Question

2. Figure 2 shows a tubular joint subjected to moment M and P. Due to these CLO4 loads, an element of the joint in xy plane ABCD has stress system shown in Figure 3. The values of the stresses are Oxx = SX , Oyy = SY, and Txy = 45 MPa. The values and units of measure for the stresses are as given in the data sheet. Construct Mohr's circle for the stress systems to scale on the graph paper provided. b) On the Mohr's circle, show the locations of principal stresses 01, 02, and maximum shear, Tmax. Predict from the Mohr's circle the values of principal stresses 01, 02 and maximum shear stress, Tmax. d) Produce the values of principal stresses locations 01 and 02 on the element for the stress system. Figure 2 Oyy Tyx D с Txy A Oxx A B Figure 3

SX= 204MPa

SY= -12MPa

Answer 1

Given,

Normal stress in horizontal direction for the given element, $\sigma_{xx}$ = 204 MPa

Normal stress in vertical direction, $\sigma_{yy}$ = -12 MPa

Shear stress, $\tau_{xy}$ = 45 MPa

The given element can be considered as combination of two different elements as shown below:

12 MPa 12 45 MPa 45 45 E 204 < 204 LG > + 204 MPa 204 (+) Point R (204,-45) 12 MPa 12 Point S 612,45) Clement

(a). Mohr's circle can be constructed by the following procedure:

1. Select the coordinate system indicating x and y axes. X - axis represents the normal stresses while Y - axis represents shear stress.
2. Select suitable scale, here 1 cm = 20 MPa.
3. On the coordinate system locate point R such that it's coordinate is (204, - 45).
4. Also locate point S such that the coordinate is (- 12, 45).
5. Join points R and S. Let RS intersect the x-axis at point C. This represents the centre of Mohr's circle.
6. With C as centre and radius as CR = CS, draw a circle. This circle obtained is called Mohr's circle.
7. Intercepts OA and OB represents the principal stresses.

Mohr's circle thus obtained is shown below:

Scale-1 cm = 20 MP て D IT max (-12,45) S 202 Р 6 X 2 B Q10 20, с A R (204,-45)

(b). In the Mohr's circle, the location of major principal stress is given by the intercept OA, while that of minor principal stress is given by OB.

To obtain the location of maximum shear stress, draw a perpendicular to x - axis at C, such that it intersects Mohr's circle at D. Then, CD represents $\tau_{max}$ .

These are represented in the Mohr's circle shown above.

(c). From the Mohr's circle, values of principal stresses and maximum shear stress can be obtained as follows:

From Mohr's circle, Major principal stress, اک ОА li scale X pi 10.6 20 ?? 212 MPa Minor principal stress, so OB X scale Х 20 22 - 20 MPa Maximum sheas streso, Imaxe C D X scale = 5.9 X 20 22 118 MPa

Hence,

Major principal stress value is 212 MPa (approx.)

Minor principal stress is - 20 MPa (approx.)

Maximum shear stress value is 118 MPa (approx.)

(d). The angle in Mohr's circle will be twice that of original angle.

The direction of major principal plane shall be determined from  $\angle RCA$ while that of minor principal plane is represented by $\angle RCB$ .

The location of principal planes is obtained as follows:

Major principal plane direction is given by, 20 - <RCA 22.6° Canti-clockwise) 11.3° Canti-clockwise). Minor principal plane disection is given by, 20, < RCB 202.6 i on 101.3° Canti'- clockwise 11

Hence, major principal plane direction is 11.3 ° (anti-clockwise) and minor principal plane direction is 101.3 ° (anti-clockwise).