A Z section is shown below Calculate Moments of inertia abou

A \"Z\" section is shown below. Calculate: Moments of inertia about the center of gravity Product of inertia Principle moments of inertia Angle of rotation between given axis and principle axis.

Solution

GIVEN DATA: a=breadth=3.5 inch b=depth=6 inch c=width=2.75 inch d=height from base=5.25 inch t=thickness=0.75 inch TO FIND: 1) MOMENT OF INERTIA ABOUT C.G: Ixc=Ixx=(ab³-c(b-2t)³)/12=3.5×6³-(6-2×0.75)³/12=42.11 inches^4. Iyc=Iyy=(b(a+c)³-2c³d-6a²cd)/12=(6×(3.5+2.75)³-2×2.75³×5.25-6×3.5²×2.75×5.25)/12=15.44 inches⁴.................... 2) PRODUCT OF INERTIA: Ixy= 2ba (a/2 - b/2) ×(c + a/2) = 2×6×3.5×(3.5/2 - 6/2)×(2.75 + 3.5/2)= -236.25 inches⁴............. 3) PRINCIPLE MOMENTS OF INERTIA: Imax/min= (Ix+Iy)/2 +/- sqrt ((Ix-Iy)/2)² + Ixy²= (42.11+15.44)/2 +/- sqrt ((42.11-15.44)/2)²+(-236.25)²=28.77 +/- 236.62.....Therefore Imax=28.77+236.62= 265.39 inches⁴ ......Imin=28.77-236.62=-207.85 inches⁴....................... 4)Angle of rotation: Alpha= 1/2 tan^-1(2×Ixy/Ixx-Iyy)= 0.5xtan^-1(2×-236.25/42.11-15.44).....Angle alpha= -43 degree= 43 degree counterclockwise.