Problem 4-141

Replace the loading by an equivalent force and couple moment acting at point O.

Units Used:

kN 103N:=

Given:

w 6kNm

:=

F 15kN:=

M 500kN m:=

a 7.5m:=

b 4.5m:=

Solution:

FR12

w a b+( ) F+:=

FR 51.0 kN=

MR M12

w a

23

a

12

w b

ab3

+

F a b+( ):=

MR 914 kN m=

Problem 4-142

Replace the loading by a single resultant force, and specify the location of the force on the beammeasured from point O.

Units Used:

kN 103N:=

Given:

w 6kNm

:=

F 15kN:=

M 500kN m:=

a 7.5m:=

b 4.5m:=

Solution:

Initial Guesses: FR 1kN:= d 1m:=

Given

FR12

w a b+( ) F+=

FR d M12

w a

23

a

12

w b

ab3

+

F a b+( )=

FR

d

Find FR d,( ):= FR 51 kN= d 17.922 m=

Problem 4-143

The column is used to support the floor which exerts a force P on the top of the column.The effect of soil pressure along its side is distributed as shown. Replace this loading by anequivalent resultant force and specify where it acts along the column, measured from itsbase A.

Given:

P 30kN:=

w1 800Nm

:=

w2 2000Nm

:=

h 3m:=

Solution:

FRx w1 h12

w2 w1( ) h+:=

FRx 4.20 kN=

FRy P:=

FR FRx2 P2+:= FR 30.29 kN=

atanP

FRx

:= 82.0 deg=

FRx y12

w2 w1( ) h h3 w1 hh2

+=

y16

h2w2 2 w1+

FRx:= y 1.29 m=

Problem 4-144

Replace the loading by an equivalent force and couple moment at point O.

Units Used:

kN 103N:=

Given:

w1 15kNm

:=

w2 5kNm

:=

d 9m:=

Solution :

FR12

w1 w2+( ) d:= FR 90 kN=

MRO w2 dd2

12

w1 w2( ) d d3+:= MRO 338 kN m=

Problem 4-145

Replace the distributed loading by an equivalent resultant force, and specify its location on the beam, measured from the pin at C.

Units Used:

kN 103N:=

Given:

w 8000Nm

:=

a 5m:=

b 5m:=

30deg:=

Solution :

FR w aw b

2+:= FR 60 kN=

FR x w aa2

w b

2a

b3

+

+=

xw a

a2

w b

2a

b3

+

+

FR:= x 3.89 m=

Problem 4-146

The beam supports the distributed load caused by the sandbags. Determine the resultant forceon the beam and specify its location measured from point A.

Units Used:

kN 103N:=

Given:

w1 1.5kNm

:= a 3m:=

w2 1kNm

:= b 3m:=

w3 2.5kNm

:= c 1.5m:=

Solution:

FR w1 a w2 b+ w3 c+:= FR 11.25 kN=

MA w1 aa2

w2 b ab2

+

+ w3 c a b+c2

+

+:=

MA 45.563 kN m= dMAFR

:= d 4.05 m=

Problem 4-147

Determine the length b of the triangular load and its position a on the beam such that theequivalent resultant force is zero and the resultant couple moment is M clockwise.

Units Used:

kN 103N:=

Given:

w1 4kNm

:=

w2 2.5kNm

:=

M 8kN m:=

c 9m:=

Solution:

Initial Guesses: a 1m:= b 1m:=

Given1

2w1 b

12

w2 c+ 0=

12

w1 b a2 b3

+

12

w2 c2 c3

M=

a

b

Find a b,( ):=a

b

1.539

5.625

m=

Problem 4-148

Replace the distributed loading by an equivalent resultant force and specify its location, measuredfrom point A.

Units Used:

kN 103N:=

Given:

w1 800Nm

:=

w2 200Nm

:=

a 2m:=

b 3m:=

Solution:

FR w2 b w1 a+12

w1 w2( ) b+:= FR 3.10 kN=

x FR w1 aa2

12

w1 w2( ) b a b3+

+ w2 b ab2

+

+=

xw1 a

a2

12

w1 w2( ) b a b3+

+ w2 b ab2

+

+

FR:= x 2.06 m=

Problem 4-149

The distribution of soil loading on the bottom of a building slab is shown. Replace this loading byan equivalent resultant force and specify its location, measured from point O.

Units Used:

kN 103N:=

Given:

w1 500Nm

:=

w2 3000Nm

:=

w3 1000Nm

:=

a 4m:=

b 3m:=

Solution :

FR w1 a12

w2 w1( ) a+ 12 w2 w3( ) b+ w3 b+:= FR 13 kN=

FR d w1 aa2

12

w2 w1( ) a 2a3+12

w2 w3( ) b a b3+

+ w3 b ab2

+

+=

d3 w3 b a 2 w3 b

2+ w1 a

2+ 2 a2 w2+ 3 b w2 a+ w2 b

2+

6FR:= d 3.76 m=

Problem 4-150

The beam is subjected to the distributed loading. Determine the length b of the uniform loadand its position a on the beam such that the resultant force and couple moment acting on thebeam are zero.

Given:

w1 40kNm

:=

w2 60kNm

:=

c 3m:=

d 2m:=

Solution:

InitialGuesses: a 1m:= b 1m:=

Given

12

w2 d w1 b 0=12

w2 d cd3

+

w1 b ab2

+

0=

a

b

Find a b,( ):=a

b

2.92

1.50

m=

Problem 4-151

Replace the loading by an equivalent resultant force and specify its location on the beam,measured from point B.

Units Used:

kN 103N:=

Given:

w1 8000Nm

:=

w2 5000Nm

:=

a 4m:=

b 9m:=

Solution:

FR12

a w112

w1 w2( ) b+ w2 b+:= FR 74.5 kN=

FR x12

a w1a3

12

w1 w2( ) b b3+ w2 bb2

+=

x

12

a w1a3

12

w1 w2( ) b b3+ w2 bb2

+

FR:= x 2.975 m=

( to the right of B )

Problem 4-152

Replace the distributed loading by an equivalent resultant force and specify where its line of actionintersects member AB, measured from A.

Given:

w1 200Nm

:=

w2 100Nm

:=

w3 200Nm

:=

a 5m:=

b 6m:=

Solution:

FRx w3 a:= FRx 1000 N=

FRy1

2w1 w2+( ) b:= FRy 900 N=

y FRx w3 aa2

w2 bb2

12

w1 w2( ) b b3=

yw3 a

a2

w2 bb2

12

w1 w2( ) b b3FRx

:= y 0.1 m=

Problem 4-153

Replace the distributed loading by an equivalent resultant force and specify where its line ofaction intersects member BC, measured from C.

Given:

w1 200Nm

:=

w2 100Nm

:=

w3 200Nm

:=

a 5m:=

b 6m:=

Solution:

FRx w3 a:= FRx 1000 N=

FRy1

2w1 w2+( ) b:= FRy 900 N=

x FRy w3 aa2

w2 bb2

+12

w1 w2( ) b 2 b3+=

xw3 a

a2

w2 bb2

+12

w1 w2( ) b 2 b3+FRy

:= x 0.556 m=

Problem 4-154

Replace the loading by an equivalent resultant force and couple moment acting at point O.

Units Used:

kN 103N:=

Given:

w1 7.5kNm

:=

w2 20kNm

:=

a 3m:=

b 3m:=

c 4.5m:=

Solution:

FR12

w2 w1( ) c w1 c+ w1 b+ 12 w1 a+:= FR 95.6 kN=

MRo12

w2 w1( ) c c3 w1 cc2

w1 b cb2

+

12

w1 a b c+a3

+

:=

MRo 349 kN m=

Problem 4-155

Determine the equivalent resultant force and couple moment at point O.

Units Used:

kN 103N:=

Given:

a 3m:=

wO 3kNm

:=

w x( ) wOxa

2:=

Solution:

FR0

axw x( )

d:= FR 3 kN=

MO0

axw x( ) a x( )

d:= MO 2.25 kN m=

Problem 4-156

Wind has blown sand over a platform such that the intensity of the load can be approximated by

the function w w0xd

3= . Simplify this distributed loading to an equivalent resultant force and

specify the magnitude and location of the force, measured from A.

Units Used:

kN 103N:=

Given:

w0 500Nm

:=

d 10m:=

w x( ) w0xd

3:=

Solution:

FR0

dxw x( )

d:= FR 1.25 kN=

d0

dxx w x( )

d

FR:= d 8.00 m=

Problem 4-157

Determine the equivalent resultant force and its location, measured from point O.

Solution:

FR0

L

xw0 sin xL

d=2w0 L

=

d0

L

xx w0 sin xL

d

FR=

L2

=

Problem 4-158

Determine the equivalent resultant force acting on the bottom of the wing due to air pressure andspecify where it acts, measured from point A.

Given:

a 3m:=

k 86kN

m3:=

w x( ) k x2:=

Solution:

FR0

axw x( )

d:= FR 774 kN=

x0

axx w x( )

d

FR:= x 2.25 m=

Problem 4-159

Currently eighty-five percent of all neck injuries are caused by rear-end car collisions. Toalleviate this problem, an automobile seat restraint has been developed that provides additionalpressure contact with the cranium. During dynamic tests the distribution of load on the craniumhas been plotted and shown to be parabolic. Determine the equivalent resultant force and itslocation, measured from point A.

Given:

a 0.15m:=

w0 200Nm

:=

k 4000N

m3:=

w x( ) w0 k x2

+:=

Solution:

FR0

axw x( )

d:= FR 34.5 N=

x0

axx w x( )

d

FR:= x 0.080 m=

Problem 4-160

Determine the equivalent resultant force of the distributed loading and its location, measured frompoint A. Evaluate the integrals using Simpson's rule.

Units Used:

kN 103N:=

Given:

c1 5:=

c2 16:=

a 3:=

b 1:=

Solution:

FR0

a b+

xc1 x c2 x2

+

12

+

d:= FR 14.9=

d0

a b+

xx c1 x c2 x2

+

12

+

d

FR:= d 2.27=

Problem 4-161

Determine the coordinate direction angles of F, which is applied to the end A of the pipeassembly, so that the moment of F about O is zero.

Given:

F 20N:=

a 8cm:=

b 6cm:=

c 6cm:=

d 10cm:=

Solution:

Require Mo = 0. This happens when force F is directed either towards or away from point O.

r

c

a b+

d

:= urr

:= u

0.329

0.768

0.549

=

If the force points away from O, then

acos u( ):=

70.8

39.8

56.7

deg=

If the force points towards O, then

acos u( ):=

109

140

123

deg=

Problem 4-162

Determine the moment of the force F about point O. The force has coordinate direction angles, , . Express the result as a Cartesian vector.

Given:

F 20N:= a 8cm:=

60deg:= b 6cm:=

120deg:= c 6cm:=

45deg:= d 10cm:=

Solution:

r

c

a b+

d

:= Fv F

cos ( )

cos ( )

cos ( )

:= M r Fv:= M

2.98

0.15

2.00

N m=

Problem 4-163

Replace the force at A by an equivalent resultant force and couple moment at point P. Expressthe results in Cartesian vector form.

Units Used:

kN 103N:=

Given:

a 4m:=

b 6m:=

c 8m:=

d 4m:=

F

300

200

500

N:=

Solution:

FR F:= FR

300

200

500

N=

MP

a c

b

d

F:= MP

3.8

7.2

0.6

kN m=

Problem 4-164

Determine the moment of the force FC about the door hinge at A. Express the result as a Cartesianvector.

Given:

F 250N:=

b 1m:=

c 2.5m:=

d 1.5m:=

e 0.5m:=

30deg:=

Solution:

rCB

c e

b d cos ( )+

d sin ( )

:= rAB

0

b

0

:= Fv FrCBrCB

:=

MA rAB Fv:= MA

59.7

0.0

159.3

N m=

Problem 4-165

Determine the magnitude of the moment of the force FC about the hinged axis aa of the door.

Given:

F 250N:=

b 1m:=

c 2.5m:=

d 1.5m:=

e 0.5m:=

30deg:=

Solution:

rCB

c e

b d cos ( )+

d sin ( )

:= rAB

0

b

0

:= Fv FrCBrCB

:= ua

1

0

0

:=

Maa rAB Fv( ) ua:= Maa 59.7 N m=

Problem 4-166

A force F1 acts vertically downward on the Z-bracket. Determine the moment of this forceabout the bolt axis (z axis), which is directed at angle from the vertical.

Given:

F1 80N:=

a 100mm:=

b 300mm:=

c 200mm:=

15deg:=

Solution:

r

b

a c+

0

:= F F1

sin ( )

0

cos ( )

:= k

0

0

1

:=

Mz r F( ) k:=

Mz 6.21 N m=

Problem 4-167

Replace the force F having acting at point A by an equivalent force and couple moment atpoint C.

Given:

F 50kN:=

a 1m:=

b 2m:=

c 1.5m:=

d 1m:=

e 3m:=

Solution:

rAB

d

c

e

:=

Fv FrABrAB

:=

rCA

0

a b+

e

:=

FR Fv:= FR

14.3

21.4

42.9

kN=

MR rCA Fv:= MR

192.9

42.9

42.9

kN m=

Problem 4-168

The horizontal force F acts on the handle of the wrench. What is the magnitude of the momentof this force about the z axis?

Given:

F 30N:=

a 50mm:=

b 200mm:=

c 10mm:=

45deg:=

Solution :

Fv F

sin ( )

cos ( )

0

:= rOA

c

b

a

:= k

0

0

1

:=

Mz rOA Fv( ) k:= Mz 4.03 N m=

Problem 4-169

The horizontal force F acts on the handle of the wrench. Determine the moment of this forceabout point O. Specify the coordinate direction angles , , of the moment axis.

Given:

F 30N:=

a 50mm:=

b 200mm:=

c 10mm:=

45deg:=

Solution :

Fv F

sin ( )

cos ( )

0

:= rOA

c

b

a

:=

MO rOA Fv:= MO

1.06

1.06

4.03

N m=

acosMOMO

:=

75.7

75.7

159.6

deg=

Problem 4-170

If the resultant couple moment of the three couples acting on the triangular block is to be zero,determine the magnitudes of forces F and P.

Given:

F1 100N:=

a 30mm:=

b 40mm:=

c 60mm:=

d 30mm:=

30deg:=Solution:

Initial Guesses: F 1N:= P 1N:=

Given

0

F c

0

0

0

P c

+F1 d

a2 b2+

0

a

b

+ 0=F

P

Find F P,( ):=F

P

30

40

N=

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