Connect 2 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics
Connect 2 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics
11th Edition
ISBN: 9780077687298
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell
Publisher: McGraw-Hill Education
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Chapter 19.5, Problem 19.147P
To determine

Show that if a periodic force of magnitude P=Pmsinωft is applied to the element, the amplitude of the fluctuating force transmitted to the foundation is Fm=Pm1+[2(ccc)(ωfωn)]2[1(ωfωn)2]2+[2(ccc)(ωfωn)]2.

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Answer to Problem 19.147P

The amplitude of the fluctuating force transmitted to the foundation is Fm=Pm1+[2(ccc)(ωfωn)]2[1(ωfωn)2]2+[2(ccc)(ωfωn)]2 and hence it is proved.

Explanation of Solution

Given information:

The magnitude of periodic force (P) is P=Pmsinωft.

Calculation:

The expression for the motion of the machine (x) as follows:

x=xmsin(ωftϕ)

Differentiate the above equation with respect to time ‘t’.

x˙=xmωfcos(ωftϕ)

Calculate the force transmitted (Fs) to the foundation for the springs using the relation:

Fs=kx

Substitute xmsin(ωftϕ) for x.

Fs=kxmsin(ωftϕ)

Calculate the force transmitted (Fd) to the foundation for the dashpot using the relation:

Fd=cx˙

Substitute xmωfcos(ωftϕ) for x˙.

Fd=cxmωfcos(ωftϕ)

Calculate the total force transmitted (Ft) to the foundation using the relation:

Ft=Fs+Fd

Substitute kxmsin(ωftϕ) for Fs and cxmωfcos(ωftϕ) for Fd.

Ft=kxmsin(ωftϕ)+cxmωfcos(ωftϕ)=xm[ksin(ωftϕ)+cωfcos(ωftϕ)]

Since Asiny+Bcosy=A2+B2sin(y+ψ).

Write the total force transmitted (Ft) to the foundation,

Ft=[xmk2+(cωf)2]sin(ωftϕ+ψ)

The expression for the natural circular frequency (ωn) as follows:

ωn=kmωn2=kmk=ωn2m

The expression for the critical damping coefficient (cc) as follows:

cc=2mωnm=cc2ωn

The expression for the amplitude of the fluctuating force transmitted (Fm) to the foundation from the above equation as follows:

Fm=xmk2+(cωf)2 (1)

The expression for the amplitude of motion to the foundation (xm) as follows:

xm=Pmk(1(ωfωn)2)2+(2(ccc)ωfωn)2

Substitute Pmk(1(ωfωn)2)2+(2(ccc)ωfωn)2 for xm in equation (1).

Fm=Pmkk2+(cωf)2(1(ωfωn)2)2+(2(ccc)ωfωn)2=Pmk(k)1+(cωfk)2(1(ωfωn)2)2+(2(ccc)ωfωn)2=Pm1+(cωfk)2(1(ωfωn)2)2+(2(ccc)ωfωn)2

Substitute ωn2m for k and cc2ωn for m.

Fm=Pm1+(cωfωn2(cc2ωn))2(1(ωfωn)2)2+(2(ccc)ωfωn)2=Pm1+(cωfωn(cc2))2(1(ωfωn)2)2+(2(ccc)ωfωn)2=Pm1+[2(ccc)(ωfωn)](1(ωfωn)2)2+(2(ccc)ωfωn)2

Thus, the amplitude of the fluctuating force transmitted to the foundation is Fm=Pm1+[2(ccc)(ωfωn)]2[1(ωfωn)2]2+[2(ccc)(ωfωn)]2 and hence it is proved.

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Chapter 19 Solutions

Connect 2 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics

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