2005 pearson education south asia pte ltd tutorial-7: stress transformation 1 problem-1 state of...

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(602

5080

2

5080 o

o0-sincos

2222 xy

yxyx'y sincos

MPa15.4

6)25(602

5080

2

5080 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(602

5080 o

o0-cossin

2005 Pearson Education South Asia Pte Ltd

TUTORIAL-7: STRESS TRANSFORMATION

4

PROBLEM-2

State of plane stress at a point on a body is represented on the element shown. Represent this stress state in terms of

1. the maximum in-plane shear stress and associated average normal stress, and

2. the in-plane principal stress and its orientation.

2005 Pearson Education South Asia Pte Ltd

TUTORIAL-7: STRESS TRANSFORMATION

5

PROBLEM-2

1. The maximum in-plane shear stress and associated average normal stress.

2xy

2yxmax

planein 2

MPa4.81

602

9020 22

xy

yxs

22tan

9167.060

29020

qs = 21.3o

2005 Pearson Education South Asia Pte Ltd

TUTORIAL-7: STRESS TRANSFORMATION

6

PROBLEM-2

MPa352

9020

2yx

avg

2. The in-plane principal stress and its orientation.

2xy

2yxyx

2,1 22

MPa)4.8135(

602

9020

2

9020 22

2005 Pearson Education South Asia Pte Ltd

TUTORIAL-7: STRESS TRANSFORMATION

7

PROBLEM-2

MPa4.46MPa)4.8135(

MPa4.116MPa)4.8135(

2

1

2yx

xyp

2tan

Principal stress orientation.

091.129020

60

qp = – 23.7o

2005 Pearson Education South Asia Pte Ltd

TUTORIAL-7: STRESS TRANSFORMATION

8

PROBLEM-3

Steel 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

zI

cM

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-3

1. 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-3

2. The in-plane principal stress and its orientation.

2xy

2yxyx

2,1 22

MPa)1279.50(

)4.116(2

086.101

2

086.101 22

MPa1.76MPa)1279.50(

MPa9.177MPa)1279.50(

2

1

2yx

xyp

2tan 285.2

2086.101

4.116

qp = – 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

14

PROBLEM-4

1. The stress state at point A has been solved in Tutorial-6 Problem-3. The results are as follows

Shear stress: txy = tA = 119.37 MPa

Normal stress: sx = sA’ + sA”

= (– 1.99 – 127.32) MPa

= – 129.31 MPa

txy

sx

2005 Pearson Education South Asia Pte Ltd

TUTORIAL-7: STRESS TRANSFORMATION

15

PROBLEM-4

2. The principal stress at point A and its orientation

2xy

2yxyx

2,1 22

MPa)75.13565.64(

37.1192

031.129

2

031.129 22

MPa40.200MPa)75.13565.64(

MPa10.71MPa)75.13565.64(

2

1

2yx

xyp

2tan 846.1

2031.129

37.119

qp = – 30.78o