Explanation Given information: The specific weight ( γ ) of the aluminum is 0.100 lb/in . 3 . Assume the acceleration due to gravity ( g ) as 32.2 ft/s 2 Calculation: Show the section 1, section 2 and section 3 of the framing anchor as in Figure 1. Calculate the density ( ρ ) of the aluminum using the relation: ρ = γ g Substitute 0.100 lb/in . 3 for γ and 32.2 ft/s 2 for g . ρ = 0.100 lb/in . 3 32.2 ft/s 2 Calculate the volume of the section 1 ( V 1 ) using the relation: V 1 = A 1 t 1 = 5.9 × 1.8 × 2.2 = 23.364 in . 3 Calculate the volume of the section 2 ( V 2 ) as below: V 2 = A 2 t 2 = π 2 ( 1.1 ) 2 × 1.8 = 3.421 in . 3 Calculate the volume of the section 3 ( V 3 ) as below: V 3 = A 3 t 3 = 1.4 × 1.1 × 1.2 = 1.845 in . 3 Calculate the mass of section 1 ( m 1 ) using the relation: m 1 = ρ V 1 Substitute 23.364 in . 3 for V 1 and 0.100 lb/in . 3 32.2 ft/s 2 for ρ . m 1 = 0.100 lb/in . 3 32.2 ft/s 2 × 23.364 in . 3 = 72.559 × 10 − 3 lb ⋅ s 2 / ft Calculate the mass of section 2 ( m 2 ) using the relation: m 2 = ρ V 2 Substitute 3.421 in . 3 for V 2 and 0.100 lb/in . 3 32.2 ft/s 2 for ρ . m 2 = 0.100 lb/in . 3 32.2 ft/s 2 × 3.421 in . 3 = 10.624 × 10 − 3 lb ⋅ s 2 / ft Calculate the mass of section 3 ( m 3 ) using the relation: m 3 = ρ V 3 Substitute 1.845 in . 3 for V 3 and 0.100 lb/in . 3 32.2 ft/s 2 for ρ . m 3 = 0.100 lb/in . 3 32.2 ft/s 2 × 1.845 in . 3 = 5.7298 × 10 − 3 lb ⋅ s 2 / ft The centroidal products of inertia of I x ' y ' , I y ' z ' and I z ' x ' are zero because the machine is symmetry. Calculate the centroid ( x ¯ 1 ) of the section 1 about x axis as below: x ¯ 1 = − ( 4.5 + 1.4 ) in . 2 = − 2.95 in . × 1 ft 12 in . = − 0.2458 ft Calculate the centroid ( x ¯ 2 ) of the section 2 about x axis as below: x ¯ 2 = − ( ( 4.5 + 1.4 ) in . ) − ( 4 3 π ( 1.1 ) ) in . = − 6.36685 in . × 1 ft 12 in . = − 0.5306 ft Calculate the centroid ( x ¯ 3 ) of the section 3 about x axis as below: x ¯ 3 = − ( 1.4 2 ) in . = − 0.7 in . × 1 ft 12 in . = − 0.05833 ft Calculate the centroid ( y ¯ 1 ) of the section 1 about y axis as below: y ¯ 1 = 1.8 in . 2 = 0.9 in . × 1 ft 12 in . = 0.075 ft Calculate the centroid ( y ¯ 2 ) of the section 2 about y axis as below: y ¯ 2 = 1.8 in . 2 = 0.9 in . × 1 ft 12 in . = 0.075 ft Calculate the centroid ( y ¯ 3 ) of the section 3 about y axis as below: y ¯ 3 = ( 1.8 − 1.1 2 ) in . = 1.25 in . × 1 ft 12 in . = 0.10417 ft Calculate the centroid ( z ¯ 1 ) of the section 1 about z axis as below: z ¯ 1 = ( 0.3 + 1.2 + 0.7 ) in . 2 = 1.1 in . × 1 ft 12 in . = 0.09167 ft Calculate the centroid ( z ¯ 2 ) of the section 2 about z axis as below: z ¯ 2 = ( 0.3 + 1.2 + 0.7 ) in . 2 = 1.1 in . × 1 ft 12 in . = 0.09167 ft Calculate the centroid ( z ¯ 3 ) of the section 3 about z axis as below: z ¯ 3 = ( 0.7 + ( 1.2 2 ) ) in . = 1.3 in . × 1 ft 12 in . = 0.1083 ft Calculate the product moment of inertia for section 1, 2 and 3 as in Table 1. section 1 2 3 m ( lb ⋅ s 2 / ft ) 72.559 × 10 − 3 10.624 × 10 − 3 5.7298 × 10 − 3 x ¯ (ft) − 0

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Moment of Inertia

Inertia can be termed as a resistance offered by a body to change its state. A rigid body initially at rest will apply resistance when acted by an external force that tends to change its state of rest or velocity, or a body initially at motion will provi…

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Chapter 9.6, Problem 9.151P

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The mass products of inertia (Ixy), (Iyz) and (Izx) of the cast aluminum machine component.

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Chapter 9.6, Problem 9.150P

Chapter 9.6, Problem 9.152P

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Determine the shear flow qqq for the given profile when the shear forces acting at the torsional center are Qy=30Q_y = 30Qy=30 kN and Qz=20Q_z = 20Qz=20 kN. Also, calculate qmax⁡q_{\max}qmax and τmax⁡\tau_{\max}τmax. Given:Iy=10.5×106I_y = 10.5 \times 10^6Iy=10.5×106 mm4^44,Iz=20.8×106I_z = 20.8 \times 10^6Iz=20.8×106 mm4^44,Iyz=6×106I_{yz} = 6 \times 10^6Iyz=6×106 mm4^44. Additional parameters:αy=0.5714\alpha_y = 0.5714αy=0.5714,αz=0.2885\alpha_z = 0.2885αz=0.2885,γ=1.1974\gamma = 1.1974γ=1.1974. (Check hint: τmax⁡\tau_{\max}τmax should be approximately 30 MPa.)

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Chapter 9 Solutions

EBK VECTOR MECHANICS FOR ENGINEERS: STA

Ch. 9.1 - 9.1 through 9.4 Determine by direct integration...Ch. 9.1 - 9.1 through 9.4 Determine by direct integration...Ch. 9.1 - 9.1 through 9.4 Determine by direct integration...Ch. 9.1 - 9.1 through 9.4 Determine by direct integration...Ch. 9.1 - 9.5 through 9.8 Determine by direct integration...Ch. 9.1 - 9.5 through 9.8 Determine by direct integration...Ch. 9.1 - Prob. 9.7P Ch. 9.1 - Prob. 9.8P Ch. 9.1 - 9.9 through 9.11 Determine by direct integration...Ch. 9.1 - Prob. 9.10P

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The mass products of inertia ( I x y ) , ( I y z ) and ( I z x ) of the cast aluminum machine component.

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The mass products of inertia (Ixy), (Iyz) and (Izx) of the cast aluminum machine component.

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