APPENDIX
Review for the Fundamentals of Engineering Examination
The Fundamentals of Engineering (FE) exam is given semiannually by the National Council of Engineering Examiners (NCEE) and is one of the requirements for obtaining a Professional Engineering License.A portion of this ex am contains problems in statics, and this appendix provides a review of the subject matter most often asked on this exam. Before solving solving any of the problems, problems, you should should review the sections sections indicated in each chapter in order to become familiar with the boldfaced definitions and the procedures used to solve the various types of problems. problems. Also, review the example problems in these sections sections.. The following problems are arranged in the same sequence as the topics in each chapter.. Besides helping chapter helping as a preparation for the FE exam, these problems problems also provide additional examples for general practice of the subject matter. matter. Solutions to all the problems are given at the back of this appendix.
Chapter 2—Review All Sections C-1. Two forces act on the hook. Determine the magnitude of the resultant force.
C-2. The force F = 450 lb acts on the frame. Resolve this force into components acting along members AB and AC , and determine the magnitude of each component.
A
C
45 450 lb 30
40
200 N B
500 N
618
Prob. C–1
Prob. C–2
30
APPENDIX C
REVIEW
FOR THE
FUNDAMENTALS
C-3. Determine the magnitude and direction of the resultant force.
OF
619
E N G I N E E R I N G E X A M I N A T I O N
C-5. The force has a component of 20 N directed along the force F as a Cartesian vector. - y axis as shown. Represent the force
y
z
250 N 3
5
400 N
4
F
30 x
300 N 150 20 N
y
70
x
Prob. C–5
Prob. C–3
5
6
determine the C-4. If F = 30i + 50 j - 45k N, magnitude and coordinate direction angles of the force.
C-6. The force acts on the beam as shown. Determine its coordinate direction angles.
z
z
y
45 x x
F
Prob. C–4
30
F 75 lb
Prob. C–6
y
620
APPENDIX C
REVIEW
FOR THE
FUNDAMENTALS
C-7. The cables supporting the antenna are subjected to the forces shown. Represent each force as a Cartesian vector.
OF
E N G I N E E R I N G E X A M I N A T I O N
C-9. Determine the component of projection of the force F along the pipe AB.
z
z F 1 160 lb
F 2 80 lb
100 ft
F 3 100 lb
B
F {20i 30 j 60k} lb
10 ft
x
2 ft
20 ft
20 ft
y
4 ft A
y
60 ft x
3 ft
30 ft
Prob. C–9 Prob. C–7
Determine the angle u between the two cords.
C-8.
Chapter 3—Review Sections 3.1–3.3 C-10. The crate at D has a weight of 550 lb. Determine the force in each supporting cable.
z C A
B 5
2m
2m 30
2m
A y
u
x
2m
4m
D
2m B
Prob. C–8
3 4
Prob. C–10
APPENDIX C
REVIEW
FOR THE
FUNDAMENTALS
C-11. The beam has a weight of 700 lb. Determine the shortest cable ABC that can be used to lift it if the maximum force the cable can sustain is 1500 lb.
OF
62 1
E N G I N E E R I N G E X A M I N A T I O N
C-13. The post can be removed by a vertical force of 400 lb. Determine the force P that must be applied to the cord in order to pull the post out of the ground.
B C
5
4 3
B
30 A
u
u
A
P
C
10 ft
Prob. C–11
Prob. C–13
C-12. The block has a mass of 5 kg and rests on the smooth plane. Determine the unstretched length of the spring.
Chapter 4—Review All Sections C-14.
Determine the moment of the force about point O.
600 lb
0.3 m
20
k 200 N/ m
0.4 m 30
5 ft
O
45
Prob. C–12
Prob. C–14
0.5 ft
622
APPENDIX C
REVIEW
FOR THE
FUNDAMENTALS
C-15. Determine the moment of the force about point O. Neglect the thickness of the member.
OF
E N G I N E E R I N G E X A M I N A T I O N
C-17. Determine the moment of the force about point A. Express the result as a Cartesian vector.
z
50 N 100 mm
B
60 F {30i 40 j 50k} N
6m A
1m 45
200 mm
2m
1m
O
5m x
100 mm
Prob. C–17
Prob. C–15
C-16.
y
1m
Determine the moment of the force about point O.
C-18. Determine the moment of the force about point A. Express the result as a Cartesian vector.
z B
500 N F 130 lb
14 ft
C
3m
2 ft 45
2 ft
O
3 ft
1 ft
x
Prob. C–16
5 ft
4 ft A
Prob. C–18
y
APPENDIX C
REVIEW
FOR THE
FUNDAMENTALS
C-19. Determine the resultant couple moment acting on the beam.
400 N
400 N
OF
62 3
E N G I N E E R I N G E X A M I N A T I O N
C-21. Replace the loading shown by an equivalent resultant force and couple-moment system at point A.
40 N
30 N 200 N m
A
200 N 0.2 m
A B
200 N 3m
5
3
3m
2m
4
3m
50 N
300 N
300 N
Prob. C–19
Prob. C–21
C-20. Determine the resultant couple moment acting on the triangular plate.
C-22. Replace the loading shown by an equivalent resultant force and couple-moment system at point A.
100 lb 200 lb
150 lb
4 ft
4 ft 4 ft
200 lb
150 lb 200 lb
A
3 ft
4 ft 300 lb
300 lb
Prob. C–20
3 ft 150 lb
Prob. C–22
624
APPENDIX C
REVIEW
FOR THE
FUNDAMENTALS
C-23. Replace the loading shown by an equivalent single resultant force and specify where the force acts, measured from point O.
OF
E N G I N E E R I N G E X A M I N A T I O N
C-25. Replace the loading shown by an equivalent single resultant force and specify the x and y coordinates of its line of action.
y
z 500 lb
200 N
500 lb
250 lb
2m 1m
x
O
100 N 3m
3 ft
3 ft
3 ft
3 ft
3m
3 m 200 N 2m 1m
100 N
y
x
Prob. C–23
Prob. C–25
C-24. Replace the loading shown by an equivalent single resultant force and specify the x and y coordinates of its line of action.
C-26. Determine the resultant force and specify where it acts on the beam measured from A.
z
150 lb/ ft
400 N 100 N 3m
B
A
4m y
500 N
6 ft
8 ft
4m
x
Prob. C–24
Prob. C–26
APPENDIX C
REVIEW
FOR THE
FUNDAMENTALS
C-27. Determine the resultant force and specify where it acts on the beam measured from A.
OF
62 5
E N G I N E E R I N G E X A M I N A T I O N
Chapter 5—Review Sections 5.1–5.6 C-29. Determine the horizontal and vertical components of reaction at the supports. Neglect the thickness of the beam.
w
160 N/ m
500 lb 5
4
w 2.5 x3
600 lb ft
3
A
A
x
B
5 ft
5 ft
5 ft
4m
Prob. C–29
Prob. C–27
C-28. Determine the resultant force and specify where it acts on the beam measured from A.
C-30. Determine the horizontal and vertical components of reaction at the supports.
500 N 200 lb/ ft
500 lb
150 lb/ ft
400 N 0.5 m A
B
B
A
6 ft
3 ft
Prob. C–28
3 ft
2m
2m
Prob. C–30
626
APPENDIX C
REVIEW
FOR THE
FUNDAMENTALS
C-31. Determine the components of reaction at the fixed support A. Neglect the thickness of the beam.
200 N
200 N
OF
E N G I N E E R I N G E X A M I N A T I O N
C-33. The uniform plate has a weight of 500 lb. Determine the tension in each of the supporting cables.
200 N
z A
30 1m
3m
1m
1m
B
C
400 N
200 lb y
2 ft 2 ft
60 3 ft
x
A
Prob. C–31
Prob. C–33
C-32. Determine the tension in the cable and the horizontal and vertical components of reaction at the pin A. Neglect the size of the pulley.
Chapter 6—Review Sections 6.1–6.4, 6.6 C-34. Determine the force in each member of the truss. State if the members are in tension or compression.
C B
6 ft
C
600 lb ft
D A
B
4 ft
3 ft
4 ft
D
4 ft 300 lb
Prob. C–32
2 ft
2 ft
300 lb
Prob. C–34
A
APPENDIX C
REVIEW
FOR THE
FUNDAMENTALS
C-35. Determine the force in members AE and DC . State if the members are in tension or compression.
OF
C-37. Determine the force in members GF, FC , and CD. State if the members are in tension or compression.
E
1000 lb
6 ft
E
F
62 7
E N G I N E E R I N G E X A M I N A T I O N
D
700 lb
F
D
3 ft G
C
A
C
500 lb
B
4 ft
6 ft
4 ft
6 ft
800 lb B
A
Prob. C–35 8 ft
Prob. C–37
C-36. Determine the force in members BC , CF , and FE. State if the members are in tension or compression.
G
F
E
4 ft
A
4 ft
C-38. Determine the force P needed to hold the 60-lb weight in equilibrium.
B
600 lb
C
4 ft
600 lb
Prob. C–36
P
D
4 ft 800 lb
Prob. C–38
628
APPENDIX C
REVIEW
FOR THE
FUNDAMENTALS
C-39. Determine the horizontal and vertical components of reaction at pin C .
OF
E N G I N E E R I N G E X A M I N A T I O N
C-41. Determine the normal force that the 100-lb plate A exerts on the 30-lb plate B.
500 lb 400 lb B
C
4 ft A
A
3 ft
3 ft
3 ft
3 ft B
Prob. C–39
4 ft
1 ft
1 ft
Prob. C–41
C-40. Determine the horizontal and vertical components of reaction at pin C .
C-42. Determine the force P needed to lift the load. Also, determine the proper placement x of the hook for equilibrium. Neglect the weight of the beam.
0.9 m
400 N 800 N m
1m
100 mm
100 mm
2m C
100 mm
C
B
P
1m B
A
1m A x
1m
6 kN
Prob. C–40
Prob. C–42
APPENDIX C
REVIEW
FOR THE
FUNDAMENTALS
OF
62 9
E N G I N E E R I N G E X A M I N A T I O N
Chapter 7—Review Section 7.1
Chapter 8—Review Sections 8.1–8.2
C-43. Determine the internal normal force, shear force, and moment acting in the beam at point B.
C-46. Determine the force P needed to move the 100-lb block. The coefficient of static friction is ms = 0.3, and the coefficient of kinetic friction is mk = 0.25. Neglect tipping.
8 kN 3 kN/ m
P 30
C
A B
1.5 m
1.5 m
1.5 m
1.5 m
Prob. C–43 Prob. C–46
C-44. Determine the internal normal force, shear force, and moment acting in the beam at point B, which is located just to the left of the 800-lb force.
800 lb
400 lb 300 lb ft
A C
B
6 ft
C-47. Determine the vertical force P needed to rotate the 200-lb spool. The coefficient of static friction at all contacting surfaces is ms = 0.4.
3 ft
3 ft
2 ft P
Prob. C–44
C-45. Determine the internal normal force, shear force, and moment acting in the beam at point B. 12 in. 6 in. A
3 kN/ m
30
A
C
B
6m
B
3m
Prob. C–45
Prob. C–47
630
APPENDIX C
REVIEW
FOR THE
FUNDAMENTALS
C-48. Block A has a weight of 30 lb and block B weighs 50 lb. If the coefficient of static friction is ms = 0.4 between all contacting surfaces, determine the frictional force at each surface.
OF
E N G I N E E R I N G E X A M I N A T I O N
C-50. The filing cabinet A has a mass of 60 kg and center of mass at G. It rests on a 10-kg plank. Determine the smallest force P needed to move it. The coefficient of static friction between the cabinet A and the plank B is ms = 0.4, and between the plank and the floor ms = 0.3.
0.2 m 0.2 m
20 lb 30
A
B
0.8 m P
10 lb
A
G
1.3 m
Prob. C–48
1m
B
C
Prob. C–50
C-49. Determine the force P necessary to move the 250-lb crate which has a center of gravity at G. The coefficient of static friction at the floor is ms = 0.4.
Chapter 9—Review Sections 9.1–9.3 (Integration is covered in the mathematics portion of the exam.) C-51. Determine the location (x, y) of the centroid of the area.
1.5 ft 1.5 ft
y
2.5 ft
P
G
4.5 ft 3.5 ft
3 ft 2 ft x
A
2 ft
Prob. C–49
3 ft
Prob. C–51
3 ft
APPENDIX C
C-52. area.
REVIEW
FOR THE
FUNDAMENTALS
Determine the location (x, y) of the centroid of the
OF
63 1
E N G I N E E R I N G E X A M I N A T I O N
C-54. Determine the moment of inertia of the area with respect to the x axis.
y y
3 in.
3 in. 8 in. 2 in.
12 in. 0.5 in. 8 in.
4 in. x
8 in.
x
1 in. 6 in.
Prob. C–52 Prob. C–54
Chapter 10—Review Sections 10.1–10.5 (Integration is covered in the mathematics portion of the exam.) C-53. Determine the moment of inertia of the crosssectional area of the channel with respect to the y axis.
C-55. Determine the moment of inertia of the crosssectional area of the T-beam with respect to the x ¿ axis passing through the centroid of the cross section.
8 in.
y
2 in.
20 mm 20 mm
100 mm
C
8 in.
20 mm 150 mm
150 mm
x
2 in.
Prob. C–53
x
Prob. C–55
x
¿
632
APPENDIX C
REVIEW
FOR THE
FUNDAMENTALS
OF
E N G I N E E R I N G E X A M I N A T I O N
Partial Solutions and Answers C–1.
C–2.
1 21 2
= 4 2002 + 5002 - 2 200 500 cos 140° = 666 N Ans.
FR
FAB
sin 105°
= =
FAC
=
sin 45° FAC = C–3.
C–7.
= 300 + 400 cos 30° - 250
FRy
= 400 sin 30° + 250 =
1 2
4 446.4
= tan-1
u
350 = 38.1° a 446.4
1 2
=
+ - 45
a b = 66.0° a b = 47.2° a b = 128°
C–5.
Fy Fy
ƒFƒ
cos g =
= F = + = g
C–6.
2
Ans.
20 a - 102.0
10 a 102.5
i -
6
5
= 73.7 N Ans.
60 a 120.4
rOA = rOB =
C–8.
Ans.
cos u =
1
Ans. Ans.
u
= 23.09 N
4 1 -
C–10.
cos2 70° - cos2 150° 68.61° From Fig. g 6 90° 23.09 cos 70°i + 23.09 cos 150° j 23.09 cos 68.61°k 7.90i - 20 j + 8.42k N Ans.
1
2
6
C–11.
Ans.
b
i
6 6
Ans.
b Ans.
5-2 + 2 + 2 6 m 52 + 4 - 2 6 m i
i
j
#
j
k
k
rOA rOB
ƒ rOA ƒ ƒ rOB ƒ - 2i + 2 j + 2k # 2i + 4 j - 2k
21
2 12 2 24
2=0
= 90° Ans.
ƒ FAB ƒ = F # uAB
2 # a- 35
1
`
b
20 100 j k 102.5 102.5 - 78.1k lb 30 100 + j k 120.4 120.4 - 83.0k lb
= - 20i - 30 j + 60k
cos 150°
100 k 102.0
i -
= 49.8i + 24.9 j
Ans.
= 75 cos 30° sin 45° = 45.93 Fy = 75 cos 30° cos 45° = 45.93 Fz = - 75 sin 30° = - 37.5 Fx
F1 = 160 lb
F3 = 100 lb
C–9.
- 20
5
= cos-1
5
= cos b
`
g
= 7.81i - 15.6 j
= - 20
ƒFƒ =
= cos-1
F2 = 80 lb
a 35 b = 350 N 2
4 302 + 502 30 a = cos-1 73.7 50 b = cos-1 73.7 - 45 g = cos-1 73.7
F
b
5
a 45 b = 446.4 N
+ 350 = 567 N
2
= cos-1
= - 31.4i - 157k lb
FRx
FR
C–4.
450 sin 30° 869 lb Ans. 450 sin 30° 636 lb Ans.
a 45.93 b = 52.2° Ans. 75 a 45.93 b = 52.2° Ans. 75 a -7537.5 b = 120° Ans.
a
+
b
4 i - j = 36 lb Ans. 5
4 FAC - FAB cos 30° = 0 5 3 + c © Fy = 0; FAC + FAB sin 30° - 550 = 0 5 FAB = 478 lb Ans. FAC = 518 lb Ans. :
© Fx = 0;
1 2 5 ft b = 10.3 ft = 2a cos 13.5°
+ c © Fy = 0; - 2 1500 sin u + 700 = 0 u = 13.5° LABC
63 3
PARTIAL SOLUTIONS AND ANSWERS
C–12.
C–13.
1 2 1 2 1 2 1
4 Fsp - 5 9.81 sin 45° = 0 5 Fsp = 43.35 N Fsp = k l - l0 ; 43.35 = 200 0.5 - l0 Ans. l0 = 0.283 m
+
C–16.
C–17.
© Fx = 0;
d + MO
e + MO
d + MO
MA
FRx
12
+ TAC sin 30° - 400 = 0 Ans. Ans.
1 2 2
2
1 1
6
C–22.
i
j
6
b
k 14 - 120
12
e + MCR
12
MAR MRA
3
=
4 1402 + 702 =
= tan-1
157 lb
70 a 140 b = 26.6°
d
Ans. Ans.
1 21 2 1 21 2
12
3 4 100 4 100 6 + 150 3 5 5 Ans. = 210 lb ft
=
#
+ T FR = © Fy ; FR = 500 + 250 + 500 = 1250 lb Ans. + bFRx = © MO ; 1250 x = 500 3 + 250 6 + 500 9 Ans. x = 6 ft
12
12
12
12
Ans.
12
12
Also, e + MCR = 300 5 - 400 2 + 200 0.2 Ans. = 740 N m
1 2#
1 2 1 2
3 100 = 140 lb 5 4 = 150 - 100 = 70 lb 5
+ bMAR = © MA ;
C–23.
= © MA = 400 3 - 400 5 + 300 5 + 200 0.2 = 740 N # m Ans.
1 2
Ans.
= © Fx ; FRx = 200 -
u
-3 -6 - 30 40 = 160i - 780 j - 300k lb # ft
6
FRx
FR
#
4 12 j k 13 13 - 30i + 40 j - 120k lb
+
;
+ T FRy = © Fy ; FRy
3
i +
1 2 35 1502162 + 200
= 470 N # m
2
2
3
5
c
= 30 3 +
MAR
= 500 sin 45° 3 + 3 cos 45° - 500 cos 45° 3 sin 45° = 1.06 kN # m Ans.
M A = rAB * F =
2
+ bMAR = © MA ;
1 1
3
4 40 2 + -1
u
= 50 sin 60° 0.1 + 0.2 cos 45° + 0.1 - 50 cos 60° 0.2 sin 45° = 11.2 N # m Ans.
5
1 2
1 2 11002 = 108 N Ans. 100 = tan a b = 68.2° Ans. 40 =
FR
a - 133
1 2
4 50 = 40 N 5 3 = 40 + 30 + 50 5
= © Fx ;FRx =
= 100 N
12
F = 130 lb
12
= 300 4 + 200 4 + 150 4 Ans. = 2600 lb # ft
- TAC cos 30° = 0
i j k = rAB * F = 1 6 5 30 40 - 50 = - 500i + 200 j - 140k N m Ans.
=
C–19.
+
:
= 600 sin 50° 5 + 600 cos 50° 0.5 = 2.49 kip # ft Ans.
5
C–18.
C–21.
12
d + MCR
+ T FRy = © Fy ;FRy
3 P 5 4 + c © Fy = 0; P 5 P = 349 lb TAC = 242 lb
C–15.
2
At A: ;
C–14.
C–20.
+Q© Fx = 0;
1 2
C–24.
+ T FR = © Fz ; FR = 400 + 500 - 100 Ans. = 800 N MRx = © Mx ; - 800y = - 400 4 - 500 4 Ans. y = 4.50 m MRy = © My ; 800x = 500 4 - 100 3 x = 2.125 m Ans.
12 12 12 12
634 C–25.
APPENDIX C
REVIEW
C–26.
FR
+ bMAR
FR
=
C–30.
12
1 21 2 1 2
C–31.
Ans.
c 12 16211502d142 + [811502]1102
= 8.36 ft
4
L12
w x dx
=
L 2.5
x 3 dx
= 160 N
12 L = L 12
=
L
C–32.
1 21 2
12
+ bMAR = © MA ;
c 1 21 2d142 + [150162]132 + 500192
1 1550d = 50 6 2 d = 5.03 ft
12
Ax
a 35 b = 0
= 300 lb Ans.
1 2
a b1 2
4 + g © MA = 0; By 10 - 500 5 - 600 = 0 5 By = 260 lb Ans. 4 + c © Fy = 0; A y + 260 - 500 =0 5 A y = 140 lb Ans.
ab
Ans.
=0 Ans.
1 2
12
3 T 12 - 300 8 - 600 = 0 5 T = 267.9 = 268 lb Ans.
+ g © MA = 0; T 4 +
© Fx = 0; A x -
a 45 b1267.92 = 0 = 214 lb Ans.
+ c © Fy = 0; A y + 267.9 + Ay
a 35 b1267.92 - 300 = 0
= - 129 lb Ans.
C–33.
© Fz = 0; TA + TB + TC - 200 - 500 = 0 © Mx = 0; TA 3 + TC 3 - 500 1.5 - 200 3 = 0 © My = 0; - TB 4 - TC 4 + 500 2 + 200 2 = 0 Ans. TA = 350 lb, TB = 250 lb, TC = 100 lb
C–34.
Joint D:
Ans.
© Fx = 0; - A x + 500
12
© Fx = 0; - A x + 400 cos 30° = 0 A x = 346 N Ans. + c © Fy = 0; A y - 200 - 200 - 200 - 400 sin 30° = 0 A y = 800 N Ans. d + © MA = 0; MA - 200 2.5 - 200 3.5 - 200 4.5 - 400 sin 30° 4.5 - 400 cos 30° 3 sin 60° = 0 MA = 3.90 kN # m Ans.
:
1 50 6 + 150 6 + 500 2 = 1550 lb Ans.
+
1 2
+
+
= 3.20 m Ans.
160
= 400 N Ans. - 500 2 = 0
Ax
:
2.5x 4 dx
0
12
Ax
+ T FR = © Fy ; FR =
:
© Fx = 0; - A x + 400 = 0; d + © MA = 0; By 4 - 400 0.5 By = 300 N + c © Fy = 0; A y + 300 - 500 A y = 200 N
4
w x dx
C–29.
+
:
Ans.
0
xw x dx
E N G I N E E R I N G E X A M I N A T I O N
1 2 1 2 1 2 1 2 1 2
Ans.
+ bMAR = © MA ; x
12
12
OF
12
1 6 150 + 8 150 = 1650 lb 2 = © MA ;
=
1650d = d
12
12
12
C–28.
FUNDAMENTALS
+ T FR = © Fy ; FR = 200 + 200 + 100 + 100 Ans. = 600 N MRx = © Mx ; - 600y = 200 1 + 200 1 + 100 3 - 100 3 Ans. y = - 0.667 m MRy = © My ; 600x = 100 3 + 100 3 + 200 2 - 200 3 Ans. x = 0.667 m
12
C–27.
FOR THE
12 12 1 2 12 12 12
+ c © Fy = 0; FCD
+
:
3 FCD - 300 = 0; 5
12
= 500 lb T
© Fx = 0; - FAD + FAD
12
= 400 lb C
Ans.
1 2
4 500 = 0; 5
Ans.
12 12
PARTIAL SOLUTIONS AND ANSWERS
Joint C :
C–38.
+R© Fy = 0; FCA = 0
P
C–39.
12
= 500 lb T Ans.
FCB
+ c © Fy = 0; FAB = 0
Ans.
+
C–40.
+ c © Fy = 0; - FDC + 400 = 0;
C–36.
12
12
= 1980 lb T Ans.
12
12
+ g © MC = 0; FFE 4 - 800 4 = 0
12
C–41.
= 800 lb T Ans.
12
12
C–37.
12
= 2200 lb C Ans.
12
1 2
+ g © MC = 0; FGF 8 - 700 6 - 1000 12 = 0 FGF
+
:
12
= 2025 lb T Ans. 4 5
© Fx = 0; - FFC + 700 + 1000 = 0 FFC
d + © MF
12
= 2125 lb C Ans.
12
12
= 0; FCD 8 - 1000 6 = 0 FCD
12
= 750 lb C Ans.
Plate A:
Plate B: + c © Fy = 0; 2T - NAB - 30 = 0
Section truss through GF, FC, DC . Use the top segment.
12
1 2 + 800
FAB sin 45° 3
+ c © Fy = 0; 2T + NAB - 100 = 0
12
T + © MF = 0; FBC 4 - 600 4 - 800 8 = 0 FBC
12-
:
+ c © Fy = 0; FCF sin 45° - 600 - 800 = 0
FFE
2
+ 400 2 = 0 FAB = 1131.37 N + © Fx = 0; - Cx + 1131.37 cos 45° = 0 Cx = 800 N Ans. + c © Fy = 0; - Cy + 1131.37 sin 45° - 400 = 0 Cy = 400 N Ans.
Section truss through FE, FC, BC . Use the right segment.
12
+ g © MC = 0;
FAB cos 45° 1
= 400 lb C Ans.
FCF
2
Ans.
Joint C : FDC
1
© Fx = 0; - Cx +
1
3 5
12
= 541.67 lb
3 541.67 = 0 5 Cx = 325 lb Ans. 4 + c © Fy = 0; Cy + 541.67 - 400 - 500 = 0 5 Cy = 467 lb Ans.
+ c © Fy = 0; - FAE + 400 = 0; = 667 lb C
FAB
:
Joint A:
FAE
a 45 b1 2192 + 400162 + 500132 = 0
FAB
= 0, Ay = Cy = 400 lb
Ax
= 20 lb Ans.
+ g © MC = 0; -
Joint A:
C–35.
+ c © Fy = 0; 3P - 60 = 0
Ans.
+Q© Fx = 0; FCB - 500 = 0;
63 5
T
C–42.
= 32.5 lb, NAB = 35 lb Ans.
Pulley C :
+ c © Fy = 0; T - 2P = 0; T = 2P Beam: + c © Fy = 0; 2P + P - 6 = 0 P
= 2 kN Ans.
12 12
+ g © MA = 0; 2 1 - 6 x = 0 x
= 0.333 m Ans.
636 C–43.
APPENDIX C
Ay
+
:
REVIEW
FOR THE
FUNDAMENTALS
= 8.75 kN. Use segment AB.
OF
C–48.
1 2
+ c © Fy = 0; 8.75 - 3 1.5 - VB = 0
:
= 4.25 kN Ans.
1 21 2
1 2
+ g © MB = 0; MB + 3 1.5 0.75 - 8.75 1.5 = 0 MB
C–44.
Ax
+
:
Block B:
+ c © Fy = 0; NB - 20 sin 30° - 50 = 0 NB = 60 lb + © Fx = 0; FB - 20 cos 30° = 0 Ans. FB = 17.3 lb 6 0.4 60 lb
© Fx = 0; NB = 0 Ans. VB
E N G I N E E R I N G E X A M I N A T I O N
= 9.75 kN # m Ans.
Blocks A and B:
+ c © Fy = 0; NA - 30 - 50 - 20 sin 30° = 0
= 0, A y = 100 lb. Use segment AB. +
© Fx = 0; NB = 0 Ans.
:
+ c © Fy = 0; 100 - VB = 0 VB
= 100 lb Ans.
C–49.
12
A x = 0, A y = 4.5 kN, wB Use segment AB.
+
:
>
:
= 2 kN m.
= 1.5 kN Ans.
C–46.
= 15 kN # m Ans.
+ c © Fy = 0; Nb - P sin 30° - 100 = 0 +
:
© Fx = 0; - P cos 30° + 0.3Nb = 0 P = 41.9 lb Ans.
C–47.
:
1 2
1 2 12
1 2
= 83.3 lb Ans.
1 2 = 588.6 N = 0; 0.41588.62 - = 0
+
:
NA
© Fx
P
P
= 235 N
P for B to slip:
1 2 1 2 = 686.7 N = 0; 0.31686.72 - = 0
+ c © Fy = 0; NB - 60 9.81 - 10 9.81 = 0 NB
© Fx
P
P P to tip A:
+ g © Fy = 0; P + 0.4NA + NB - 200 = 0
d + © MC
= 98.2 lb Ans.
1 2
= 0; - P 4.5 + 250 1.5 = 0
+ g © MB = 0; 0.4NA 12 + NA 12 - P 6 = 0 P
= 100 lb
+ c © Fy = 0; NA - 60 9.81 = 0
:
© Fx = 0; 0.4NB - NA = 0
1 2
P for A to slip on B:
+
+
= 250 lb
© Fx = 0; P - 0.4 250 = 0
P
C–50.
c 12 162122d122 - 4.5162 = 0
NC
d + © MA
1 21 2
MB
1 1 22 Ans.
= 27.3 lb 6 0.4 90 lb
P If tipping occurs:
1 + c © Fy = 0; 4.5 - 6 2 + VB = 0 2
+ g © MB = 0; MB +
© Fx = 0; FA - 20 cos 30° - 10 = 0
If slipping occurs:
+
© Fx = 0; NB = 0 Ans.
VB
= 90 lb
+ c © Fy = 0; NC - 250 lb = 0
= 600 lb # ft Ans.
MB
NA
FA
+ g © MB = 0; MB - 100 6 = 0
C–45.
1 1 22
= 206 N
1 2
1 21 2
= 0; P 1.3 - 60 9.81 0.2 = 0 P
= 90.6 N Ans.
63 7
PARTIAL SOLUTIONS AND ANSWERS
C–51.
' © xA = x = ©A
C–53.
1 21 21 2 1 21 2 a b132132 = 1.57 ft 1 2 1 2 1 21 2 1 - 1 2 2 + 1.5 3 3 + 4 2 1 2 2 + 3 3 + 3 3 2
Ans.
Iy
-
1 21 2
1 100 260 12
2 1 1 22 1 21 2 12 12
Ans. = 0 symmetry ' © yA 4 1 8 + 9 6 2 y = = = 7 in. 1 8 + 6 2 ©A
3
4
Ans.
I
= © I + Ad2 =
3
3
1 21 2 1 21 2 a 12 b132132 = 1.26 ft 1 2122 + 3132 + 132132 2 1
3
1 2 c 121 1821122 + 1821122162 d 1 + c 1621122 + 16211221 - 22 d = 5760 in Ans. 12
C–54.
2
2
4
1 21 2 1 21 2 12 12
' © xA 4 8 2 + 9 2 8 = = 6.5 in. x = ©A 8 2 + 2 8
1 2 2 + 1.5 3 3 + 1
x
1 120 300 12 6
' © yA = y = ©A
C–52.
1 21 2 = 124 110 2 mm =
C–55.
Ans.
1 2 c 121 122182 + 18212216.5 - 42 d 1 Ans. + c 182122 + 218219 - 6.52 d = 291 in 12
Ix¿
Ans.
= © I + Ad2 = 3
3
2
2
4
