I
The Fundamentals of Engineering (FE) exam is given semiannually b}
the National Council of Engineering Examiners (NCEE). and is one 01
the requirements for obtaining a Professional Engineering License. A
portion of this exam contains problems in mechanics of materials. and
this appendi.'t provides a review of the subject matter most often asked
on this exam.
Before solving any of the problems. you should review the sections
indicated in each chapter in order to becomefamiliar with the boldfaced
definitions and the procedures used to solve various types of problems.
Also, review the example problems in thesesections.The following problems are arranged in the same sequenceas the topics in each chapter.
Partial solutions to all the problems are given at the back of this appendi.'t.
Reference: .Mechanics of Materials. by R. C. Hibbeler, 3rdEdition
AP. D REVIEW FOR THE FUNDAMENTALS OF ENGINEERING EXAMINATION
Ch.pter l-Re\1e" All Sct.-tiun!i
0-1 Detenninethe resultantinternalmomentin the memo
ber of the frame at point F.
807
D-6 The ban of the truss each have a cross-sectional area
of 2 in.2 Determine the average normal stress in member
CB.
£
0-7 The frame supportsthe loadingshown.The pin at A
hasa diameterof 0.25in. If it is subjectedto doubleshear.
detenninethe averageshearstressin the pin.
0.75ft
Prob. 0-1
0-2 The beam is supported by a pin at A and a link BC.
Determine the resultant internal shear in the beam at point
D.
0-3 The beamis supportedby a pin at A and a link BC.
Determinethe averageshearstressin the pin at B if it has
a diameterof 20 mm and is in doubleshear.
D-4 The beamis supponedby a pin at A and a link BC.
Determinethe averageshearstressin the pin at A if it has
a diameterof 20 mm and is in singleshear.
0-5 How many independentstresscomponentsare there
in three dimensions?
Prl/b. 0-7
D-8 The uniform beam is supported by two rods AS and
CD that have cross-sectionalareas of 10 mm2 and 15 mm2.
respectively. Determine the intensity w of the distributed
load so that the average normal stress in each rod does not
exceed 300 kPa.
808
AP. D
REVIEW FOR THE FUNDAMENTALS
OF ENGINEERING
0-9
The bolt is used to support the load of 3 kip.
its diameter d to the nearest t in. The allowable
normal stress for the bolt is UalkJw
- 24 ksi.
Determine
3 kip
E.XAMINATION
Chapter 2-Review All Sections
D-13 A rub~r band hasan unstretchedlength of 9 in. If
it is stretchedarounda pol~ havinga diameterof 3 in.. determineth~ averagenormalstrain in the band.
0-14 The rigid rod is supported by a pin at A and wires
BC and DE. (f th~ maximum allowable nonnal strain in each
wire is E;lJ~w= 0.003. detennin~ the maximum vertical displacement of the load P.
Prob. 0-9
D-I0 The two rods support the vertical force of P = 30 kN.
Determinethe diameterof rod AD if the allowable tensile
stress for the material is Uallow= 150MPa.
0-11 The rods AS and AC have diameters of 15 mm and
12 mm, respectively. Determine the largest vertical force P
that can be applied. The allowable tensile stress for the rods
is UaUow= ISO MP:l.
D-15 The load P causesa normal strain of 0.0045 inJin. in
cable AB. Determine the angle of rotation of the rigid beam
due to the loading if the beam is originally horizontal beforc:
it is loaded.
0-12
The allowable bearing stress for the material under
=
the supports A and B is UalkJW 500 psi. Determine the max-
imum unifonn distributed load w that can be applied to the
beam. The bearing plates at A and B have square cross sections of 3 in. x 3 in. and 2 in. x 2 in.. respectively.
"-6ft
'!';.'i""~t.-"
Pruh. 0-12
AP. D
REVIEW FOR THE FUNDAMENTALS
0-16 The squarepiece of materia!is deformedinto the
dashedposition.Determinethe shearstrain at comer C.
OF ENGINEERING
EXAMINATION
809
D-23 A l00-mm long rod has a diameter of 15 mm. If an
axial tensile load of 100 kN is applied. determine its change
in length. E = 200 GPa.
0-24 A bar has a length of 8 in. and cross-sectional area
of 12 in2. Oetennine the modulus of elasticity of the material if it is subjected to an axial tensile load of 10 kip and
stretches 0.003 in. The material has linear-elastic behavior.
0-25 A 100mm-diameter
brassrod hasa modulusof elasticity of E
- 100 GPa. If it is 4 m long and subjected to an
axial tensileload of 6 kN, determineits elongation.
0-26 A lOO-mmlong rod hasa diameterof 15 mm. If an
axial tensile load of 10 kN is applied to it, determineits
change in diameter. E
= 70 GPa, v = 0.35.
Chapter 4-Review Sections 4.1-4.6
0-27 What is Saint-Venant'sprinciple?
D-28 What are the tWo conditions for which the principle
of superposition is valid?
Chapter 3-Revie,,' Sections 3.1-3.7
0-17 Define homogeneous
material.
0-18 Indicate the points on the stress-strain diagram which
represent the proportional limit and the ultimate stress.
D-19 Detenninethe displacement
of endA with respectto
end C of the shaft. The cross-sectional
area is 0.5 in2 and
E =29(103) ksi.
0'
D
A.J--.d' ""'~~--~oO
2 kip
6 kip
4 kip
,\
B C
\£
A~2ft '! B6ft
~
Prob.0-29
Prob. D-18
0-19 Define the modulusof elasticityE.
D-20 At room temperature,mild steelis a ductile material. True or false.
0-30 Determine the displacement of end A with respect to
C of the shaft. The diameters of each segment are indicated
in the figure. E = 200 GPa.
D-2l Engineering stressand strain are calculated using the
actual cross-sectionalarea and length of the specimen.True
or false.
0-22. If a rod is subjected to an axial load. there is only
strain in the material in the direction of the load. True or
false.
Prl.h. 1)-.'0
~
810
AP. D
REVIEW FOR THE FUNDAMENTALS
OF ENGINEERING
0-31 Oeterminc: the angle of tilt of the rigid beam when it
is subjected to the load of 5 kip. Before the load is applied
the beam is horizontal. Each rod has a diameter of 0.5 in..
and E = 29( Io!) ksi.
EXAMINATION
0-34 The column is constructed from concrete and six steel
reinforcing rods. If it is subjected to an axial force of 20 kip,
determine the force supported by the concrete. Each rod
has a diameter
= 4.20(103)
of 0.75 in. Ecorr<
ksi. Esr
=
29(103) ksi.
6 in
2ft
0-32 The unifonn bar is subjected to the load of 6 kip.
Detennine the horizontal reactions at the supports A and B.
~:.:
D-33 Thecylinderis madefrom steelandhasan aluminum
core.If its endsare subjectedto the axial force of 300kN,
determinethe averagenormalstressin the steel.The cylinder hasan outer diameterof 100mm and an inner diameter of 80 mm. Est= 200GPa,Eat= 73.1GPa.
.Prob.0-34
0-35 Two bars. each made of a different material. are connected and placed betWeentwo walls when the temperature
is Tl - lSoC.Determinethe forceexertedon the (rigid) supports when the temperature becomes T2 = 25°C. The material properties and cross-sectionalarea of each bar are given
in the figure.
Brass
E.,- IOOGPa
Steel
Ell . 200 GPa
a.,' 12(lo-6)rC
A., = 175mm:
Proh. 0-33
A., = 300mm:
c
B
A
~
,,'~~
a., - 11( loo.6)fOC
Inn
1\N
mm
.or
Pn)b. 0-35
:
2OOmm---i
811
D-36 The aluminum rod has a diameter of 0.5 in. and is
lttached to the rigid supports at A and B when TI - gooF.
If the temperature becomes T~ = 100°F.and an axial force
){
D-4O The solid 1.5-in.-diameter shaft is used to transmit
the torques shown. Determine the shear stressdeveloped in
the shaftat point B.
p s 1200Ib is appliedto the rigid collar asshown.deter-
mine the reactions at A and B. ae/ = 12.8(10-b)rF.
Eel
=
10.6(1Q6) psi.
frob. 0-40
frob. 0-36
0-41 The solid shaft is used to transmit the torques shown.
Determine the absolute maximum shear stressdeveloped in
the shaft.
Prob. 0-37
Chapter 5-Re,ie'" Sections5.1-5.5
0-38 Canthe torsionformula,l' = TclJ,be usedif the cross
sectionis noncircular?
0-39 The solid O.75-in.-diametershaft is used to transmit
the torques shown. Determine the absolute maximum shear
stress developed in the shaft.
frllb. 0-:\9
812
AP. D
REVIEW FOR THE FUNDAMENTALS
OF ENG(NEER(NG EXAMINATION
0-43 Dl:tl:rmin\: thl: angll: of twist of thl: l-in.-diameter
shaft at end A wh\:n it is subjl:cted to thl: torsional loading
shown. G = 11(10-')ksi.
0-47 Thc: shufl is mudt: from u stt:el tube having a b
core. If it is fixcd lO tht: rigid support. determine the al
of twist thut occurs ul its c:nd. Cst
= 75
GPa and G
37 GP3.
(
~
Pruh. D~3
I'rl,h. O-J7
D-44 The shaft consistsof a solid section AS with a diameter of 30 mm. and a tube B D with an inner diameter of 25 mm
and outer diameter of 50 mm. Determine the angle of twist
at its end A when it is subjected to the torsional loading
shown. G = 75 GPa.
D-48 Determinethe absolutemaximumshearstressin
shaft.JG is constant.
Do.i,~:-
Prl.h. ~
Prllb. D~
0-45 A motor delivers 200 hp to a steel shaft. which is tubular and has an outer diameter of 1.75 in. If it is rotating at
150 rad/s. determine its largest inner diameter to the nearest ~ in. if the allowable shear stress for the material is
"allow= 20 ksi.
D-46 A motor delivers 300 hp to a steel shaft. which is tubular and has an outer diameter of 2.5 in. and an inner diameter of 2 in. Determine the smallest angular velocity at which
it can rotate if the allowable shear stress for the material is
T"uuw= 20 kosi.
Chapter 6-Review Sections 6.1-6.5
0-49 Detennine the internal moment in the beam a~
functionof .\',where2 m S .\'< 3 m.
~ kN/m
~"..,
2m
:
Prclh. O~9
1m
Im--:
AP. D
REVIEW FOR THE FUNDAMENTALS
0-50 Determine the internal moment in the beamas a
functionof x. ~'here0 s x s 3 m.
0-54
OF ENGINEERING
813
EXAMJII;ATION
Determinc the maximum moment in the beam
.x.N
!
4m
I~
-"- m
I
;2.--;
Prob. D-5.a
D-SI Determinethe maximummomentin the beam
D-S5 Determinethe absolutemaximumbendingstressi
the beam.
8 kip
8 kip
D-S2 Determinethe absolutemaximumbendingstressin
the beam.
1-4ft
I
-2Olin.
-.
in.
-I14fi--l
6 ft
Prob. 0-55
0-56 Determine the maximum bendingstressin the SO-mmdiameter rod at C.
16kN
400 N/m
Prob. D-52
D-S3 Determinethe maximummomentin the beam.
A.
...L.B
L~~___~
,"
-~m
.
I
-,
I
Prob. D-56
Prnb. 0-53
0-57
What is the strain in a beam at the neutral axis?
AP. D
(
REVIEW FOR THE FUNDAMENTALS
OF ENGINEERING
D-S8 Determinethe momentM that shouldbe appliedto
the beamin order to createa compressive
stressat point D
of to ksi.
0-61
EXAMINATION
Determine the maximum stress in the beam's cross
s~ction.
-"!"
~ in.
...1..
.,---
Chupter 7-Revie" St.'cticm~7.1-7..&
0-62 Determinethe maximumshearstressin the beam.
Pnlh. I)-~~
Determine the ma.~imumbending stre$ in the beam.
11
V.20tN'
200 mm
!
i
t--
~
:,A'
l50mm"
I
Proh.
0-62
~3
The beamhasa rectangularcrosssectionand is sub= 2 kN. Determinethe ma.~imum
shearstressin the beam.
jected to a shear of V
D-6O Determin~the maximumload P that can be applied
to th~ beamthat is madefrom a material havingan allow-
=
able bending str~ss of 0',,11ow 12 MPa.
'
150mm
.1-
p
I
'--
0.5 in.
~
..1
20 mm
T
110.5 in.
o.s~
;0 mm
-:-T 20 mm
IOOmm
~
I
c
,:i
...;;:r;i~ii
2m
;,.
PTllh. 1)-iu}
-i-"'--~~~~:::
'" m
-
--
3.
1ft.
~
1'rllh. 1>-6.;
.
IjI..5 1ft.
~
AP. D REVIEW FOR THE FUNDAMENTALS OF ENGINEERING EXAMINATIO~
D-64 Delerminethe absolutemaximumshearstressin the
shafthavinga diameterof ~ mm.
815
0-67 The beam is made from four boards fastened to~ether
at the top and bottom with two rows of nails space:d ever)'
4 in. If an internal she:ar force of V = 400 lb is applied to the
boards. determine
the shear force resisted by each nail.
(
!
3
.
.
-:
3 ..
.
11
Prob. D~
0-65 Determinethe shearstressin the beamat point A
whichis locatedat the top of the web.
Chapter 8-Revie,,' All Sections
D-68 A cylindricaltank is subjectedto an internalpressure
of 80 psi. If the internal diameterof the tank is 30 in.. and
the wall thicknessis 0.3in.. detenninethe maximumnonnal
stressin the material.
Prob. 0-65
~
A pressurized
sphericaltank is to be madeof 0.2S-in.thick steel. If it is subjectedto an internal pressureof p s
ISOpsi. determineits inner diameterif the maximumnormal stressis not to exceed10 ksi.
0-70 Determine the magnitude of the load P that will
cause a maximum nonnal stressof Umax- 30 ksi in the link
D-" The beamis madefrom two boardsfastenedtogether
at the top and bottom with nails spacedevery 2 in. If an
internal shearforce of V = 150Ib is appliedto the boards,
detenninethe shearforce resistedby eachnail.
along section a-Q.
I
t-2
PrClIl. 1)-711
I
:T
-L
.
O.5m.
816
AP. D
REVIEW FOR THE FUNDAMENTALS
OF ENGINEERING
0-71 Determineth~ maximumnormal stressin the horizontal portion of th~ bracket.The bracket hasa thickness
of 1 in. and a width of 0.75in.
EXA~"NATI0N
0-74
The the
solidcomponents
cylinder is subjected
the loading
shown.
Oetennine
of stressattopoint
B.
0.75in.
Prl)b. 0-71
0-72 Determinethe maximumload P that can be applied
to the rod so that the normal stressin the rod does not
exceed O'max
Z 30 MPa.
Prub. 0-7.&
20mm
frob. 0-72
D- 73 The beam has a rectangular cross section and is subjected to the loading shown. Determine the components of
stress O'x.O'yand T:t.V
at point B.
Chapter 9-Review Sections 9.1-9.3
0-75 Whenthe stateof stressat a point is representedby
the principalstress,no shearstresswill act on the element.
True or false.
0-76 The state of stressat a point is shown on the element.
Determine the maximum principal stress.
6ksi
~
4ksi
.5
.
17::-- In,1.5in.
Prllh. 1)-7.;
Pn)/). 0-76
AP. D REVIEW FOR THE FUNDAMENTALS OF ENGINEERING EXAMINAnON
0-77 The state of stressat a point is shown on the element.
Determine the maximum in-plane shear stress.
817
D-8O The beam is subjected to the loading shown.
Determine the principal stress at point C.
IISOpsi
-r7Smm
,
.
-r-
100psi
7Smmi
-1-
200 psi
Prob. D-77
Pr(tb. D-80
0-78 The state of stressat a point is shown on the element.
Determine the maximum in-plane shear stress.
130MPa
Chapter
50MPa
12-Review Sections12.1-U.2. U.5
0-81 The beam is subjected to the loading shoWl!.
Determinethe equationof the elasticcurve.£1 is constant.
2kip/ft
f
Prob. D-78
)
0-79 The beam is subjected to the load at its end.
Oetenninethe maximumprincipalstressat point B.
~B
~
-3ft
Prob. 0-81
D-82 The beam is subjected to the loading shown.
Determine the equation of the elastic curve. £1 is constant.
3~
-'!'--W--1.-:T
'!I".~.,
".-r--'--
A
~~.'!;:ii,'
101\,
Prclh. D-S2
c'.i!
i
818
AP. D
REVIEW FOR THE FUNDAMENTALS
OF ENGINEERING
D-83 Determin~ the displacem~nt at point C of the beam
shown. Us~ th~ method of superposition. EI is constant.
EXAMINATION
D-87 A 12-rt wooden rectangular column has the dimen.
sions shown. Determine the critical load if the ends are as..
sumed to be pin-connected. E - 1.6(103)ksi. Yielding does
not occur.
p
,."
1.1'
3ft
'I
Pruh. 0-83
T
4in.
1
D-84 Determin~the slopeat point A of the beamshown.
Usethe methodof su~rposition. £1 is constant.
~
kN/m
Prob. 0-87
D-88 A steel pipe is fixed-supported at its ends. If it is 5 m
long and has an outer diameter of SOmm and a thickness of
10 mm. determine the maximum axial load P that it can carry
without buckling. Est = 200 GPa. O'y- 250 MFa.
Chapter 13-Review Sections 13.1-13.3
D-85 Thecritical loadis the maximumaxialloadthat a column can supportwhen it is on the vergeof buckling.This
loading representsa caseof neutral equilibrium. True or
false.
D-86 A 50-in.-long rod is made from a 1-in.-diameter steel
rod. Determine the critical buckling load if the ends are fixed
supported. E 29(103) ksi, O'y = 36 ksi.
-
0-89 A steel pipe is pin-supported at its ends. If it is 6 ft
long and has an outer diameter of 2 in.. detennine its smallest thicknessso that it can support an axial load of P = 40 kip
without buckling. En - 29(103)ksi. O'y- 36 ksL
0-90 Oetennine the smallest diameter of a solid 4O-in.-Iong
steel rod, to the nearest tr;in., that will support an axial load
of P - 3 kip without buckling. The ends are pin connected.
ESt = 29(103)ksi, O'y= 36 ksi.
AP. D
REVIEW FOR THE FUNDAMENTALS
D-8
PARTIAL SOLlTTIONS AND ANS\I\TERS
CD is a two-forcemember
MemberAE:
= 0;
FCD
- 600 Ib
D-9
=
w
n' )
1\1"
W W
10'
-
-'
= 3 MN/m
0'
- ~;24 - -/4I; d - 0.3989m.
use d - 0.5 in.
- 0; A,. = 6 kN
SegmentAD:
1:F,.= 0; V - 2 kN
0., T AS
AM.
D-1. BC is a two-forcemember.
BeamAB:
1:Ms
=
Rod AB:
P
us -' 300(
A'
Rod CD:
= 0; A>, = 800 Ib
SegmentACF:
IMF = 0; MF = 600 lb. ft
819
Beam:
I.F.v
IME
EXAMINATION
I.MA-o.,TcD=2w
0-1 Entire frame:
IMs
OF ENGINEERING
Ans.
Ans.
D-4 SC is a two-forcemember
BeamAS:
-
Ans.
-
1:MA 0; Tsc 4 kN
1:Fx = 0; Ax 3.464 kN
-
- 0; A" = 6 kN
FA - V(3.464)"l+ (6)"l- 6.928kN
'fA - fA - 6.928 - 22.1kPa
A f(O:m)'i
1:F"
D-S
6: O'z,0'7' O'~,TzyoTy:, T:z
Ans.
Ans.
Ails.
1.8 w w -'(2)(2)'
D-6 Joint C:
:t.I.Fx - 0; Tc. - 10kip
Ta
u
A
10
2
.
Sksi
11(3)- 9
Ans.
9
0-14
- 0.0472in.lin.
-
.4ns.
(8DE)max ~masIDE 0.003(3) 0.009m
By proportionfrom A,
8sc 0.009 (t)
0.0036m
(8sc)max
- EmasIsc = 0.003(1)- 0.003m < 0.0036m
Use 8sc - 0.003m. By proportion from A,
8p = 0.003
0.00525
m - 5.25mm
Ans.
S
-
Ans.
1.11kip/it
S
-
(¥) -
820
AP. D
REVIEW FOR THE FUNDAMENTALS
D-lS
lAB = \rf(4)~+ (3)~= 5 ft
/' ,IB S 5 + 5(0.(x~5) = 5.0225 ft
The angleBCA wasoriginally 8
OF ENGINEERING
EXAMINATION
D-27 Stressdistributionstl:nd to smoothout on sections
- 9()0.Using the co-
sine law. the new angle BCA (8')
-
5.0225='\1(3):' + (4)Z 2(3)(4) cas 8
is
further rl:movl:d from thl: load.
0-28
An.r.
1.) Linear-~lastic material.
2.) No large deformations.
An.r.
8 = 1,X).53Mo
Thus
~8' - 90.5)80
- 9()0
- 0.538°
= 0.0166in.
Ans.
0-18 Proportionallimit is A.
Ultimate stressis D.
Ans.
Ans.
+
0-19
The initial slope o( the 0'
- E diagram.
Ans.
Am.
0-10 True
0-2l False. Use the original cross-sectionalarea and
length.
AIlS.
f
0.5(29(103»
O.S(29(1f}J»
-f
0-17 Material hasuniform propertiesthroughout.
+ 4(6)(12)
-2(2)(12)
Am.
Ans.
12(103)(0.5)
(0.02)2
200(10'1)
27(loJ)(0.3)
(0.05)2 200( 10'1)
- 0.116mrn
Ans.
0-31 BeamAS:
~M.. 0: FSD 2 kip
~Fy 0: F"c = 3 kip
-
-
~" _!!::..
3(8)(12)
.
AE = of(0.5)2
29(1OJ)- 0.0506In.
~s
PL - Ai
2(3)(12)
of(0.5)2
29( 103)
- 0.01264in.
0-22 False.There is also strain in the perpendiculardirection due to the Poissoneffect.
AIlS.
Ans.
Ans.
(1(xx)()(8)
F..
2£
Ans.
=!.!S!:l
= 4 kip. FB= 2 kip
AE
Ans.
0-33 Equilibrium:
P$l + Pill - 300(loJ)
Compatibility:
6(103)4
-
P",L
..tn
[f (O.08)Z]73.1 (IQY)
MPa
Ans.
64.3
f (O.OI):!100(10"') - 3.06 mm
-
-
D-32 Equilibrium:
F",+FB-6
Compatibility:
F.. (1)
6c'", = 3aB: AE
AP. D
...14
REVIEW FOR THE FUNDAMENTALS
821
D-J8 No, it is only valid for circular cross sections. Noncircular cross sections will warp.
Ans.
Equilibrium:
PCQfIC+ Pst
OF ENGINEERING EXAMINATION
= 20
Compatibility:
- 40 Ib . ft
D-.J9T- = ~CD
P~
(2)
~ = 8$'; ['IT (6)2 - 6(f)(O.7S)2]
4.20(103)
-J -
Pit (2) (103)
- 6(f) (0.75)229
P- - 17.2kip
Pit - 2.84kip
Ans.
D-.-
,.35 cSr-p= Ia4TL
4
40(
f 12)(0.375)
(0.375)
IC
T-
- 5.79ksi
Equilibrium of segmentAS:
TB =30Ib ft
T = I:;. = 30(12)(0.75) 543 .
B
J
f (0.75)"
psi
.
-
Ans.
B.o8d=I.fb.
AE
Compatibility:~
+~ =0
12(10-6)(25- 15) (0.4)+ 21(10~)(25 - 15)(0.2)
- £(0.4)
£(0.2)
175(10-6)(200(109» - 300(10~)(100(109» = 0
F = 4.97kN
Ans.
)-36 Equilibrium:
FA + Fs = 1200
Compatibility
Removesupportat B. Require
8s (8B1A)telnp+ (8B1A)\oad 0
-
D-4%
- of200(12)(3)(12)
(0.75)411(1<1')
+ 0 + of300{12)(2)(12)
(0.75)411(1<1')
-
a4 TL + "'"
!..!::..
=
=0
LAE
12.8(10-6)(100
- 80)(14)+
1200(6)
F.(14)
f (0.5)2
10.6 (1C1') - f (0.5)2
10.6 (1C1') -
-
Ans.
Am.
a4TL
P = 1.86kip
~
:.t
='
I!:.
£... fG
-
600(12)(3)(1~)
f (0.5)411(106)
-
+ f200(12)(2)(12)
(0.5)411(106) f 100(12)(3)(12)
(O.5t 11 (106)
Am.
-f
40(0.3)
(0.OLS)4
75(10"')
=0
+ f [(O.025~-20(0.2)
(O.Oi25~J
75(10")
+.fb. = 0
AE
12.8 (10-6) (SO- SO)(14) +
Ans.
- 0.253 tad counterclockwise when viewed from A.
Compatibility
Removesupportat B. Require
+ (8B1A)load
- 0.0211fad clockwise when viewed
tad
0
0-37 Equilibrium:
FA+Fs=P
SinceFs = O.FA = P
8s = (8B1A)remp
-0.0211
from A.
D-43
Fs 1.05kip
FA -1531b
TL
-400(12)(2)(12)
= 2: 1G - f (0.75)411(1<1')
.AIB
P(6)
t (O.5:)21Q6
(106) - 0
Ans.
- f [(0.025)4-30(0.3)
(O.OI25).J
75(10")
= -0.209 (10-3) fad - 0.209(10-3) fad clockwise
whenviewedfrom A.
.4ns.
822 AP.D REVIEW FOR THE FUNDAMENTALS OF ENGINEERING EXAMINAll0N
D-!G A" = 3 kN
Use sectionof length~.
P - 110(XXJ
T- -;;;
1SO - 733.33tb . ft
T.lkN
rj
dl
= j;Tc
20(103)
-
Intensity of w -
t~ at~.
L+ 1:M= ~ -3~ + (t~)[t.r)(t.r)] + At. 0
f 733.33(0.875)
[(0.875)4
- rj4]
M=3.r-- 9
- 0.867in.
- 1.73 in. use dj - 1.625in. = It
in.
.~ns.
D-5l
By
Ay
",
=2.6 kip
-
4.6 kip
Draw M-diagram
Mmax- 7.80k. ft (at C)
Ails
T - .f. - 165~
OJ
D-S2 Draw M-diagram
Mm.. = 20 kN . m (at C)
ill.!!!!
. (12)(1.25)
-
An$
- (1)4]
T [(1.25)4
.-tns.
(II a 54.7 rod/s
D-47 Equilibrium:
Tst + Tbr = 950
Compatibility:~t
-
-
Ay = 2.33 kN
By 6.617kN
-
Draw M-diagram
Mmaxc 11 kN' m (at C)
=~
(0.6)
f [(0.03)4 -Tst
(0.015)4]
75 (109)
T b,
0-53
-
Tbr (0.6)
f (0.015)4
37(1Q9)
D-S4 Ay
,.\n,f
= By =800N
DrawM-diagram
Mmax= 1600N . m (within CD)
Tn 919.8N. m
30.25(0.6)
~ ~ = f (0.015)4
37(109)- 0.00617
tad
-
Ans.
D-SS Ay = By = 8 kip
M- c 8 (4) - 32kip. ft
I
0'
D-48 Equilibrium:
T,4+Tc-600
Compatibility:
D-S6
T,4 (2)
881c= 881,4;!£.ill.
JG =~
-
Mc
T
32(12)(3)
if
(2)(6)3
=
-
=
32
D'
An...
J
Ay z By
100>N
Mmu
1250N m
-
.
l?qvO.025\ , 5509MPa
O'mu
- -Mc
I - .--,
f (0.025)4 .
T,4=200N.m
Tc=400N.m
Tc - 400(0.025)
Tmax= T
f (0.02S~- 16.3MPa
A. = 5.5kN
Usesectionof length%.
L+ !M 0; -5.5 % + 4 (2)(%
,\{ = 8 2.5%
-
-
-
0-58
1) + M
.-\n.f
An.\'.
0-57 . - 0
0-49
.\ns
30.25 N . m
-0
\n\
0'
M
M
=~
\n.\
-
M(1)
10(103) [i'f (4)(4)3
-"i'I (3)(3~]
- 145.8kip .in. - 12.2kip'
ft
\11\
AP. D REVIEW FOR mE FUNDAMENTALS OF ENOINEERINO EXAMINATION
-59 From bottomof crosssection
- - ~~ = 4«80)(20)
+ 95(30)(100) - 7,S.870
Y
80(20) + 30(100)
DIm
I
ft (20)(80)' + 2O(S»(7's.870-
-
40)2
-
+ -!i (100~)' + 1~X9.5 - 75.m)2 4235(10-6)
m
- TMc -
0'-
10(103)(0.075870)
4~S'(10-o) . - 179MPa AIlS.
-
I.
D-67 I-.l. 12 (6)(6}'
- 34.66(10-6) m4
~Mc
(1)(6.5
. AM
,.- XR.
/1 - 4(103)(8
84.667
(1)- 5») - 567pst
F. q.r. (56.2SIb/in.) (2 in.) . 112.5Ib
(0.02)(0.150>'
+ 2 [-& (0.1)(0.02)3
+ (0.1)(0.02)(0.(MS)2]
0"
/- -k (1)(6)' + 6(1)(5 - 3)' + -k (8)(1)' +
8(1) (6.5 - 5)Z- 84.667 in.
(6)(4)3
. 32in.
12
15O(1)(6)(2)J
- 56'tc Ib/iIn.
q . ~.J
32
- PI2 (2) - P (at C)
ft
-Y - ~I.yA = 3(6)(1)
+ 6.5(1)(8)
.
6(1) + 1(8)
- 5 In.
D-66 J.l
...a Ay- PI2
M-
D-'5 From the bottom:
M
12(Iv-)
q - ~1
P CO.(W5)
- 34.66
(10-6)
p - 4.38tN
Ans.
(0'_)0
-
!
400(2.5)(6)(1)
- qs - 34.62Ibfan.(4 in.) - 138lb
t
.
! (6)(4)J - 404
. pst
+ (SOsin JOO)12 (2)
- 69.23Ib/im.
- 69.23/2 - 34.62 Ib/in.
D-a 0'- l!.. - ~
(4) (6)J
86.67
Foronenail
q
F
).41 Maximum stress~
at DorA.
(SO~ JOO)12 (3)
-
- .l.12 (4)(4)3- 86.67in4
0.3
- 4(XX)
psi- 4 ksi
Q is upperor lower half of crosssection.
Tmu-~-
-lMPa
v-a
AIlS.
0-63 [=,', (3)(4)3-,', (2)(3)3- 11.5in4
Q is upperor ~
half of ~
secUon.
- (0.75)(1.5)(2) - 3.75
~~
XR.
[t - ~
11.5(1) - 0.UJ~
tsi.
T- D-64
Ay
- 4.5tN. By -
~
'rmu - It -
2(o:s
Ans.
+
-&
(0.5)(2)3
Ans.
P-3kip
0-71 At 8 sectionthrou8h the center of bracketon centroidal axis.
N-700lb
V-o
1.5tN
section.
M - 700(3 + 0.375) - 2362.5 Ib . in.
¥ t (0.06)
... (0.03)2]
£t'r (O.03)C]
4..5(10')[(
= 2.12MPa
M - (2 + l)P - 3P
P Mc
0'--+A
I
P
(3P)(1)
30 )
in3
V- - 4.5tN (atA)
Q is upperhalfof ~
Ans.
D-70 At the sectionthroulh centroidaJaxis
N-P
It
Q - (1) (2) (3)
Ans.
150,
2 (0.25)
Ans.
r - 33.3in.
d - 66.7in.
~2
Am.
v- AP + TMc
.-tn.Oj
- 26.1ksi
700 + [1'r
2362.5(0.375)
(1)(0.75)3J
-0:7S'('i)
Ans.
824
AP. D
REVIEW FOR THE FUNDAMENTALS
OF ENGINEERING
EXAMINATION
0-75 True
0-76 0'..=4 ksi. u.y- -6 ksi.1'.., -8 ksi
Apply Eq. 9-5.0'1 8.43ksi. O'l -10.4 ksi
0-'77 0'%
- 200psi.0', = -ISOpsi.1'..,- 100psi
Apply Eq. 9-7.1'max = 202psi
in.".a...
0-78 0'..- -SO MPa.0', - -30 MPa, 1'..,- 0
Use Eq. 9-7. 1'max = 10MPa
in-pla...
0-79 At the crosssectionthroughB:
N 4 kN. V = 2 kN. M 2 (2) 4 kN . m
-
-
{
.-tns.
P-23.5N
0-73 At sectionthroughB:
= 500 lb.
N
V
- 400 lb. M
= 400
(10)
- 4(XX) lb.
in.
-
0'8--+---
Axial load:
0'..
Shear load:
1'..,
= ~.
It
=I
I
'TB
.
4
-
-
4/0.03)
.,
-+
,
0:03{0:()6) *<0.03)(0.06)3
224 kPa (T)
- 0 sin<:e Q -
O.
Thus
0'\
pst
An.J.
An.J.
224 kPa
~=O
D-8O Ay - By - 12 kN
Bending moment:
My
4000(1)
0'..
Note
400(1.5)(3)(1)1.37.5.
£,'I(3)(4)']3
Mc
A
- .f.
A -~
4 (3) - 41.667psi (T)
-
-
P
.
Segment
S ,::&(3)(4)1- 250pst (C)
A 1I.r.
A C:
Vc.O. Mc= 24 kN. m
'Tc - 0 (since V c = 0)
Thus
- 41.667
- 2500', - 0
'Txy = 37.5 psi
O'x
~
psi (q
Ans.
Ans.
Ans.
D-74 At sectionB:
N; - SOO N. V, = 400N. Mx = 400 (0.1) = 40 N. m
M, = 500(0.05)- 25 N . m. T; = 30 N . m
Axial load:
p
0'; = A
SOO - 63.66 kPa (C)
= -;"(O:O'SV
0-81 Ay - 3 kip
Use section having a length x.
M = 3.t - x2
El-:"TJ2v = 3x -
0':
1';"- 153kPa
Ans.
-
Use section having a length .x.
Intensityof w
- (-to\r
L+IM=O:
r
at .x.
-IS.\' + 100 + (t.x) [t(ii-.x1.x>]
153 kPa
- 63.66 + 407.4 = 471 kPa
1:.:-c 0
1':. - 0
x2
D-82 A.y= IS kip
MA 100kip . ft
M
0:.- 0
-0
-1 (- x4+ 0.5.r;3
- 2.25x)
-
0'..- 0
+M
v= EI'
Moment about y axis:
0';
0 since B is on neutral axis.
Thus
(f)
L + IM = 0; -3x + ~
Integrate tWice. use
v = 0 at x = O. v - 0 at x - 3 m
-
Torque:
Tc -1(0:05)4
30(0.05)Tzz = T
Am.
u.:
Shearload:
'T;, 0 sinceat B. Q O.
Momentabout.\' axis:
Mc = T(O.05)4
40(0.05) c 407.4kPa (C)
0'; = T
-
O'c = 0 (since C is on neutral axis)
0'\ c 0': c 0
Am.
Ans.
..\m.
Ans.
Ans.
Am.
-
IS.x
- 0.05.r3 -
Integrate tWice, use
v = 0 at .\' = 0, dvldx
-
-0
- O.OS
.r3- 100
tfl11= IS.x
EI "d.'t!
v
+ At'
100
= 0 at .x - 0
11 (2.5.0c3
- 0.0025r - SOr)
.-\ns.
AP. D
REVIEW FOR THE FUNDAMENTALS
D-88
A
-
OF ENGINEERING
'Ir(0.02S)2- (0.015)2)= 1.257(10-3) m2
1- ~11'«0.02S)4 p
_:!!!:.!:!-
c
(KL)Z
- 84.3kN
-34 From Appendix C, consider distributed load and couple moment separately.
woL3
ML
8A = 45E/ + 6Ei
4 (3)-'
20 (3)
= 45E/+~-
12.4kip. ((2
£1
0'
D-89 p
A ns.
J
-85 True
Ans.
-
- ~(KL)Z - r
(29 (loJ» (-1(0.5)4)
[0.5(50)]2
= 225.
k'
'P
0-90 P =~.
--
ksi < 36 ksi OK
,= 1 - 0,528 = 0.472 in.
3
Ans.
- .,,229(10-') G ,4)
(KL)2'
0.382in.
P = ". (0.382)2
3
A
d = 2r
AM.
- rz4)1
'[I (6)(12)f
- ..
IT
0'
Ans.
29( 10-')[7 (14
rz = 0.528in,
= Ap = 11' [(1)2 -40(0.52S)Z] = 17.6
,z
m4
r (200(109»(262.04)(10-9)
(0.5(5)]2
- ..1!!..!:!-.
(KL)Z' 40 -- r
Ans.
= 2.03 kip
(0.015)4) = 267.04 (10-')
84.3(10-3)
(1(}1) - 67.1 MPa < 2SOMPa OK
AP - 1.257
Thus
-86 P
825
EXAMINATION
[1 (40)]2
- 6.53ksi < 36 ksi OK
- 0.765
Ans.