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.



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



3

PROBLEM-1

222 xy

yx'y'x cossin

MPa8.68

6)25(602

5080 o

o0-cossin



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.



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



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



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



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:





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



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



11

PROBLEM-3


2xy

2yxmax

planein 2

MPa127

)4.116(2

086.101 22

MPa9.502

086.101

2yx

avg



12

PROBLEM-3


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



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



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



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

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

Documents

stress transformation

state of stress

stress distributions

stress state of point

state of plane stress

plane principal stress

maximum shear stress

plane shear stress