D-483
~~@
®M-~
~JA\'iffi@~&;
STRUCTURAL DESIG,N and CONSTRUCTION '
NOV. 2019
Max. flexural stress :
1. CE Board Nov. 2019
f = 6M
A temporary earth retaining wall consists of
planks driven vertically into the ground. The
wall is designed to resist 2.4 m. height of soil.
Given:
Plank dimension: 300 mm wide x 75 mm thick
Plank allowable stresses:
Bending =.10.4 MPa
Shear =0.8 MPa
Unit weight of soil = 17 .3 kN/m3
Active earth pressure coefficient Ka = 1/3
Which of the following gives the maximum
flexural stress?
: Fi nd the maximum shear stress.
What should be the thickness (mm) of the
planks to prevent failure?
f = 6(3.986)1°0
b
300(75) 2
Solution:
M.
aximum flexural stress
w=17.3(2.4X0.3{
® Maximum shear stress
F
i)
F = (3)( 4980)
y'
2(300)(75)
F
V
@
=0.33MPa
'
Min. thickness of planks to prevent failure
By bending:
6(3.986)10
10.4 = 300 d2
r-
II
p =4.98 kN
I
•'
''
''
''
''
'
''
.
I
I
''
''
'
'
I
fl
.
8
d =88 say 90 mm
By shear:
'
2.4 1
M::~
3
M.:: 3,986 k
N.m.
2bd
fb = bcf'
2
M:p( n2.4)
=!:!_ .
v
6M
W=4.152
P::~
6
fb =·14.2 MPa
<D
<D
bd2
b
.
''
- -- -
w
''
'
'
p
3V
F =v
2bd
3(498Ql
O.B = 2(300)d
d =31
mm
used=90mm
D-484
CMI En11naar1nu ucensure Examinations
LMo=O
2. CE Board Nov. 2019
35.6(4.3) + 142.4(0) =178x
X = 0.86 m.
A simply supported girder of a bridge spans
25 m. The standard truck load (H load)
consists of 2 moving loads, 4.3 m. apart. The
loads are as follows:
P2= 35.6 kN
P1 = 142.4 kN
To obtain max. moment, place the lo
su?h _a war that -the center of the span ~
comc1de·with the center of the biggest 10Will
and the resultant load as shown in :
e
figure. _
35.6 kN
142.4 kN
ft. :
®
}:Me=O
25 R1 = 178(12.07)
R1 = 85.94 kN
'
CD Calculate the maximum support reaction.
@ Calculate the maximum moment in the ·
girder.
@ Calculate the maximum shear at midspan.
142'.4 kN
35.6kN
_,R=l78 kN
max . moment
occurs here
Solution:
CD Maximum support reaction.
Placed the biggest load at point B where
max. reaction occurs.
R1
12.5
125
Ri
35.6 kN
142.4 kN
4.3 m
Max. moment: .
Mmax = R1 (12.07)
Mmax = 85.94(12.07)
Mmax = 1037.3 kN.m
X
R=l78 kN
LMA=O
~SR1 = 142.4(25) + 35.6(20.7)
Rt= 111.9 kN
@
Maximum shear at mid-span.
To obtain max. shear at the mid-span use
influence line.
142.4.kN 35.6 kN
® Maximum moment in the girder.
142.4 kN
+0.5
35.6 kN
B
12.5
4.3 m··- ..,. _ -
R1
- ·20.7 m
~0.5
D-485
CIVIi Enatneerlnu ucensure Examlnadons
h
-0.5
CD Find the minimum width of beam ·b".
required to satisfy on cover requirements
8,2=125
h=-0.328
@
Find the minimum width of beam ·b"
adequate for a factor shear force
Vu = 600 kN if the spacing of 12 mm
diameter ties is 50 mm.
@
If Vu = 450 kN and spacing of 12 mm
diameter ties is 70 mm what is the required
minimum width of the beam b" mm?
Max. shear at mid-span:
Vmax =142.4(0.5) +35.6(0.328)
Vmax =82.BBkN
1
3. CE Board Nov. 2019
Solution:
Given:
G)
As=8-28 mm0
As' =4- 28 mm0
Minimum width ·b" required for cover
requirements
Jlmm0
ds = 12 mm diam. ties
h1 =125 mm
h2=475mm
a=55 mm
fc' =28 MPa
fyb =415 MPa (longitudinal bars)
fyv =275 MPa (ties)
Shear strength reduction factor= 0.75
Clear concrete cover = 40 mm
Specified maximum aggregates size in the
concrete mixture = 20 mm
b
Vaiue of x : should .be the larger value of the
following.
o bar diameter = 28
f) 25mm.
C) .
. · . ... .,,
d
.
:..
...
.
. .' . -'¼ .. . t
·. :.•.. __..
~-:•. ·,t~::· ., a=55
.
.
.--~---'
. .
.
b
½
4
max. aggregate size = 3(20) =26.67 .
usex =28 mm.
b =40(2) + 12(2) + 28(4) + 3(28) =300 mm.
..
. · ·. :·· -... I
1
l1 1=415
'
D-486
CMI En11neerln1 ucensure Examinations
® Value of b if Vu= 600 KN and spacing of
tie wires = 50 mm
28 55
d =600-40-12- - - -
2
2
d =506.5mm
V
....!!.
0
=VS +VC
vs
2
50 = (12) (2)(275)(506.5)
vs
V
:i_v
e
0 .
@
I
Which of the following gives the maximum
moment at D?
600000 .
0.7S -630121 =169879
Ve =0.17'A.ft:b.d
169879 =0.17(1)✓28b(506.5)
b =373 say 370 mm
@
Which of the following gives the maximum
reaction at D?
shear at D?
V
=
G)
® Which of the following gives the maximum
Vs =630121
Ve
Beam ABCDE is to be analyzed for maxim
forces at ultimate conditions. The beamu~
15
simply supported at A, B, C, D.
Factored loads:
Dead load = 15 kN/m
Live load = 19 kN/m
S = A}>td
f
4. CE Board Nov. 2019
Value of b if Vu= 450 kN and spacing of
tie wires is 70 mm
·
.' 1,1 I I ~ I l I l
6m
f
DL = 15 kN/m
LL "' 191:N/m
F V I·
6m
I II
I''
6m
6m
Both Dead Load and Live Loads Pattern:
A f...d
S=.2..L
vs
1t {12)
2
(2)(275)(506.5)
4
70=---------
vs
-
v. =450087
V
VC =.....M..-VS
0
V=
C.
450000
0.75 -45Q087
OJ 93wL
SHE,ti'>-
0536wl
~
0.464wL
0 .b07wl
:dw
L.
(IJ'/;.L
MOMEN'.f-.-.r, ~-r-f---,-~r--~
I
J
-0.107 1wL
149912 =0.17(1)f2ab(506.5)
b =329.03 say 330 mm.
l'>
~ '::J
VC =149912
Ve =0.171i.ft:bwd
Otll!•·L
·. ~
o.~wL
,
,_
j
-- ~
0 ..5.\1,L
,:
I -0.0114wL' i ,0.1()11"
I
'- ~ ~
· o~l(il.,
···
LI
CIVIi Enatn11r1na ucensure
D-487
Examinations
LL= 1.143 wL
Live Load Patterns:
LL= 1.143(19)(6)
wL
i
1
OJll•L
SHE-tb
Max. Ro ·= 102.87 + 130.30
Max. Ro= 233.17 kN
E
D
C
B
A
LL = 130.30 kN/m
. wL
wL
0.603w L
I'--.
® Max. shear at D
0.55llwL
~
I~
DL = 0.607wL
~~--o_,l_~ '--J
" "'-.II
OJ<l7wL
DL = 0.607(15)(6)
0 .442wL
062••L
DL = 54.63 kN
LL= 0.607wL
LL= 0.607(19)(6)
LL= 1144.61 kN
0.603L
Max. shear Vo= 114.61 + 54.63
· Max. shear Vo = 169.24 kN
Live Load Patterns:
@
Max. moment at D
DL = 0.1071wL2
DL = 0.1071(15)(6)2
DL = 57.83 kN.m.
· LL=0.1071wL2
LL= 0.1071(19)(6)2
LL = 73.26 kN.m.
Mo= 73.26 + 57.83
Mo= 131.09 kN.m.
+O.!J9Y6.,,.z
MOM£N?;\-- ..
~Ll
~
0.446t -01)5~
-0 .0357wL l
I
-l) .OH6wt l
I
0.518/,
5. CE Board Nov. 2019
CD
Solution·
•
Max. reaction at D
DL :: 1.143 wL
DL- ◄1
DL = -143(15)(6)
- 102.87 kN/m
From the figure below, has the following data:
a =1 m
0
=30·
B= 45·
P1 = 1.8 kN
P2 =0.90 kN
PJ =0.45 kN
D-488
CIVIi Enulnearlna ucensure ExamiRadons
6.. CE Board Nov. 2019
Determine the vertical reaction at B.
® Determine the horizontal reaction at B.
@ Determine the moment where P1 is acting.
Given the following properties of angular
.
section:
mm_
16
x
mm
100
x·
mm
100
One Angle
A= 2974 mm 2
rx = 30.48 mm
rv = 30.48 mm
rz = 19.79 mm
G)
. Solution:
G)
Double Angle 100 mm x 100 mm x 16 mm
A= 5948 mm 2
rx = 60.96 mm
rv = 43.69 mm
Vertical reaction at B
K= 1.0
Fy = 348 MPa
E = 200000 MPa
<D Ftnd the· capacity of a single angle 100
mm x 100 mm x 16 mm if it is used asa
column having a height of 4.5 m.
® Find the capacity of the double angle 100
mm x 100 mm x 16 mm if it is used asa
column having a height of 4.5 m.
@
I:MA=0
Rey (2) + 0.318(0.707)-0.318(1.707)
-0.90(1)-0.90(0.134)-1.56(0.5) = 0
Rey= 1.059
® Horizontal reaction at B
LFh=0
1.56-0.318- Reh= 0
Rah= 1.242
Find the length of a single angle if based '
on the .design Of compressive force in the
limiting slenderness ratio.
Solution:
<D Capacity of single angular column section
KL/ r = (1)( 4500)
19.79
KL/ r =227.39
C =
@
Moment where P1 is acting
LFv=O
RAy +Ray= 0.90 + 0.90 + 0.318
RAy = 1.059 kN
M= 1.059(0.134)
· M = 0.1419 kN.m.
C
✓2nF E
2
y
2
C = 21t (200000)
348
C
CC =126.17 < KUr
I
CIVIi 111a1naartn1 Ucensure Exa111nauons
use:
@
121t2E
F, = 23(KL / r) 2
D-489
Length of single angle
KL
-=20 0
r
121t2(200000)
F, = 23(227.39) 2
~=20 0
F =19.92 MPa
L=200(19.79)
Capacity of a single angular section:
L=3958mm
P =Fa A
L = 3.958 m
rz
•
p =19.92(2974)
P=59235 N
P =59.235 kN
7. CE Board Nov. 2019
® Capacity of double angle
KL/ r = 1(4500)
43.69
KL/r=103
cc =126.17
F=[1- (Kl/r-)2] ~
1
2 CC2
F.S.
F.S. = ~ + 3(KL/r) _ (KL/r)3'
3
8 CC
8 CC3
KL/r
103
= 126.17 =0.82
c:-
F.S. =~ + ~(0.82) - (0.82)3
3 8
F.S. =1.91
Fa= ( 1- (0.82)
·
Properties of channel
A =4529 mm 2
Ix =35.4 x 106 mm•
d = 229 mm
ly =3.0 x 106 mm•
tr= 14 mm
b1 =87 mm
tw= 10 mm
Radius of gyration, rx =88.4 mm
Radius of gyration, ry 2 25.S mm
Distance from the back of the web to the y-
axis, x =2.49
2opooo MPa
Modulus of elasticity, E =
8
· oout the
CD Compute the slenderness ratio a
2
]
~
2
1.91
Fa :: 120.94 MPa
Capacity 0f d
ouble angle column·
p::: Fa A
.
p::: 120.94(5948)
p ~ 719373
p..
N
.. 719.37 l<N
Two channels welded at the tip of their flanges
form a built-up column. Unsupported column
height is 9 m. The column is braced against
sideway in both direction. Column ends are
fixed, K = 0.50.
x-axis.
. abollt the
® Compute the slenderness ratio
y-axis.
. . . ted to a concentriC
® If the column is subJec . e the thickness.
determtn
load of 1400 kN • . a width of 15Omnl
of cover plates having and t,ottom of ~he
be placed on top
compressive
t
o .
section.
Allowable
P
stress= 110.6 M a
D-490
CMI En11near1n1 ucensure Examlnauons
Solution:
KL _ 0.5(9000) .
67.22
<D Slenderness ratio about the x-axis
y
KL= 66.94mm
ry
- - - - ,I- - -
i
124.9 1 6:? .I
1
87
1
62.1
..,
1
24 .9
X
I
@
Thickness of cover plates placed on both
top and bottom of the section to carry a
concentric load of 1400 kN
P=AF8
1400000 = [ 2(4529) + 2(150)t ]110.6
87
t =12mm
Ix= 2(35.4)106
Ix= 70.2 x 106 mm4
ly = 2(3 X 106) + 2(4529)(62.1 )2
ly= 40.93 x 106 mm4
r
=
X
fi:"
VA
A= 2(4529)
70.2 X 106
r =
2(4529)
X
rx
=88.41 mm
•
KL _ 0.5(9000)
--
--'-~
88.41
rx
KL
-=50.90mm
rX
®
:l:dr,ess ratio about the y-axis
y
vt
r :
40.93 X 106
1
2(4529)
rr =67.22mm
8. CE Board Nov. 2019
The butt connection shows bolts spaced as
follows:
S1 =·40 mm
S3 = 50 mm
S2 = 80'mm
S4 = 100 mm v
t1 = 16mm
ti= 12mm
Steel strength and stresses are:
Yield strength, Fy = 248 MPa
Ultimate stren-gth, Fu= 400 MPa
Allowable tensile
stress on gross
area = 0.60 Fy
Allowable tensile stress on the net
area= 0.5 Fu
Allowable bolt shear stress = 120 MPa
Allowable bearlng stress, Fp = 240 MPa
Bolt hole diameter = o + 3 mm
Tensile force T = 352 kN
CD Determine the diameter of bolt due to
shear stress of bolt.
00
® Determine the diameter of bolts based
the net area capacity of the plate.
th
@ Determine the bolt diameter based on e
bearing capacity of bolts.
CIVIi 1a11n11rtna ucensure Examtnaaons
D-491
9. CE Board Nov. 2019
~ l~ad W is to be lifted using the crane which
·~ ~ '.l ~ . ~
C, ; ~ ~- --
. . 100 - ~o.. r50- -100
- - ~o--
! S ,--40
➔
T
1s hinged at B as shown in the figure.
ls,= so
- I
_
,S, =40
~50
Solution:
Diameter of bolt due to shear stress of bolt
G)
T =Ab FP n
352000 = ; (d} 2(120}(2} . double shear
d=15.3 say 16 mmr,
@
Diameter of bolts based on the net area
capacity of the plate
S1 = 0.5 Fv
Dh=D+3
T=,StAnet
<D When "W" is being lifted, find the force in
member AC.
® Find W in 'terms of AB.
@ If an uplift force is acting at C equal to 30
kN, find W.
A
Solution:
<D Force in member AC
4
352000 =0.5(400)(160- 2 Dh)(16)
tan0= -
160 - 2 Dh = 110
e=29.os·
AC
7.2
D = 160-110
.
h
2
Oh= 25
tanB=
4
w
A
3 ·.
B =53.13°
Dh=0+3
25 =D+ 3
D= 22mm e
® Bolt diameter based on the bearing
capacity of bolts
T=Sb At, n
352000
=240(d)(t)(8)
352
000 =240(d)(16)(8)
d=11 .45 say 12 mm
C
. 36 87" =AC Sin 60.95'
ABSm ·
AB= 1.457 AC
Ifv =0
5' - AB COS 36.87'
W + AC Cos 60.9 • : 1 457(AC) Cos 36.87'
W + AC Cos 60.95 - .
W= Q.486AC
AC= 2.06W
D-492
CIVIi EnD1naer1na ucensure Examlnauons
W in terms of AB
(2)
CD Find the normal shear stress.
® Find the torsional stress.
@ Find the bending stress.
AB= 1.457 AC
AC=2.06W
AB = 1.457(2.06}W
W= 0.333AB
@
Solution:
CD Normal shear stress
4V
u=3A
W if there is an uplift of 30 kN at C
u = 4(5000}
3(: }100)
36.87'
2
u =0.85MPa
30
® Torsional stress
-.-c /;
- -- - - - - - -60
AC Sin 29.05 = 30
AC= 61 .78 kN
/
,
',
'
:
/
:
" y
:
X
"'
I:Fh=0
: R := 0 .80
" 60"~·'
":
'
Y
AB Sin 36.87° = 61.78 Sin 60.95"
AB= 0.6 kN
LFv=0
X =R -
W+ 0.6 Cos 36.87" = 61.78 Cos 60.95°
W= 29.52kN
X = 0.80(1
X =0.107
R Sin
so·
- Sin 60")
T= 5x
T = 5(0.107}
T = 0.535 kN.m.
10. CE Board Nov. 2019
Torsional stress:
~ curve quarter-circular beam having a
diameter of 100 mm as shown in the figure.
Radius of circular beam= 0.8 m
16T
Jtd3
t=-
t
= 16(0.535)(10)6
7t(100} 3
t
@
J
P=5kN
5KN
=2.72 MPa
Bending stress
Y=0.80 Cos 60"
y = 0.4 m
M=Py
M = 5(0.4)
M=2kN.m.
CIVIi En11n11rln11 Llcansura Exa111nauons
,ro•
Solution:
l=-
64
G)
4
7t(100)
I=--=--
Allowa.ble uniform load
w
1{
64
1=4.91
D-493
6
x 10 mm
4
J J
t
3m
C=~
2·
f
= MC
I
'b
9m
RA
C= 100 =50
LMs=O
~ (9) = w(12)(6)
RA=Bw
,
- f = 2(10)6(50)
b
4.91(10) 6
LFv= 0
~+Rs= 12w
Rs= 12w-8w
Rs=4w
fb =20.37 MPa
5w
11. CE Board Nov. 2019
A WF 360 x 382 beam is simply supported at
Aand,is hinged at B. Column AC is a 300 mm
square .solid section.
Properties of W 360 x 382
bt =406 mm
Ix = 1410 x 106
d=416mm
ly=536x106
f l l l
3m
A
-4.5w
9m
6m
MC
f =b
I
C
Find th
, e allowable uniform load (w) that
can
be
1·
·
st
. app 1ed if the allowable flexural
® _ress ts 165 MPa?·
Find the b It d'
she . 0 rameter at B if it is in double \
11
@
:a~le shear stress is 130 MPa.
·size of Plat a_nng stress of plate at A if the
CD
Fin:~:
e IS 200 X 250. .
165
''
l l
M(T)
= 1410 X 106 ·
M=1118.5 kN.m.
M=8w
1118.5 =Sw
w= 139.8125 kN/m
t
RB
CMI Ena1near1n1 llcensure Examlnalions
@
Bolt diameter at B
Vmax =5w
Vmtrx = 5(139.8125)
Vmax = 699.1 kN
12. CE Board Nov. 2019
· A Fink
roof truss is used for a residential
building and loaded as shown:
fV =FV
V
-=F
2A v
699.1 X 103 =
2(: )d')
1.5 m
130
d =58.5 say 60 mm
@
Bearing stress on plate A
1-
3m
3m
3 Ill
~
<D If P = 85 kN, find the force in member BC,
neglecting H1, H2, and H3.
® When H1·= 11 kN, H2 = 16 kN, H3 = 27 kN,
find the reaction of the roller at A if P=O.
@
When ti1 = 11 kN, H2 = 16 kN, H3 = 27 kN,
find the vertical reaction of Gif P=0.
Solution:
<D Force in member FC, FcF
P=85kN
~=Bw
~ = 1118.5 kN
A
D
. f = RA
P,. Ap1as
f = 1118.5 X 10 3
P
200(250)
fp = 22.37 MPa
A,,
3m
I
F
3m
· 3m
G,.
In a truss, if th.ere are three members on a
joint .and two members are collinear, th8
third member is a zero force member,
· are
At point B, members AB and BC _o.
collinear, so member BD =0, and CD .,
EF =qand CF = 0.
D-495
CIVIi Enuln11r1nu Ucensure Examlnauons
At joint C:
LMA=O
Gv (9) = 11(0) + 16(2.37) + 27(4.74)
Gv = 18.43 kN
P=85kN
R= 11 + 16 + 27
R =54 kN
FBC
FCE
Fae= FcE
tane=~
1Jv= 0
e = 18.43'
Foc(
4.5
¾a)+ J¾o =85
2Foc(
.
'
LFv=O
Av + Gv = R Cos 0
Av ·= 54 Cos 18.43-18.43
¾a) =85
Av= 32.8 kN
Fec =134.4kN
@
Vertical reaction of G
Gv = 18.43 kN
® Reaction of roller at A
H,=27 kN
13. CE Board Nov. 2019
1.5 m
A steel column with unsupported length equal
to 4.5 m is to support an elevated floor.
Available section is L100 x 100 x 16 mm. Use
Fv =248 MPa.
y
length ~ member AC (~c):
LAC :: ✓~5)2 +(1.5)2
LAc
=4.74 m
__
LAB::: L
- 4.74
ec-2
LAS == 2.37 m
:. - - - X
. ylL--4-- --f
IJ-496
=:.::==~===~--------.."'.
CMI Ena1ne1r1nu ucensure Examlnadons
Properties of L100 x 108 x 16
A= 2974 mm2
Ix= 2.76 x 1Q6 mm 4
ly =2.76 x 106 mm 4
lxy = 1.60 X 106 mm4
rx =30.48 mm
ry = 30.48 mm
rz = 19.75 mm
= 2[1y + Ad 2]
lyT = 2[2.76 x 106 + 2974(31.24)~
lyT
lvr = 11.32 x 106 mm
lxT < lyT
x =31.24mm
s:s2 x 106
y =31.24mm
rX =
Kl S200
rX = 30.48
r
CD Find the allowable load for double angle
welded together.
4
2974(2)
kl _ 1(4500)
rx
30.48
® Find the allowable load for single angle.
@
Find the length of a single angle if based
on the limit of slenderness ratio for a
compression member.
kl = 147.64
rX
C=t1t'FE
C
Solution:
y
<D Allowable load for double angle
.r
2
C = 21t (200000)
C
248
cc ~126.17
kl
r
.
> CC (long column)
.f= 31.24 l'= 31.24
2
IXr =21x
F = 121t (200000)
8
23(147.64) 2
lx1 =2(2.76 X 106)
lxT =5.52 X 1~ mm4
.
Fa= 47.25 MPa
D.497
CIVIi Enalneerlna ucensure Examinations
@
p =Fa AT
p =47.25(2)(2974)
P= 281 kN
.
Length of a single angle based
on the hmit
of slenderness ratio fo
r a compression
member
kl
-S200
CV Allowable load for single angle
The angle bends on the z-axis.
r
1(L)
200
19.75 =
L= 3.95m.
14. CE Board Nov. 2019
z
Beam A~C is supported by spandrel beams at
the exterior edges and by a column at B.
For beams b x h = 300 mm x 400 mm
Column = 400 mm x 400 mm
Dead load= 6 kN/m (all weights included)
Live _load = 4 kN/m
rz =19.75 mm
kl = 1(4500)
19.75
r
kl
- =227.85
Co"""•
I
L,=6m
L,=6.8 m.
r
kl
7 >Cc
(long column)
F = 121t2 E
. 23(~ )'
CD Determine moment at end Bfor beam AB.
® Determine shear force at B for beam BC.
@ Find the positive moment at midspan of
beam AB.
Solution:
CD Moment at end B for beam AB
2
F1 =!31t (200000)
23(227.85)2
F. =19.85 ~Pa
P:::FaA
p::: 19,85(2974)
p .. 5
.. 9,03 kN
w=6+4
w= 10kN/rn
D-498
CMI Enu1naar1nu Llcensure ba111nadons
t
CD Determine the max. punching shear
.
:
stress.
® Determine the wide beam shear stress.
@ Determine the number of 12 mm 0 bars
parallel to the length of footing within the
,length of d1 ..
wl2
M =-9
s
l= 5 + 5.a =6.4m
2
M = _ 10(6.4)
•
2
6m
9
B
l
3.9 m
MB = - 45.51 kN,m.
® Shear force at B for beam BC
·. .,,A-
V = 1.1'5wl
2
B
I
.
.:B:
-
l
I
·. ·
-·· • .
I
156.8
VB= 39.10kN
@
l .
I .· ·
761.f>
I
I
I
I
I
2
B
· I .
I
I
I
V = 1.15(10)(6.8)
4m
· 0.4
0.4
Vu
'
Positive moment at midspand of beam AB
wL2
M=-
d=0.5
1j
M= 10(6)
11
-509.60
2
M = 32.73 kN.m
15. CE Board Nov. 2019
A combined footing is shown on the figure.
Dimension = 6 m x 4 m
Column = 400 mm x 400 mm
Net soil pressure in ultimate
condition = 98 kPa
columns= 3.9 m
between
Distance
.
'
Effective depth of foo~ing = 500 mm
Reduction factor for shear 0 = 0.75
Reduction factor for moment 0 = 0.90
fc' = 27 .5 MPa
fy = 413 MPa
Solution:
CD Max. punching shear stress
Net ultimate uniform soil pressure in kN/m ·
along the length of footing:
qu = 98(4)
qu = 392 kN/m
0.4 ·
· 0.4 ·
:
·■ 0.4
· ■ o.4
_
-==i
~-~- ----=--6m
I
1
I 4111
II
J.,
CIVIi pg1n11r1nu ucensure Examlnauons
Oet. The column load Pu2.
761 _6 + 392(0.4) - Pu2 = - 50~.60
pu 2 = 1428 kN
Net upward soil pressure in kPa = 98 kPa
Vu =1428 - 98(0.9) 2
Vu= 1348.62 kN (punching force)
V
u
up - b d
"
0
b0 =4(900)
@
D-499
No. of 12 mm 0 bars along d1=1.30 m
M = 509.60(1.3)
u
2
Mu =331.24 kN.m.
Mu= 0 fc' b d2 oo(1 - 0.59co)
331.24 X 106 =0.90(27.5)(4000)(500)2
oo(1 - 0.59co)
ro-0.59ro2 =IA01338
.crJ. -1.6949co + 0.02268 =0
ro = 0.0133
b0 =3600
d=500
1348620
up= 0.75(3600)(500}
up =1.0MPa
® Wide beam shear stress
Vu+ 392(0.5) = 761.6
Vu= 565.6 kN
_ _I
900
·
Wide beam shear stress:
V
0bd
\)::: - L
\) :: _ 565600
0.75(4000)(500)
'U ::: 0.38 MPa
Steel ratio:
(I) fc'
p=fy
0.0133(27.~
P= - 413
p =0.000884
, ,-.,uu
CMI Enalnearlnu ucansure 11an11n1taoaus
P=0P
u
n
1.4
pmil = fy
pu =0.65(5940)
1.4
pmin = 413 -
Pu = 3861kN
Pm =0.00339
Use pmil = 0.00339
@
Design moment
Rn = 0.175
As= pmin b d
_
As= 0.00339(4000)(500)
As= 6780 mm 2
(from interaction diagram)
R = _P~n_e_
n
f.'A
h
C
g
No. of 12 mm 0 bars:
0.175 ~
: (12)2N = 6780
e = 175 mm
5940000e
27.5(600)(600)(600)
N = 16.78 say 17 bars
Mu= Pue
Mu =_3861 (0.175)
16. CE Board Nov. 2019
A tied column 600 mm x 600 mm ·is subjected
to a nominal load Pn at an eccentricity e.
Concrete compressive strength, fc' = 27.5 MPa
Reinforcing steel yield strength, fy=413 MPa
Strength reduction factor 0 = 0.65
From the interaction diagram, Kn = 0.60 and
Mu = 675~68 kN.m.
@
Value of "h" so that p = 0.04 when e =300
mm
·
Rn = 0.21 (from interaction diagram)
p = 0.03
.
Pe
CD Determine the ultimate axial load Pu.
R = --"-"" fC I Ag h
® Determine the design moment.
@ Determine "h" so that p = 0.04 when
e = 300 mm
0.21 = 3861000(300)
27.5(600)(60Q)h
•
Solution:
h =557.14mm
CD Ultimate axial load Pu
p
K :: ---lL_
n
f, A
C
g
p
0.6Q =
n
27.5(600)(600)
Pn =5940kN
CIVIi 1■1naar1n1 Ucensura Examlnauons
2.0
D-s01
~-- r"T T-- ,--, r--r -!T ~=- ---- --r- --INTERACTION DIAGRAM
fc' = 27.5 MPa
i::
1.8
···············
····· ······· ·
1
1
,.,..
I
I
fy=413 MPa
y = 0.7
1.6
m"'; ··························-f-··········· ····· .. ······f················.. ,-......
l.4
.i
I
·----.. r-················ ... ······r·--············ ······~---·
h---.i
...
yh
I
.. 1
I
I
•• t •
•
• ••
♦
i·
1
I
1
i
s
If., =0
••····· ,····v··•······"•······; , ....., ·-········ ...
l
······-·. ••f--..............................~•-···············
1.0
5 .i
i
i
0.8
-----·.-.•············
0
0.4
.
~ ..J.
.!
0.2
1
i
t:t==~~~=~~-4+-~~-f~-t"T-1 . . .
•,=0.~
L""'---'--~~_LL~4~~~~~~
·1
o.o
i ..
···)······· •·······---~···
---t ---- -1
j
:'
l
:
0,0()
.
r
I
D-502
000
~CM~I~En~•~•n~••~r~ln!u~ll~c•~n~su~r:•:Exa::m:1°:•:::=__• ___.---- --........,---....."'
solution:
17. CE Board Nov. 2019
CD Effective length factor of column BC
The frame shown has the following values of
moment of inertia of beams and columns.
1
3m
.......
DD
B
E
.
n
IJ
4m
I
AB
BC
DE
EF
GH
HI
BE
CF
EH
Fl
r:
l
I
11.5 m
Members
,
fl,,
'
Moment of
Inertia (mm4)
221.85 x 106
221.85 x 106
' 300.94 x 1()6
300.94 x 106
275.96 x 1Q6
275.96 x 1Q6
757.54 x 106 ·
· 557.75 x 106
986.47 x 106
757.54 x 106
11.5 tn
Length
(mm}
At the top at C:
4000
· 3000
4000
3000
G = L IC /LC
I Ig /Lg
A
4000
221.85 X 108
3000
· 11500
11500
11500
11500
Using the appropriate alignment chart,
determine the effective length factor of
column BC. Assume this to be uninhibited.
® Using the appropriate alignment chart,
determine the effective ~ngth factor of
column AB. Assume this to be uninhibited.
@
557.75 X 108
A
11500
GA= 1.52
Using the appropriate , alignment chart,
determine the effective length factor of
column AB. Assume this to be inhibited.
1
•
At the bottom at B:
G
(D
3000
G=
= I: IC /LC
I
B
IG /LG
221.85 X 106 + ~
G
=
8
3000
400P--
6
557.75 X 10
11500
GB= 2.67
Using the alignment chart:
K = 1.6 .
D-503
clVII E1111neer1nu ucansura ExamlnaUon$
00
Git
-
At the top at B:
K
00
20.0
to .0
""
100.0
5(l.0
100.0
50.0
30.0
20 .0
5.0
4.0
3().0
20.0
10.0
9.0
8.0
221.85 X 106 + 221.85 X106
10.0
9.0
8.0
7.0
3.0
j
G = L IC /LC
A
LIg /L g
7.0
GA= - ~
30=00~-=-~ 4~0~00[__
557.75 X 106
6.0
5.0
6.0
5.0
4.0
11500
4.0
2.0
GA= 2.67
At the bottome at A:
Gs = 1.0 (fixed support)
From the alignment chart:
(b)
K = 1.65
Sidesway unhibited
G.4
® Effective length factor of column AB
3 rn I
I
i
4 rn
OD
B
E
I
I
K =t.65 ,
ll.5 m
00
100.0
50 .0
30.0
20.0
10.0
9.0
8.0
7 .0
6.0
K
00
20.0
10.0
100.0
50.0
5.0
30.0
4.0
20 .0
3.0
10.0
5.0
4.0
Gs
00
2.0
9.0
8.0
7 .0
6.0
5.0
4.0
3.0
,. __
2.0
11.5 m
1.0
0
1.0
(b)
SiJcsway unhibitl!<l
0
D-504
i;IVII Engineering llcensare Examlnauons
@
Effective length factor of column AB,
inhibited
DC]
3m
E
B
4m
,
k =0.825
11.5 m
11.5 m
. Gs
K
GA
00
00
500
10.0
5.8
4.
3.0
2.0
1.0
0.9
0.8
0 .7
0 .6
0.5
;
0.3
0.3
· 0.6
0.2
0.1
At the top of column AB:
G =
A ·
r 'C /LC
I,I /L
g
(a)
Sidesway inhibited
g
221.85 X 106 221.85 X 1.06
----+--4000
3000
G =
6
557.75 X 10
A
11500
GA= 2.67
\
At the bottome at A:
Gs = 1.0 (fixed support)
)..
From the alignment chart:
. k 0.825
=
0
0.5
.