Question

This is the non-multiple choice version of online Quiz 4. The questions in D2L will be the same but each sub-question will be multiple choice in the online format. It is recommended that you complete problems in the standard given-find-solution method for your reference. Please use the beam section tables provided on the Google Drive. 1. Given a plane element in a body is subjected to a normal tensile stress in the x-direction of 120 MPa, a normal stress in the y-direction of-75 MPa and shear stresses of 50 MPa, as shown. Determing a. What is the maximum principal stress? b. What is the minimum principal stress? 75 MPa What is the maximum shear stress? 50 MPa c. d. What is the angle to the principal plane, 0p? e. What is the angle to the plane of maximum shear, 0,? If a normal stress of 125 MPa direction. What would the resulting maximum 120 MPa applied in the z- f. was shear stress be?

Answer 1

(a)

$\sigma$_x=120 Mpa

$\sigma$_y=-75 Mpa

$\tau$_xy=50 Mpa

$\sigma$₁=0.5($\sigma$_x+$\sigma$_y)+0.5{[($\sigma$_x-$\sigma$_y)²+4$\tau$_xy²]}^0.5

$\sigma$₁=0.5[120-75]+0.5{[(120-(-75)²+4(50)²)}^0.5

$\sigma$₁=22.5+109.57

$\sigma$₁=132.07 Mpa

(b)

$\sigma$₂=0.5($\sigma$_x+$\sigma$_y)-0.5{[($\sigma$_x-$\sigma$_y)²+4$\tau$_xy²]}^0.5

$\sigma$₂=0.5[120-75]-0.5{[(120-(-75)²+4(50)²)}^0.5

$\sigma$₂=22.5-109.57

$\sigma$₂=-87.07 Mpa

(c)

$\tau$_max=[($\sigma$_x-$\sigma$_y)/2]²+$\tau$_xy²]^0.5

$\tau$_max=[120+75/2]²+50²]^0.5

$\tau$_max=109.57 Mpa

(d)

Tan2$\theta$_p=2$\tau$_xy/($\sigma$_x-$\sigma$_y)

Tan2$\theta$_p=2x50/(120+75)

2$\theta$_p=27.14

$\theta$_p=13.57

(e)

Tan2$\theta$_S= -($\sigma$_x-$\sigma$_y)/2$\tau$_xy

2$\theta$_S= -(120+75)/2x50

$\theta$_S= -31.425

Answer 2

(a)

$\sigma$_x=120 Mpa

$\sigma$_y=-75 Mpa

$\tau$_xy=50 Mpa

$\sigma$₁=0.5($\sigma$_x+$\sigma$_y)+0.5{[($\sigma$_x-$\sigma$_y)²+4$\tau$_xy²]}^0.5

$\sigma$₁=0.5[120-75]+0.5{[(120-(-75)²+4(50)²)}^0.5

$\sigma$₁=22.5+109.57

$\sigma$₁=132.07 Mpa

(b)

$\sigma$₂=0.5($\sigma$_x+$\sigma$_y)-0.5{[($\sigma$_x-$\sigma$_y)²+4$\tau$_xy²]}^0.5

$\sigma$₂=0.5[120-75]-0.5{[(120-(-75)²+4(50)²)}^0.5

$\sigma$₂=22.5-109.57

$\sigma$₂=-87.07 Mpa

(c)

$\tau$_max=[($\sigma$_x-$\sigma$_y)/2]²+$\tau$_xy²]^0.5

$\tau$_max=[120+75/2]²+50²]^0.5

$\tau$_max=109.57 Mpa

(d)

Tan2$\theta$_p=2$\tau$_xy/($\sigma$_x-$\sigma$_y)

Tan2$\theta$_p=2x50/(120+75)

2$\theta$_p=27.14

$\theta$_p=13.57

(e)

Tan2$\theta$_S= -($\sigma$_x-$\sigma$_y)/2$\tau$_xy

2$\theta$_S= -(120+75)/2x50

$\theta$_S= -31.425

Answer 3

(a)

$\sigma$_x=120 Mpa

$\sigma$_y=-75 Mpa

$\tau$_xy=50 Mpa

$\sigma$₁=0.5($\sigma$_x+$\sigma$_y)+0.5{[($\sigma$_x-$\sigma$_y)²+4$\tau$_xy²]}^0.5

$\sigma$₁=0.5[120-75]+0.5{[(120-(-75)²+4(50)²)}^0.5

$\sigma$₁=22.5+109.57

$\sigma$₁=132.07 Mpa

(b)

$\sigma$₂=0.5($\sigma$_x+$\sigma$_y)-0.5{[($\sigma$_x-$\sigma$_y)²+4$\tau$_xy²]}^0.5

$\sigma$₂=0.5[120-75]-0.5{[(120-(-75)²+4(50)²)}^0.5

$\sigma$₂=22.5-109.57

$\sigma$₂=-87.07 Mpa

(c)

$\tau$_max=[($\sigma$_x-$\sigma$_y)/2]²+$\tau$_xy²]^0.5

$\tau$_max=[120+75/2]²+50²]^0.5

$\tau$_max=109.57 Mpa

(d)

Tan2$\theta$_p=2$\tau$_xy/($\sigma$_x-$\sigma$_y)

Tan2$\theta$_p=2x50/(120+75)

2$\theta$_p=27.14

$\theta$_p=13.57

(e)

Tan2$\theta$_S= -($\sigma$_x-$\sigma$_y)/2$\tau$_xy

2$\theta$_S= -(120+75)/2x50

$\theta$_S= -31.425