Vector Mechanics for Engineers: Statics and Dynamics
Vector Mechanics for Engineers: Statics and Dynamics
12th Edition
ISBN: 9781259638091
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Textbook Question
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Chapter 16.2, Problem 16.138P

Solve Prob. 16.137 when θ = 90°.

16.137    In the engine system shown, l = 250 mm and b = 100 mm. The connecting rod BD is assumed to be a 1.2-kg uniform slender rod and is attached to the 1.8-kg piston P. During a test of the system, crank AB is made to rotate with a constant angular velocity of 600 rpm clockwise with no force applied to the face of the piston. Determine the forces exerted on the connecting rod at B and D when θ = 180°. (Neglect the effect of the weight of the rod.)

Chapter 16.2, Problem 16.138P, Solve Prob. 16.137 when  = 90. 16.137In the engine system shown, l = 250 mm and b = 100 mm. The

Fig. P16.137

Expert Solution & Answer
Check Mark
To determine

The forces exerted on the connecting rod at B and D when θ=90°.

Answer to Problem 16.138P

The forces exerted on the connecting rod at B is 525N_.

The forces exerted on the connecting rod at D is 322N_.

Explanation of Solution

Given information:

The length of the rod BD is l=250mm.

The length of the rod AB is b=100mm.

The mass of the rod BD is mBD=1.2kg.

The mass of the piston P is mP=1.8kg.

The angular velocity of AB is ωAB=600rpm.

Calculation:

Consider the acceleration due to gravity g=9.81m/s2.

Calculate the angular velocity in rad/s as shown below.

ωAB=600rpm×2π60rad/s1rpm=62.832rad/s

Sketch the Free body Diagram of the system as shown in Figure 1.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.2, Problem 16.138P , additional homework tip  1

Refer to Figure 1.

Calculate the distance (x) as shown below.

x2=l2b2x=l2b2

Substitute 250mm for l and 100mm for b.

x=25021002=52,500=229.1mm×1m1,000mm=0.2291m

Calculate the position vectors as shown below.

Position of B with respect to A.

rB/A=(100mm×1m1,000mm)j=(0.1m)j

Position of D with respect to B.

rD/B=(0.2291m)i(100mm×1m1,000mm)j=(0.2291m)i(0.1m)j

Position of mass center G with respect to D.

rG/D=12((0.2291m)i+(100mm×1m1,000mm)j)=12((0.2291m)i+(0.1m)j)=(0.11455m)i+(0.05m)j

Calculate the velocity at B (vB) as shown below.

vB=ωAB×rB/A

Substitute (62.832rad/s)k for ωAB and (0.1m)j for rB/A.

vB=62.832k×0.1j=(6.2832m/s)i

Calculate the velocity at D (vD) as shown below.

vD=vB+ωBD×rB/D

Substitute (6.2832m/s)i for vB and (0.2291m)i(0.1m)j for rB/D.

vDi=6.2832i+ωBDk×(0.2291i0.1j)=6.2832i0.2291ωBDj+0.1ωBDi=(6.2832+0.1ωBD)i0.2291ωBDj

Resolving the i and j components as shown below.

For j component.

0.2291ωBD=0ωBD=0

For i component.

6.2832+0.1ωBD=vD

Substitute 0 for ωBD.

vD=6.2832m/s

Consider that the angular acceleration as αAB=0.

Calculate the acceleration at B (aB) as shown below.

aB=αAB×rB/AωAB2rB/A

Substitute 0 for αAB, 62.832rad/s for ωAB, and (0.1m)j for rB/A.

aB=0(62.832rad/s)2×(0.1m)j=(394.786m/s2)j

Calculate the acceleration (aD) and the angular acceleration (αBD) using the relation as shown below.

aD=aB+αBD×rB/DωBD2rD/B

Substitute (394.786m/s2)j for aB, 0 for ωBD, and (0.2291m)i(0.1m)j for rB/D.

aDi=394.786j+αBDk×(0.2291i0.1j)0=394.786j0.2291αBDj+0.1αBDi=(394.7860.2291αBD)j+0.1αBDi

Resolving i and j components as shown below.

For j component.

394.7860.2291αBD=00.2291αBD=394.786αBD=1,723rad/s2

For i component.

0.1αBD=aD

Substitute 1,723rad/s2 for αBD.

aD=0.1×(1,723)=172.3m/s2

Calculate the acceleration of mass center G of bar BD (aG) as shown below.

aG=aD+αBD×rG/DωBD2rG/D

Substitute (172.3m/s2)i for aD, (1,723rad/s2)k for αBD, 0 for ωBD, and (0.11455m)i+(0.05m)j for rB/D.

aGi=172.3i+(1,723)k×(0.11455i+0.05j)0=172.3i197.3695j+86.15i=86.15i197.3695j

Resolving the components as shown below.

aG=86.15m/s2

Calculate the mass moment of inertia for BD (I¯BD) as shown below.

I¯BD=112mBDl2

Substitute 1.2kg for mBD and 250mm for l.

I¯BD=112×(1.2kg)×(250mm×1m1,000mm)2=0.00625kgm2

Sketch the Free Body Diagram of the piston with the bar BD as shown in Figure 2.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.2, Problem 16.138P , additional homework tip  2

Refer to Figure 2.

Apply the Equilibrium of forces along x direction as shown below.

Fx=maxBx=mPaD+(mBDaG)x

Substitute 1.8kg for mP, 172.3m/s2 for aD, 1.2kg for mBD, and 86.15m/s2 for aG.

Bx=1.8×(172.3)+1.2×(86.15)=413.52N

Apply the Equilibrium of moment about B as shown below.

MB=IGα+mad(xN)k=I¯BDαBDk+rD/B(mPaD)+rG/B(mBDaG)

Substitute 0.2291m for x, 0.00625kgm2 for I¯BD, 1,723rad/s2 for αBD, (0.11455m)i(0.05m)j for rG/B, 0.1m for rD/B, 1.8kg for mP, 172.3m/s2 for aD, 1.2kg for mBD, and (86.15i197.3695j)m/s2 for aG.

(0.2291N)k=[(0.00625×(1,723))k+0.1×(1.8×(172.3))k+(0.11455i0.05j)(86.15i197.3695j)×1.2]0.2291Nk=10.76875k31.014k+27.1304k5.169k0.2291N=19.82135N=86.518N

Apply the Equilibrium of forces along y direction as shown below.

Fy=maxN+By=(mBDaG)y

Substitute 86.518N for N, 1.2kg for mBD, and 197.3695m/s2 for aG.

86.518+By=1.2×(197.3695)By=323.36N

Calculate the force acting at B as shown below.

B=Bx2+By2

Substitute 413.52N for Bx and 323.6N for By.

B=(413.52)2+(323.6)2=275,715.7504=525N

Hence, the forces exerted on the connecting rod at B is 525N_.

Sketch the Free Body Diagram of the piston as shown in Figure 3.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.2, Problem 16.138P , additional homework tip  3

Refer to Figure 3.

Calculate the force acting on the rod at D as shown below.

FD=maD+Nj=mPaDi

Substitute 86.518N for N, 1.8kg for mP, and 172.3m/s2 for aD.

D+86.518j=1.8×(172.3)iD=310.14i86.518j

Calculate the magnitude of force at D as shown below.

D=(310.14)2+(86.518)2=103,672.1839=322N

Therefore, the forces exerted on the connecting rod at D is 322N_.

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Chapter 16 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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