Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics
Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics
11th Edition
ISBN: 9781259639272
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 9.1, Problem 9.20P
To determine

The moment of inertia (Iy) and radius of gyration (ky) of the shaded area with respect to y axis.

Expert Solution & Answer
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Answer to Problem 9.20P

The moment of inertia (Iy) and radius of gyration (ky) of the shaded area with respect to y axis is 0.61345a3h_ and 1.299a_ respectively.

Explanation of Solution

Given information:

The equation of the upper segment is y=mx+b.

The equation of the lower curve is y=csinkx.

Calculation:

Draw the subject area with vertical differential strip element as in Figure (1).

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 9.1, Problem 9.20P

Consider the curve equation as y=csinkx.

Apply the boundary condition at y1.

Condition 1:

At x=2a and y=0.

0=csink(2a)

Since sinθ will be zero when angle at 180°(π).

2ak=πk=π2a

Condition 2:

At x=a and y=h.

h=csink(a)

Substitute π2a for k.

h=csinπ2a(a)h=csinπ2h=c(1)(sinπ2=1)c=h

Consider the upper curve equation y=mx+b.

Apply the boundary condition at y2.

Condition 1:

At x=a and y=2h.

2h=ma+b (1)

Condition 2:

At x=2a and y=0.

0=m2a+b2ma+b=0 (2)

Subtract the equation (1) and (2).

2h0=ma+b2mab2h=mam=2ha

Substitute 2ha for m in equation (2).

2(2ha)a+b=04h+b=0b=4h

Rewrite the lower equation.

y=csinkx

Substitute y1 for y, h for c and π2a for k.

y1=hsinπ2ax

Rewrite the upper equation.

y=mx+b

Substitute y2 for y, 2ha for m and 4h for b.

y2=2hax+4h=2ha(x+2a)

Calculate the area of the given figure (A) using the formula:

A=dA=(y2y1)dx

Substitute hsinπ2ax for y1 and 2ha(x+2a) for y2.

A=[2ha(x+2a)(hsinπ2ax)]dx=a2ah[2a(x+2a)(sinπ2ax)]dx=h[(2a(x+2a)22(1))(cosπ2axπ2a)]a2a

=h[(1a(x+2a)2)+2aπcosπ2ax]a2a=h[(1a(2a+2a)2+1a(a+2a)2)+(2aπcosπ2a2a(2aπcosπ2aa))]=h[(1a(a+2a)2)+2aπ(1)0]

=h[(1a(a+2a)2)2aπ]=ah[(11(1+2)22π)]=ah[12π]=0.36338ah

Calculate the moment of inertia about y axis using the formula:

Iy=dIy=x2dA=a2ax2(y2y1)dx

Substitute hsinπ2ax for y1 and 2ha(x+2a) for y2.

Iy=a2a[x2(2ha(x+2a))(hsinπ2ax)]dx=a2ah[2a(x3+2ax2)x2sinπ2ax]dx (3)

Take u=x2 from the equation (3).

Differentiate the u equation with respect to x.

du=2xdx

Take dv=sinπ2axdx from the equation (3).

Integrate the above equation with respect to x.

dv=sinπ2axdxv=cosπ2axπ2av=2aπcosπ2ax

Consider the second term from the equation (3).

x2sinπ2axdx

Integrate the above equation by UV method with respect to x.

Substitute 2aπcosπ2ax for sinπ2ax.

Since uv=udvvdu.

x2sinπ2axdx=x2(2aπcosπ2ax)(2aπcosπ2ax)(2xdx)=2ax2πcosπ2ax+4aπ[x(2aπsinπ2ax)+2aπsinπ2ax(1)]=2ax2πcosπ2ax+8a2π2xsinπ2ax+4aπ(2aπ×2aπcosπ2ax)=2ax2πcosπ2ax+8a2π2xsinπ2ax+16a3π3cosπ2ax

Substitute 2ax2πcosπ2ax+8a2π2xsinπ2ax+16a3π3cosπ2ax for x2sinπ2axdx in the equation (1).

Iy=h[2a(x44+2ax33)(2ax2πcosπ2ax+8a2π2xsinπ2ax+16a3π3cosπ2ax)]a2a={h[2a((2a)44+2a(2a)33)(2a(2a)2πcosπ2a2a+8a2π22asinπ2a2a+16a3π3cosπ2a2a)]h[2a(a44+2aa33)(2aa2πcosπ2aa+8a2π2asinπ2aa+16a3π3cosπ2aa)]}=h{[2a(16a44+16a43)(8a3π(1)+16a3π3)]}h[(2a(a44+2a43))8a3π2]=0.61345a3h

Calculate the radius of gyration (ky) along y axis using the formula:

ky=IyA

Substitute 0.36338ah for A and 0.61345a3h for Ix.

ky=0.61345a3h0.36338ah=1.299a

Thus, the moment of inertia (Iy) and radius of gyration (ky) of the shaded area with respect to y axis are 0.61345a3h_ and 1.299a_ respectively.

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Chapter 9 Solutions

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics

Ch. 9.1 - Prob. 9.11PCh. 9.1 - Prob. 9.12PCh. 9.1 - Prob. 9.13PCh. 9.1 - 9.12 through 9.14 Determine by direct integration...Ch. 9.1 - Prob. 9.15PCh. 9.1 - Prob. 9.16PCh. 9.1 - Prob. 9.17PCh. 9.1 - Prob. 9.18PCh. 9.1 - Prob. 9.19PCh. 9.1 - Prob. 9.20PCh. 9.1 - Prob. 9.21PCh. 9.1 - Prob. 9.22PCh. 9.1 - Prob. 9.23PCh. 9.1 - 9.23 and 9.24 Determine the polar moment of...Ch. 9.1 - Prob. 9.25PCh. 9.1 - Prob. 9.26PCh. 9.1 - Prob. 9.27PCh. 9.1 - Prob. 9.28PCh. 9.1 - Prob. 9.29PCh. 9.1 - Prove that the centroidal polar moment of inertia...Ch. 9.2 - Prob. 9.31PCh. 9.2 - Prob. 9.32PCh. 9.2 - Prob. 9.33PCh. 9.2 - Prob. 9.34PCh. 9.2 - Prob. 9.35PCh. 9.2 - Prob. 9.36PCh. 9.2 - Prob. 9.37PCh. 9.2 - Prob. 9.38PCh. 9.2 - Prob. 9.39PCh. 9.2 - Prob. 9.40PCh. 9.2 - Prob. 9.41PCh. 9.2 - 9.41 through 9.44 Determine the moments of inertia...Ch. 9.2 - Prob. 9.43PCh. 9.2 - Prob. 9.44PCh. 9.2 - 9.45 and 9.46 Determine the polar moment of...Ch. 9.2 - Prob. 9.46PCh. 9.2 - Prob. 9.47PCh. 9.2 - 9.47 and 9.48 Determine the polar moment of...Ch. 9.2 - 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