kN·m. Determine:1. The maximum in-plane shear stress and associated average normal
stress.2. The in-plane principal stress and its orientation.
2005 Pearson Education South Asia Pte Ltd
TUTORIAL-7: STRESS TRANSFORMATION
9
Investigate a point on pipe that is subjected to a state of maximum critical stress.
• Torsional and bending moments are uniform throughout the pipe’s length.
• At arbitrary section a-a, loadings produce the stress distributions shown.
• Point A undergoes maximum compressive stress and point B undegoes maximum tensile stress.
• Thus, the critical point is at B.
PROBLEM-3
2005 Pearson Education South Asia Pte Ltd
TUTORIAL-7: STRESS TRANSFORMATION
10
PROBLEM-3
Maximum tensile stress at point B:
MPa86.101)0.03(0.04
4)(3500)(0.044
4π
zIcM
Maximum shear stress at point B:
MPa4.116)0.03(0.04
04)(-8000)(0.44
2π
J
cTτ
2005 Pearson Education South Asia Pte Ltd
TUTORIAL-7: STRESS TRANSFORMATION
11
PROBLEM-31. The maximum in-plane shear stress and associated average
normal stress.
2xy
2yxmax
planein 2
MPa127
)4.116(2
086.101 22
MPa9.502
086.101
2yx
avg
2005 Pearson Education South Asia Pte Ltd
TUTORIAL-7: STRESS TRANSFORMATION
12
PROBLEM-32. The in-plane principal stress and its orientation.
2xy
2yxyx
2,1 22
MPa)1279.50(
)4.116(2
086.1012
086.101 22
MPa1.76MPa)1279.50(MPa9.177MPa)1279.50(
2
1
2yx
xyp
2tan 285.2
2086.1014.116
p = – 33.1o
2005 Pearson Education South Asia Pte Ltd
TUTORIAL-7: STRESS TRANSFORMATION
13
PROBLEM-4
Stress in Shafts Due to Axial Load, Bending Load and Torsion
A shaft has a diameter of 4 cm. The cutting section shows in the figure is subjected to a compressive force of 2500 N, a bending moment of 800 Nm and a torque of 1500 Nm.
Determine: 1. The stress state of point A. 2. The principal stresses and its orientation
2005 Pearson Education South Asia Pte Ltd
TUTORIAL-7: STRESS TRANSFORMATION
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PROBLEM-41. The stress state at point A has been
solved in Tutorial-6 Problem-3. The results are as follows
Shear stress: xy = A = 119.37 MPa
Normal stress: x = A’ + A”
= (– 1.99 – 127.32) MPa
= – 129.31 MPa
xy
x
2005 Pearson Education South Asia Pte Ltd
TUTORIAL-7: STRESS TRANSFORMATION
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PROBLEM-42. The principal stress at point A and its orientation
2xy
2yxyx
2,1 22
MPa)75.13565.64(
37.1192
031.1292
031.129 22
MPa40.200MPa)75.13565.64(MPa10.71MPa)75.13565.64(
2
1
2yx
xyp
2tan 846.1
2031.12937.119
p = – 30.78o