Question

The bar has a diameter of 42.8 mm. If it is subjected to the loadings shown (P = 825 N, T = 38.1 Nm, L1 = 205 mm, and L2 = 103 mm, determine the normal stress (in kPa) at point A. Note: Your answer should include requested units and be to 3 decimal places. L₂ B x y 30° P

Normal and shear stress please on point A

Answer 1

Normal stress at point A is only due to the normal load (P_y = P sin30^o).

The normal stress at point A due to the bending moment(M_X) due to the force (Pcos30^o) is 0. This is because the point A lies on the neutral axis of the bending plane(xz-plane).

The normal stress at point A is " $\sigma$ _A = 286.712 kPa ".

1 MPa = 1 N/mm² = 10⁶ N/m²

Giver, P=825N Z d = 42.8mm 205mm 38. IN.M De Psin 30 -- L=205mm Poso normal stres at A Ny + Area Iz JA = Dintace from X = 825 (sin 36) Mn=666130) x 205 GA= 0.28 6712 MPa Iz = idy 64 2) CA = 286.712 kpa Shear stres TA Ples 30' TR + who J = TdY Area 32 R=21.4mn 2.971 Mpa
The shear stress is also calculated just for additional information, please ignore if you don't need.