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

3. PROBLEM-1

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
the maximum in-plane shear stress and associated average normal stress, and
the in-plane principal stress and its orientation.

5. PROBLEM-2
The maximum in-plane shear stress and associated average normal stress.
qs = 21.3o

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

7. PROBLEM-2
Principal stress orientation.
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:
The maximum in-plane shear stress and associated average normal stress.
The in-plane principal stress and its orientation.

9. PROBLEM-3
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.

10. PROBLEM-3 Maximum tensile stress at point B:
Maximum shear stress at point B:

11. PROBLEM-3
The maximum in-plane shear stress and associated average normal stress.

12. PROBLEM-3
2. The in-plane principal stress and its orientation.
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 Nm and a torque of 1500 Nm.
FG09_17c.TIF
Notes:
Procedure for Analysis
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
txy sx
Shear stress: txy = tA = MPa
Normal stress: sx = sA’ + sA”
= (– 1.99 – ) MPa
= – MPa
FG09_17c.TIF
Notes:
Procedure for Analysis

15. PROBLEM-4 2. The principal stress at point A and its orientation
FG09_17c.TIF
Notes:
Procedure for Analysis
qp = – 30.78o