Answer:
=> In-plane principal stresses at point A = 14.18 MPa and -8.02 MPa.
=> maximum in-plane shear stress= 11.1 MPa.
Explanation:
Okay, we are given the following Important parameters or information or data which is going to assist us in order to solve this question effectively and efficiently.
Jy = 500 Nm, Jb = 10 KNm, wy = 60 KN, wx = -80 KN.
Thus, I(bb) = 10/12 × 100 × (200 + 2 × 20)^3 - 1/2 (100 - 20) × 200^3.
l(bb) = 6.19 × 107 mm^4.
l(yy) = 1/12 × 20 × 240^3 × 2 + 1/2 × 200 × 20^3.
l(yy) = 4.62 × 10^7 mm^4.
Thus, the bending stress due to Jy = Jy × y / l(bb)
=> 500 × 10^ 3 × 0 /l(yy) = 0 MPa.
The bending stress due to Jb;
=> 500 × 10^ 6 ×( 200/2)/ 6.19 × 10^7 = 16.16 MPa.
The bending stress due to wx = wx/A= -80/800 = 10 MPa.
Transverse share stress = 60 × 10^3 × 220000/ 6.19 × 10^ 7 × 20 = 10.66 MPa.
Principal stresses= T1 - T3/ 2 +/- ( T1 - T3/)^2 + c^2.
=> 16.16 - 10/2 +/- (16.16 -10)^2 + 10.66^2.
=> 3.08 +/- 11.1.
=> = 14.18 MPa and -8.02 MPa.
=> maximum in-plane shear stress = ( T1 - T3/)^2 + c^2 = 16.16 -10)^2 + 10.66^2.
= 11.1 MPa.
