Copy of ps4 solutions PDF

Title	Copy of ps4 solutions
Course	Biomechanics I
Institution	University of California Irvine
Pages	4
File Size	303 KB
File Type	PDF
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Summary

Solutions to the fourth problem set....

Description

Problem Set #4

BME 110A

1. . Locate the centroid of the plane area shown.

Then

A, mm2

x , mm

y , mm

xA, mm3

yA, mm3

1

200  150  30000

100

250

 30000000

6750000

2

400  300  120000

200

150

24000000

18000000



150000

21000000

24750000

X 

xA 21000000 mm  150000 A

or X  140.0 mm

Y 

 yA 24750000 mm  150000 A

or Y  165.0 mm

Page 1 of 4

Problem Set #4

BME 110A 2.

Locate the centroid of the plane area shown.

1 2

3



 4 

A,in 2

x ,in.

y ,in.

xA,in3

yA,in3

17  9  153

8.5

4.5

1300.5

688.5

  4.5  15.9043 8  2

 4



 6  2   28.274

4  4.5 3

 6.0901 9 

4  4.5

10.5465

3

 7.0901  96.857

112.761

 298.19

182.466

905.45

393.27

6.4535

108.822

Then

X 

xA 905.45  A 108.822

and

Y 

yA 393.27   A 108.22

or or

Page 2 of 4

X  8.32 in. Y  3.61 in.

Problem Set #4

BME 110A 3.

For the beam and loading shown, determine (a) the magnitude and location of the resultant of the distributed load, (b) the reactions at the beam supports.

(a) Note that in the free-body diagram: R1 

1 4.2 m 600 N/m   1260 N, 2

1 2

and R2   4.2 m 240 N/m  504 N

Then for the equivalence of the systems of forces: Fy :

R  R1  R2  1260  504  1764 N

M A :

1   2      x 1764 N    2 + 4.2  m 1260 N  + 2 + 4.2 m 504 N   3.8000 m 3 3        

R  1764 N

or x  3.80 m  (b) Equilibrium:

Fx  0:

Ax  0

Fy  0:

Ay  1764  0

A  1764 N     0:

M A  3.80 m 1764 N   0  M A  6.70 kN m



Page 3 of 4



Problem Set #4

BME 110A

4. Determine the reactions at the beam supports for the given loading.

lb   R1   240  4.8 ft   1152 lb ft   R2 

1 lb   180   3.6 ft   324 lb 2 ft 

Equilibrium:

 Fx  0:

Ax  0

 F y  0:

A y  1152 lb + 324 lb  0

A  828 lb 

A y  828.00 lb

M A  0:

M A   2.4 ft 1152 lb   6 ft  324 lb   0

M A  820.80 lb ft

M A  821 lb  ft

Page 4 of 4

...