Problem Set 4Problem 4.1 Moment of inertia by integrationDetermine the moment of inertia of the shaded area about the y-axis.Figure P4.1: Problem 4.1Problem 4.2 Moment of inertia of composite cross sectionDetermine ¯y which locates the centroidal x0-axis, for the cross section ofthe beam, and then find the moments of inertia Ix0x0and Iyy.Figure P4.2: Problem 4.2Problem 4.3Determine the moment of inertia about the x-axis of the shaded areashown in Fig. P4.3 where m = h/b and b = h = 1 m. Use integration.y = m xbyxhy = m xbyxhydy(b, y)(x, y)Figure P4.3: Problem 4.3Problem 4.4Determine the moments of inertia and the products of inertia about thecentrodial axes of the shaded area shown in Fig. P4.4, where a = 1 in. Findthe centroid polar moment of inertia. The mass center of the shaded area isat C.Caaaa3axy123Figure P4.4: Problem 4.4Solution% ICxx = b h^3/12; ICyy = h b^3/12% upper rectagle 1A1 = 3*a^2;IC1xx = (3*a)*a^3/12; IC1yy = a*(3*a)^3/12;Ixx1 = IC1xx + (a^2)*A1; Iyy1 = IC1yy + 0*A1;% Ixx1 = (13*a^4)/4 = 3.250 (in^4)% Iyy1 = (9*a^4)/4 = 2.250 (in^4)% square 2A2 = a^2;IC2xx = (a)*a^3/12; IC2yy = a*(a)^3/12;Ixx2 = IC2xx + 0*A2; Iyy2 = IC2yy + 0*A2;% Ixx2 = a^4/12 = 0.083 (in^4)% Iyy2 = a^4/12 = 0.083 (in^4)% lower rectagle 3A3 = 3*a^2;IC3xx = (3*a)*a^3/12; IC3yy = a*(3*a)^3/12;Ixx3 = IC3xx + (a^2)*A3; Iyy3 = IC3yy + 0*A3;% Ixx3 = (13*a^4)/4 = 3.250 (in^4)% Iyy3 = (9*a^4)/4 = 2.250 (in^4)ICxx=Ixx1+Ixx2+Ixx3; ICyy=Iyy1+Iyy2+Iyy3;% ICxx = (79*a^4)/12 = 6.583 (in^4)% ICyy = (55*a^4)/12 = 4.583 (in^4)% products of inertia% x-axis and y-axis are symmetry axes => ICxy=0% z=0 => ICxz=ICyz=0% centroidal polar momentIC=ICxx+ICyy;% IC = ICzz = (67*a^4)/6 = 11.167 (in^4)Problem 4.5Determine the moment of inertia of the area about the y-axis. Solve theproblem in two ways, using rectangular differential elements: a) having athickness dx and b) having a thickness of dy. Given: a = 1 in and b = 4 in.Figure P4.5: Problem 4.5Solution% a = 1; b = 4; iny = b*(1-(x/a)^2);% a) using a thickness dx% area of the differential element parallel to y-axis% dA = y dx = b*(1-(x/a)^2) dx% Iyy = int(x^2 dA) = int(x^2*y dx) where -a<x<aIyy = int(x^2*y,x,-a,a);% a) Iyy = (4*a^3*b)/15 = 1.067 (in^4)% b) using a thickness dyx = a*sqrt(1-y/b);% moment of inertia of the differential element about y-axis% dIyy = (1/12) (dy) (2 x)^3 = (1/12) (dy) (2*a*sqrt(1-y/b))^3% Iyy = int(dIyy) where 0<y<bIyy = int((1/12)*(2*x)^3, y, 0, b);% b) Iyy = (4*a^3*b)/15 = 1.067 (in^4)Problem 4.6Determine the polar moment of inertia of the area about the z-axis passingthrough point O.OFigure P4.6: Problem 4.6SolutionIO= Ixx+ Iyy=π r08+π r08=π r04.Problem 4.7Locate the centroid yCof the cross-sectional area for the angle. Then findthe moment of inertia ICx0about the Cx0centroidal axis.Locate the centroid xCof the cross-sectional area for the angle. Thenfind the moment of inertia ICy0about the Cy0centroidal axis.Given: a = 2 in, b = 6 in, c = 6 in, and d = 2 in.Figure P4.7: Problem