4. 17 3) 98x10 N/m m 4) 99x10°N/m A wooden bar is subjected to a tensile stress of 5MP a. The value of normal stress across a seciton, which makes an angle of 25° with the direction of the tensile stress is (cos'0.6428=509) 1) 0.9 MPa 2) 1.9 MPa 3) 2.9 MPa 4) 4.9 MPa
Answers
Answered by
0
Answer:
you can see my explanation
Explanation:
Correct option is
A
tanθ
Consider the equilibrium of the plane BB'.
A force F must be acting on this plane making an angle (90
o
−θ) with the normal ON. Resolving F into two components, along the plane and normal to the plane.
Component of force F along the plane, F
P
=Fcosθ
Component of force F normal to the plane,
F
N
=Fcos(90
o
−θ)=Fsinθ
Let the area of the face BB' be A'. Then
A
′
A
=sinθ
∴A
′
=
sinθ
A
∴ Tensile stress =
A
′
Fsinθ
=
A
F
sin
2
θ
Shearing stress =
A
′
Fcosθ
=
A
F
cosθsinθ=
2A
Fsin2θ
Their corresponding ratio is
Shearingstress
Tensilestress
=
A
F
sin
2
θ×
Fcosθsinθ
A
=tanθ
Similar questions