Explanation Given information: The thickness of steel sheet is t , the specific weight of steel is γ , the point O is origin, the radius of the semicircle section 2 is a , the side dimension of square plate section 2 is 2 a , the centroidal axis for square section 1 is x ′ , y ′ and z ′ , the centroidal section for semicircular section 2 is x ″ , y ″ and z ″ , the perpendicular distance between the centroidal point P and y -axis is 0 , the perpendicular distance between centroidal point Q and x -axis is 0 . The following figure illustrates different centroidal axis. Figure-(1). Write the expression for the mass of section 1. m 1 = γ g ( 2 a × 2 a × t ) = 4 γ a 2 t g ..... (I) Here, the mass of the square section 1 is m 1 , the specific weight of the steel sheet is γ , the acceleration due the gravity is g , the length of square plate is 2 a , the thickness of square plate is t . Write the expression for the mass of section 2. m 2 = γ g ( π r 2 2 × t ) = γ π r 2 t 2 g ..... (II) Here, the mass of the square section 2 is m 2 , the radius of semicircular plate is r . Write the expression for hypotenuse side A B using Pythagoras theorem in right angled triangle A B C . A B = ( A C ) 2 + ( B C ) 2 ..... (III) Here, the hypotenuse side in right angled triangle A B C is A B , the height of the right angled triangle A B C is A C , the base length of right angled triangle A B C is B C . Write the expression for hypotenuse side O A using Pythagoras theorem in right angled triangle O B A . O A = ( A B ) 2 + ( O B ) 2 ..... (IV) Here, the hypotenuse side in right angled triangle O A B is O A , the height of the right angled triangle O B A is O B , the base length of right angled triangle O B A is A B . Write the expression of position vector of the line joining O A . O A → = x 1 i ^ + y 1 j ^ + z 1 k ^ ..... (V) Here, the vector along the line joining the points O and A is O A → , the unit vector along x direction is i ^ , the unit vector along y direction is j ^ , the unit vector along z direction is k ^ , the coordinate on x -axis is x 1 , the coordinate on y -axis is y 1 , the coordinate on z -axis is z 1 . Write the expression for unit vector along O A → . η ^ OA = O A → O A ..... (VI) Here, the unit vector along O A is η ^ OA . Write the expression of mass moment of inertia of section 1 about x -axis. ( I x ) 1 = m 1 12 ( ( L 1 ) 2 + ( L 1 ) 2 ) + m 1 ( ( L 1 ) 2 + ( L 1 ) 2 ) ..... (VII) Here, the mass moment of inertia of section 1 about x -axis is ( I x ) 1 , the dimension of square plate section 1 is L 1 . Write the expression of mass moment of inertia of section 2 about x -axis. ( I x ) 2 = m 2 4 ( r 2 ) + m 2 ( ( L 1 ) 2 + ( r ) 2 ) ..... (VIII) Here, the mass moment of inertia of section 2 about x -axis is ( I x ) 2 . Write the expression of total mass moment of inertia about x -axis. I x = ( I x ) 1 + ( I x ) 2 ..... (IX) Here, the total mass moment of inertia about x -axis is I x . Write the expression of mass moment of inertia of section 1 about y -axis. ( I y ) 1 = m 1 12 ( L 1 2 ) + m 1 ( r 2 ) ..... (X) Here, the mass moment of inertia of section 1 about y -axis is ( I y ) 1 . Write the expression of mass moment of inertia of section 2 about y -axis. ( I y ) 2 = m 2 ( L 1 2 ) 2 ( 1 2 − 16 9 π 2 ) + m 2 ( ( 4 r 3 π ) 2 + L 1 2 2 ) ..... (XI) Here, the mass moment of inertia of section 2 about y -axis is ( I y ) 2 . Write the expression of total mass moment of inertia about y -axis. I y = ( I y ) 1 + ( I y ) 2 ..... (XII) Write the expression of mass moment of inertia of section 1 about z -axis. ( I z ) 1 = m 1 L 1 2 12 + m 1 ( L 1 2 ) 2 ..... (XIII) Here, the mass moment of inertia of section 1 about z -axis is ( I z ) 1 . Write the expression of mass moment of inertia of section 2 about z -axis. ( I z ) 2 = m 2 ( L 1 2 ) 2 ( 1 4 − 16 9 π 2 ) + m 2 ( ( 4 r 3 π ) 2 + ( 2 a ) 2 ) ..... (XIV) Here, the mass moment of inertia of section 2 about z -axis is ( I z ) 2 . Write the expression of total mass moment of inertia about z -axis. I z = ( I z ) 1 + ( I z ) 2 ..... (XV) Here, the total mass moment of inertia about z -axis is I z . From, the symmetry in above figure about x ′ , y ′ and z ′ . ( I x ′ y ′ ) 1 = ( I y ′ z ′ ) 1 = ( I z ′ x ′ ) 1 = 0 ..... (XVI) Write the expression for product of mass moment of inertia in x - y plane. ( I xy ) = ∑ i = 1 i = 2 ( ( I x ′ y ′ ) i + m i x i ¯ y i ¯ ) ( I xy ) = ( ( I x ′ y ′ ) 1 + m 1 x 1 ¯ y 1 ¯ ) + ( ( I x ′ y ′ ) 2 + m 2 x 2 ¯ y 2 ¯ ) ..... (XVII) Here, the product mass moment of inertia is ( I xy ) , the subscript limit is i , the perpendicular distance between x ′ -axis and x axis is x 1 ¯ , the perpendicular distance between y ′ axis and y -axis is is y 1 ¯ , the perpendicular distance between x ″ -axis and x axis is x 2 ¯ , the perpendicular distance between y ″ axis and y -axis is is y 2 ¯ . Write the expression for product mass moment of inertia in y - z plane. ( I yz ) = ∑ i = 1 i = 2 ( ( I y ′ z ) i + m i y i ¯ z i ¯ ) ( I yz ) = ( ( I y ′ z ′ ) 1 + m 1 y 1 ¯ z 1 ¯ ) + ( ( I y ′ z ′ ) 2 + m 2 y 2 ¯ z 2 ¯ ) ..... (XVIII) Here, the product mass moment of inertia is ( I yz ) , the subscript limit is i , the perpendicular distance between y ′ axis and y -axis is y 1 ¯ , the perpendicular distance between z ′ -axis and z -axis is z 1 ¯ , the perpendicular distance between y ″ axis and y -axis is y 2 ¯ , the perpendicular distance between z ″ -axis and z -axis is z 2 ¯ , Write the expression for product mass moment of inertia in z - x plane. ( I zx ) = ∑ i = 1 i = 2 ( ( I z' x ′ ) i + m i x i ¯ z i ¯ ) ( I zx ) = ( ( I z ′ x ′ ) 1 + m 1 x 1 ¯ z 1 ¯ ) + ( ( I z ′ x ′ ) 2 + m 2 x 2 ¯ z 2 ¯ ) ..... (XIX) Here, the product mass moment of inertia in z - x plane is ( I zx ) , the perpendicular distance between z ′ -axis and z -axis is z 1 ¯ , the perpendicular distance between z ″ -axis and z -axis is z 2 ¯ , the perpendicular distance between x ′ -axis and x axis x 1 ¯ , the perpendicular distance between x ″ -axis and x axis x 2 ¯ . Write the expression of mass moment of inertia about O A . I OA = I x λ x 2 + I y λ y 2 + I z λ z 2 − 2 I xy λ x λ y − 2 I yz λ z λ y − 2 I xz λ x λ z ..... (XX) Here, the mass moment of inertia about O A is I OA , the coefficient of i ^ in unit vector η O A is λ x , the coefficient of j ^ in unit vector η O A is λ y , the coefficient of k ^ in unit vector η O A is λ z . Calculation: Substitute a for x 1 , 2 a for y 1 , a for z 1 in Equation (V). O A → = a i ^ + 2 a j ^ + a k ^ Substitute a for A C , a for B C in Equation (III). A B = ( a ) 2 + ( a ) 2 = 2 a 2 = 2 a Substitute 2 a for A B , 2 a for O B in Equation (IV). O A = ( 2 a ) 2 + ( 2 a ) 2 = 2 a 2 + 4 a 2 = 6 a Substitute a i ^ + 2 a j ^ + a k ^ for O A → , 6 a for O A in Equation (VI). η ^ OA = a i ^ + 2 a j ^ + a k ^ 6 a = a ( i ^ + 2 j ^ + k ^ ) 6 a = 1 6 i ^ + 2 6 j ^ + 1 6 k ^ Substitute 2 a for L 1 in Equation (VII). ( I x ) 1 = m 1 12 ( ( 2 a ) 2 + ( 2 a ) 2 ) + m 1 ( ( a ) 2 + ( a ) 2 ) = 2 m 1 a 2 3 + 2 m 1 a 2 = 8 m 1 a 2 3 Substitute 2 a for L 1 , a for r in Equation (VIII)

12th Edition

ISBN: 9781260265521

Author: BEER

Publisher: MCG

See similar textbooks

Concept explainers

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…

Videos

Question

Chapter B, Problem B.58P

To determine

The mass moment of inertia of the component with respect to the line joining the origin O and point A.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter B, Problem B.57P

Chapter B, Problem B.59P

Students have asked these similar questions

الثانية Babakt Momentum equation for Boundary Layer S SS -Txfriction dray Momentum equation for Boundary Layer What laws are important for resolving issues 2 How to draw. 3 What's Point about this.

R αι g The system given on the left, consists of three pulleys and the depicted vertical ropes. Given: ri J₁, m1 R = 2r; απ r2, J2, m₂ m1; m2; M3 J1 J2 J3 J3, m3 a) Determine the radii 2 and 3.

B: Solid rotating shaft used in the boat with high speed shown in Figure. The amount of power transmitted at the greatest torque is 224 kW with 130 r.p.m. Used DE-Goodman theory to determine the shaft diameter. Take the shaft material is annealed AISI 1030, the endurance limit of 18.86 kpsi and a factor of safety 1. Which criterion is more conservative? Note: all dimensions in mm. 1 AA Motor 300 Thrust Bearing Sprocket 100 9750 เอ

Chapter B Solutions

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS

Ch. B - A thin plate with a mass m is cut in the shape of...Ch. B - Prob. B.2P Ch. B - Prob. B.3P Ch. B - Prob. B.4P Ch. B - A piece of thin, uniform sheet metal is cut to...Ch. B - Prob. B.6P Ch. B - Prob. B.7P Ch. B - Prob. B.8P Ch. B - Prob. B.9P Ch. B - Prob. B.10P

Ch. B - Prob. B.11P Ch. B - Prob. B.12P Ch. B - Determine by direct integration the mass moment of...Ch. B - Prob. B.14P Ch. B - A thin, rectangular plate with a mass m is welded...Ch. B - A thin steel wire is bent into the shape shown....Ch. B - Prob. B.17P Ch. B - Prob. B.18P Ch. B - Prob. B.19P Ch. B - Prob. B.20P Ch. B - Prob. B.21P Ch. B - Prob. B.22P Ch. B - Prob. B.23P Ch. B - Prob. B.24P Ch. B - Prob. B.25P Ch. B - Prob. B.26P Ch. B - Prob. B.27P Ch. B - Prob. B.28P Ch. B - Prob. B.29P Ch. B - Prob. B.30P Ch. B - Prob. B.31P Ch. B - Determine the mass moments of inertia and the...Ch. B - Prob. B.33P Ch. B - Prob. B.34P Ch. B - Prob. B.35P Ch. B - Prob. B.36P Ch. B - Prob. B.37P Ch. B - Prob. B.38P Ch. B - Prob. B.39P Ch. B - Prob. B.40P Ch. B - Prob. B.41P Ch. B - Prob. B.42P Ch. B - Prob. B.43P Ch. B - Prob. B.44P Ch. B - A section of sheet steel 2 mm thick is cut and...Ch. B - Prob. B.46P Ch. B - Prob. B.47P Ch. B - Prob. B.48P Ch. B - Prob. B.49P Ch. B - Prob. B.50P Ch. B - Prob. B.51P Ch. B - Prob. B.52P Ch. B - Prob. B.53P Ch. B - Prob. B.54P Ch. B - Prob. B.55P Ch. B - Determine the mass moment ofinertia of the steel...Ch. B - Prob. B.57P Ch. B - Prob. B.58P Ch. B - Determine the mass moment of inertia of the...Ch. B - Prob. B.60P Ch. B - Prob. B.61P Ch. B - Prob. B.62P Ch. B - Prob. B.63P Ch. B - Prob. B.64P Ch. B - Prob. B.65P Ch. B - Prob. B.66P Ch. B - Prob. B.67P Ch. B - Prob. B.68P Ch. B - Prob. B.69P Ch. B - Prob. B.70P Ch. B - For the component described in the problem...Ch. B - Prob. B.72P Ch. B - For the component described in the problem...Ch. B - Prob. B.74P

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

The mass moment of inertia of the component with respect to the line joining the origin O and point A .

Solution Summary: The author explains the mass moment of inertia of the component with respect to the line joining the origin O and point A.

Concept explainers

Videos

The mass moment of inertia of the component with respect to the line joining the origin O and point A.

Want to see the full answer?

Chapter B Solutions