Page 17 Macaulay’s Method,
description
Transcript of Page 17 Macaulay’s Method,
-
1
Hints to Problems
Chapter-1
1.18
150 mm 200 mm550 mm
40 kN 10 kN 20 kN 50 kN
Elongation is given by, AEPL=
= ( ) 224.020000050550000301500004000020530)4/(
12 =++ XXXX mm
1.22
Area of column = 3002465604
2 =X mm2; Area of steel = 4072364
4 2 =XX mm2 Area of concrete = 246 300 4072 = 242 228 mm2
We have, c
c
s
s
EE = or
00015000210cs = or cs 14=
s X As + c X Ac = 800 000 14 c X 4072 + c X 242 228 = 800 000 or c = 2.673 MPa s = 2.673 X 14 = 37.429 MPa
Average compressive stress = 248.3300246000800 = MPa
1.23
P
1 2 3 P 1P 2 P 1
P
21
-
2
On welding, tensile stresses will be developed in 1 and 3 whereas compressive stress in 2. From equilibrium equation, 2 P1 cos = P2 .(assuming negligible change in ) (i) From compatibility equation, 1 = 2 cos
or
=
22
22
11
11
EALP
EALP cos or
=
22
22
11
21
cos EALP
EALP
cos
or 222
22
2
111 cos
=
EALP
LEAP
If all rods are of same material and section,
222
222
21 coscos
=
= P
LAE
AELP
LAEP
From (i), 23
22
cos2 PPLAE =
or
)cos21(cos2
32
3
2
+=
LAEP and
)cos21(cos
32
2
1
+=
LAEP
1.24 Here, Lt= ;
222
226-2
22
222
2
111 cos40120010 X 191200
0008542cos
=
=
EALPXXX
EALPtL
LEAP
From (i), 23
22
222
2
11 cos2 PEALPtL
LEA =
or 23
22
222
2
11 cos2 PEALPtL
LEA =
or ( ) ( )
+
=+
=
311
222
222
3
22
11
32
2
11
2
cos21cos
21
cos2
EAEAL
tLEA
EAEA
tLL
EA
P
1.26 Aa = 80420 =X mm2 , As = 3201620 =X mm2 Total compression in Al = Total tension in steel
32080 XX sa = or sa 4=
c
cccc
s
ssss E
LtLE
LtL =+ or c
cc
s
ss E
tE
t =+ 1.28 Let 3 be the stress in the direction of restraint. Then,
1 = 1/E 3/E (i) 2 = 3/E 1/E (ii)
-
3
and 3 = 3/E 1/E (iii) If there is no restraint in the third direction, 13 ' E= However, for the given restraint, 133 2
2/' E
==
113 2 EE
= or 13 2 =
Substituting in (i), 1 = ( )212111 222
12
. =
=
EEEE
Modified modulus of elasticity, E = 21
1
22
=E
Chapter-2
2.11 Both stresses are compressive,
tan = 816.0180120 ==
y
x
or = 39.23o
r = 2222 sincos yx +
129598640
23.39sin18023.39cos120 2222
+=+= oo
= 147 MPa compressive 2.14
OL
R
FMCE
S
P
N60
5
10
10.952.5
2.5
2
10.9
5
0.9
-
4
2.15
(i)Principal stress }4){(21)(
21 22 ++= yxyx
8.5380
1000016002180
})50(4)60100{(21)60100(
21 22
=+=
++=
= 133.8 MPa (tensile) and 26.2 MPa (tensile)
tan2 = yx
2 = 5.2
60100502 =
X
or 2 = 68.2 or 1 = 34.1o and 2 = 34.1o + 90o = 124.1o max = )2.268.133(
21 = 53.8 MPa at 34.1o + 45o =79.1o
2.16
30o240 MPa
180 MPa
A B
C
30o240 MPa
180 MPa
A B
C
120 MPa
120 MPa
207.8 MPa
On plane BC, x = 8.20730cos240 =o MPa 12030sin240 == o MPa The equivalent state of stress on the panes is shown in the figure. Resultant stress on plane AB = 22 120180 + =216.3 MPa
5.1120180tan == or = 56.3o
Principal stress }4){(21)(
21 22 ++= yxyx
8.1208.193
1000025600219.193
})120(4)1808.207{(21)1808.207(
21 22
=+=
++=
-
5
= 314.6 MPa (tensile) and 73 MPa (tensile)
tan2 = yx
2 = 63.8
1808.2071202 =
X
or 2 = 83.4o or 1 = 41.7o and 2 = 131.7o max = )736.314(
21 = 120.8 MPa at 41.7o + 45o = 86.7o and 176.7o
2.18
pq r
40080200
P
R'
Q
R
CO
28oO
r60o q
60op80
400200
60o 60
E F(a)
(b)
O'
32o
OF = 449 MPa inclined at 28/2 or 14o to 400 MPa stress OE = 251 MPa inclined at 32/2 or 16o to 200 MPa stress 2.19
p q r
1.6 X 10-6
P
R'
QC
O
10oO
r45o
q
45op
45oE F
(a)
O'
10o
45o
3 X10-64 X10-6
1.6 X 10-6
3 X10-64 X10-6
Major principal strain 1= OF = 4.04 X 10-6 at 10/2 or 5o clockwise of plane p
-
6
Minor principal strain 2 = OE = 1.58 X 10-6 at 95o clockwise of plane p.
Chapter-3 3.13 A = (/4)X242 = 144
Diameter of the reduced section = 24/2 = 12 mm Area of the reduced section = (/4).122 = 36
When the bar is turned down to half the diameter along half of its length, let P be the equivalent load to induce the same maximum stress.
=000205144
60000020536
600X
PXXPX
+ = 0.000 03235 P = kP ..... (Taking k =0.000 03235)
Using the energy equation
W (h+ ) = 21 P.
W (h+ kP) = 21 P. kP
P2 2WP 2Wh/k = 0 (multiplying throughout by 2/k)
3.15 As = ( /4)242 = 144 ; Ab = ( /4)(362 - 242) = 180 Let x = Extension of bar in mm
L
xEss
.= and L
xEbb
.=
Strain energy of the bar LAELxELA
ELxELA
ELA
E bbb
ss
sb
b
bs
s
s
2.2.22 222
2
2222
+=+=
)(222
222
bbssbb
ss AEAE
LxA
LxEA
LxE +=+=
)18000098144000205(30002
2
XXX
x += = 24693 x2 N.mm Potential energy lost by the weight = W(h + x) = 13500 (6 + x) N.mm
Chapter-4
-
7
4.11
B
(b) SF Diagram
(c) BM Diagram
C A
(a)6m1m 3m
6 kN 8 kN
D
8
36
246
Taking moments about end A = 016986 =+ XXXRb or kNRb 11= ; and kNRa 31114 ==
4.14
40 kN 20 kN40 kN 20 kN
1 m 1 m2 m2 m2 m
A BC D E F
40
40
20
80
20
Taking moments about end A = 2404206202405 XXXXXRb ++=+ or kNRb 40= ; and 804020204040 =+++=aR kN Shearing force diagram Portion AC: Fx = kNRa 40= (constant ) Portion AD: Fx = kN408040 =+ (constant ) Portion DE: Fx = 04040 = (constant )
-
8
Portion BE: Fx = 20 kN (constant ) Portion BF: Fx = kN204020 =+ (constant ) Bending moment diagram Portion AC: Mx = - x40 (linear )
At A, x = 2 m and Ma = -80 kN.m; Portion AD: Mx = )2(8040 + xx (linear )
At D, x = 4 m and Md = 0 Portion DE: Mx = 0 Portion EB: Mx = - x20 (x from E) (linear )
At B, x = 1 m and MB = -20 kN.m Portion BF: Mx = - )1(4020 + xx (x from E) (linear )
At F, x=2m, Mf = 0
4.17 The loading on the beam is shown in the figure
10 m
w
A BDE
R a R ba 4-a6 mC
Let distance of pier at A be a m from the end D, the other will be at B, (6-a) m from end E. Distance CB = CE BE = 5- (4-a) = 1+ a
Taking moments about B, )1(106 awRa += or 3)1(5 awRa
+= Bending moment
Portion DA: Mx = 22wx (parabolic); Md = 0; Ma = 2
2wa (negative value)
Portion AB: Mx = )(3
)1(52
2
axawwx ++
It is maximum when dM/dx = 0 or 03
)1(5 =++ awwx or 0)1(53 =++ ax or x = 5(1+a)/3 Maximum bending moment =
++++ aaawaw )1(
35
3)1(5
2)]1)(3/5[( 2
= )25(9
)1(5)1(18
25 2 aawaw ++++
= [ ])41055)(1(185 aaaw +++
-
9
= [ ])5)(1(185 aaw +
= )45(185 2aaw +
The maximum bending moment will be as small as possible if the magnitudes of the
sagging and the hogging bending moment are equal.
4.19
8 kN/m
B
l = 15 m8 m
CA
16 kN
5 m D
25 kN 18.33 kN2.33 kN
51 kN
32 kN
113.9 kN.m 114.9 kN.m
85 kN.m
SF
BM
5.81 m
13.05 m
(a)
(b)
(c) Taking moments about B,
38
288
315
2815101615 XXXXXXRa +=
or 25=aR kN and 8325282316 =
+= XRb kN Intensity of distributed loading at any cross-section at a distance x from A = (w/l).x=wx/l Shear force diagram
Portion AD: Fx = 75.3
25152
82521 22 x
Xxx
lwxRa == (parabolic)
Fa = 25 kN; Fd = 18.33 kN
Portion DB: Fx = 1675.3
252
x ..(parabolic); Fd = 2.33 kN; Fb = -51 kN Between DB, shear force is zero when Fx = 1675.3/25 2 x =0 or x= 5.81 m
-
10
Portion BC: It is convenient to deal this portion by using variable x from end C.
Fx = 2828
21 22 x
Xxx
lwx == (parabolic); Fc = 0; Fb = 32 kN
Shear force diagram is shown in Fig. 4.21b.
Bending moment diagram
Portion AD: Mx = 25.11
253
.75.3
2532 xxxxx = (cubic); Fa = 0; Fd = 113.9 kN.m
Portion DB: Mx = 25.11
)5(16253xxx (cubic)
Fd(x=5) = 113.9 kN.m; Fb(x=15) = - 85 kN Mmax(x=5.81) =
9.11425.11
81.5)581.5(1681.52525.11
)5(162533
== Xxxx kN.m
Bending moment is zero at 25.11
)5(16253xxx = 0 or 090025.1013 = xx
Solving by trial and error, x = 13.05 m Portion BC: x from end C.
Mx = 632
32 xxx = (cubic) ; Fc = 0; Fb = -85 kN.m 4.20
w'xx' w
x
C
AB
dx
wx
67503600
3600
Let w = rate of loading at the midspan
Then total load = wwX 4632 =
or w47200 = or w = 1800 kN/m Taking C as origin of the parabola, At a distance x from A or x from C,
-
11
2'' kxw x = , when x =3 m, 1800' =xw kN or k = 200 22 )3(200)1800('200' xworxw xx == or 2)3(2001800 xwx = or 22 2001200)69(2001800 xxxxwx =+= ( ) 1322 3200600360020012003600 CxxdxxxFx ++== At x = 0, 3600=xF , C1 = 0
32006003600
32 xxFx +=
0 1 2 3 4 5 6 Fx 3600 3067 1733 0 -1733 -3067 -3600
2
43
32
3502003600
32006003600 CxxxdxxxFM xx ++=
+==
At x = 0, Mx =0 C2 = 0
3502003600
43 xxxM x +=
0 1 2 3 4 5 6 Mx 0 3417 5867 6750 5867 3417 -3600
4.21
4 kN4 kN
2 m2 m3 m3 m
A BC D
8 kN.m
1
3
4 kN
E
3
8 kN.m5
2
Chapter-5
5.13
-
12
64444
1045.3664
)200220(64
)( XdDIxx === mm4 3
6
max
104.331110
1045.36 XXyIZ xxx === mm3
Internal area of pipe = 01.0)2.0(4
2 = m2 Weight of water /m run of pipe = 3082.081.9101.0 =XX kN Area of cross-section of pipe = 2
22
3.65974
)200220( mm= Weight of pipe /m run = 0.0065973 X 70 X1 = 0.4618 kN Total weight/m run = 0.3082 + 0.4618 = 0.77 kN 5.14
D
b
d
A B
CD
sinDBC = and cosDAB = ;6
cos.sin6
222 DbdZ == If beam is to be strongest, Z must be maximum.
i.e. 0=ddZ or ( ) 0cossin2.cossin
63
2
=+ D or 22 cos2sin = or 2tan2 = or 2tan = or 3/2sin = or 3/1cos =
DDXBC 816.03/2 == and DDAB 577.03/1 ==
5.16
-
13
80 mm
20 mmx x
y
y
120 mm
643 10394.11)40(64
12080121 XXXIxx == mm4
10503
5.135002450 =+= XXXRa N ; 90010501950 ==bR N Bending moment between CB,
22
2504506002
500)1(4501050 xxxxxM x +==
For maximum value, 0=dx
dM x or 0500600 = x or x =1.2 m 810)2.1(2504502.1600 2max =+= XM N.m
5.17
2030 30
30 2504/110004/max === XWlM N.m
31907)2030(64
44 == xI mm4 1.2127
1531907
max
===yIZ xxx mm
3; 5.1171.2127
000250/max === ZM MPa For composite section, Moment of inertia about neutral axis,
( ) 4810561520304
319074 222 =
+ X mm4
-
14
5.18
Neglecting self weight W of the beam, ''.44
'maxmaxmax
max kWWyIly
IlWy
IM =
===
If W is considered,
+=
+=
+=2
'4
.2
'84
'' maxmaxWWky
IlWW
IyWllW
%Error = 1002/'
2/100
2'
'2
'100
'' X
WWWX
WWk
kWWWkX +=
+
+=
or 1002/'
2/ XWW
We += or eWeWW += '2100 or '2)100( eWWe =
or e
eWW
2100' =
5.21
b
t
t'
b/2
t
t'
Modular ratio is 2. The equivalent steel section is shown in the figure. As the neutral axis of the section is to be at the dividing line,
2''.
22. ttbtbt = or 707.0
21
'. ==tt
5.22
2028
If the stress in steel reaches to maximum value, the stress induced in brass
= 1410
75.1150 X = 61.22 MPa
brasssteelr MMM += = 20028720.
3222.61
282028
32150 3
44
=
+
XX N.mm
-
15
5.25
10 m
x
16080
At a section x-x, at a distance x from the top, side of the square,
xs .10000
8016080 += or )80(125 = sx mm
Moment of inertia about the diagonal or the neutral axis = 12
4s mm4
Bending moment at the section = 1000x =1000 X 125 (s-80) N.mm
3
6
4
3 )80(.2105.1
212/)80(10125
ssXsX
ssX == MPa
For bending stress to be maximum, 0=dsd
Chapter-6
6.12 Area = 2bt + t(2b-2t) = 4bt - 2t2 = 4bt (Neglecting higher values t)
[ ] [ ] tbtbbbtbbtbtbbbIx 33444433 38)4(32)(3212 )22)((12)2( ==== shear stress (maximum) at neutral axis,
zIyAF=
23
22)(
2
222
2 tbtbtbtbttbbtyA =+=+
= neglecting higher powers t.
btF
tbttb
FzIyA
F169
)3/8.(2/3
3
2
=== Average shear stress = F/4bt
25.2=
AverageMax
6.15
-
16
z
h
b
t
Refer Fig., 3233
101.5075.2755512
555212
605 XXXXXIx =
++= mm4 The shear stress in the flanges,
At a distance z from the tip, zFXXtX
XztFtI
yAF .102.54101.5075.27... 6
3===
Total force in each flange,
FzXFXdzFztXtdzFf 448.025102.54.102.54.
5.27
0
266 =
===
Chapter-7
7. 9
==lxpw
dxdF sin
Integrating, dx
dMClxplF =+
= 1cos
Integrating, 2122
.sin CxClxplM ++
= At x =0, M=0 , C2 = 0; At x =l, M=0 , C1 = 0
Maximum bending moment = 22
pl
Reaction at support (Value of F at x=0) = pl
7.10
P
l/2l
W=wlA BC
-
17
Taking moments about A, 2/.2/3 lWlPXXlRb += or 2/)3( WPRb += Consider a cross-section at distance x from C,
2
2
2
222
+= lxwlxRPxdx
ydEI b
Integrating, 1322
26222ClxwlxRxP
dxdyEI b +
+=
Integrating again, 21433
.224266
CxClxwlxRxPEIy b ++
+= 7.15
4 kN/m
A BC
8 m
6 kN
D3 m 3 m 3 m
E
2 m
4 kN/m
A
B
C
8 m
6 kND
3 m 3 m 3 m
E
2 m
To apply Macaulays method, uniformly distributed load has to be continuous upto the end of the beam. To compensate the same an upward uniformly distributed load has to be considered from D to E as shown in the figure. Taking moments about A, 5.3341168 XXXXRb += or 5.13=bR kN and 5.45.1318 ==aR kN At any section x from A, )8(5.13
2)5(4
2)2(45.4
22
2
2
+= xxxxdx
ydEI
Integrating, 2
)8(5.136
)5(46
)2(425.4 233
1
2 ++= xxxCxdxdyEI
Integrating again, 6
)8(5.1324
)5(424
)2(4.65.4 344
21
3 +++= xxxCxCxEIy At A, (x=0), y = 0, 02 =C (consider only first part)
At B, (x=8m), y = 0, 24
)58(424
)28(486
85.4044
1
3 ++= CX ; C1 = 22.69 slope and deflection equations are as follows:
-
18
2)8(5.13
3)5(2
3)2(27.2225.2
2332 ++= xxxx
dxdyEI
6)8(5.13
6)5(
6)2(7.22
65.4 3443 ++= xxxxxEIy
Deflection at E, x=11m
6
)811(5.136
)511(6
)211(117.226
115.4 3443 ++= XXEIy =68.2 kN.m3
mmXX
Xy 7.221015000200
102.686
12
== For maximum deflection at distance x, making slope equal to zero,
mxxxorxx 27.301.6875.6)2(23
)2(27.2225.20 233
2 ==+++=
4.486
)227.3(27.37.226
27.35.4 43 =++= XXEIy kN.m3
mmXX
Xy 1.161015000200
104.486
12
== 7.16
xA B
Mb
l
Ma
MaMb
BM diagram
ymaxzab
A' B'
D'
C
As the intercepts on a given line ( on BD) between the tangents to the elastic curve of a beam at any two points is equal to the net moment taken about that line of the area of the bending moment diagram between the two points divided by EI.
( ) [ ]baabaab MMEIlllMMllM
EIz +=
+= 2
63..
21
2..1
2
Slope of A, [ ] ( )babaa MMEIlMM
lEIl +=+= 2
621.
6
2
-
19
Now, the difference of slopes between any two points on an elastic curve of a beam is equal to the net area of the bending moment diagram between these two points divided by EI. At C, the section at which deflection is maximum, slope is zero,
Slope of A, ( ) ( )
+=
+= abaabaa MMlxxM
EIxMM
lxxM
EI 2.1
2.1
2
Equating the two,
( ) ( )baabaa MMEIlMM
lxxM
EI+=
+= 2
62.1
2
( ) ( )
( )( ) 03
2.2
23.62
2
22
=++
+=+
ab
ab
ab
a
baaba
MMMMlx
MMlMx
MMlMMxxlM
or ( )( ) 0810382108.
810882 22 =
++XxXXx or 03.277642 =+ xx ; x= 4.07 m
Deflection at A= ( ) ( )
+=
+= abaabaa MMlxM
EIxxxMM
lxxxM
EI23
632.
22..1
2
( ) mmormXXXXXXXX
76.51076.5810108
07.42108310126
07.4 3336
2=
+=
7.18
6 kN/m
A BC
5 m
4 kN.m4 kN.m
4 kN.m4 kN.m -+ wl2/8
As the loading is symmetrical, area of half the bending moment diagram can be considered. Moments of the bending moment diagram about A
= 8384
54
.2
.165.
8.
2.
32 242 MlwlllMlwll =
Deflection at A
= mmmXXXXX
MlwlEI
363.010363.08
54000384
56000510100
18384
51 3246
24
==
=
7.20
-
20
Let l be the length. Then maximum bending moment is 3000l at the fixed end (the strongest section of 40 mm diameter).
Thus mmlorXXl 1572040)64/(
300075 4 == Let d be the diameter at a distance x from the free end of cantilever. Then for uniform strength throughout,
xdordXXdx 407
2)64/(300075 34 ==
Moments of the bending moment at x about A= xx.3000
Deflection = == 1570
3/4
2
3/4
157
0
24
0
2 .407
291.0.3000.)64/(210000
1.30001 dxxxdxx
dXdxx
EI
l
mmXxXdxxX 265.0)4569(109.573/5
105.96.105.96 6157
0
3/56
157
0
3/26 ==
==
7.22
W
A B
B.M.
4l
Wl
l
4I
EIWl
54
ll
EIWl
45.2.
5
EIWl5
C D
EIWl
54
I
The simply supported beam is shown in the figure a. Reaction at A = Wl/5; Reaction at B = 4Wl/5 Bending moment diagram for the real beam is shown in Fig. b. Now, in the conjugate beam method, this diagram is to be considered as loading diagram and a new bending moment diagram is to be drawn which will give the deflection of the beam. For the same, first we need to find the reaction on the supports.
EIWlRorlXl
EIWllll
EIWllXR bb
2
56.0324.
24.
534.
2.
545 =
+
+=
EIWl
EIWll
EIWll
EIWlRa
22
24.056.024.
52.
54 =+=
Slope at A= shear force at A for conjugate beam
-
21
= radXX
XXEI
Wl 0014.01020205000
2000600024.024.0 622
== Deflection at midspan = bending moment at C for conjugate beam
mmXX
XEI
WlllXll
EIWllX
EIWl 5.5
10202050002000600047.047.0
35.2.
25.2
45.2.
55.224.0 6
332
===
For maximum deflection, the bending moment is to be maximum or shear force is to be zero. 7.23
W
A B
B.M.
l/2
Wl
l/2
Wl/2
I16I
EIWl2
EIWl
32EI
Wl16
C
Let moment of inertia of portion CB be 4)2/(
64dI =
For portion AC, IdIac 16644 == ,
The cantilever is shown in the figure a. Bending moment diagram for the real beam in Fig. b. the conjugate beam in Fig. c. The fixed end is transformed into a free end and the free end into a fixed end.
Slope at C= shear force at C for conjugate beam = EI
WllEI
WllEI
Wl1283
21.
2.
322.
32
2
=+
Slope at B= shear force at B for conjugate beam = EI
WllEI
WlEI
Wl12819
21.
2.
21283 22 =+
Deflection at C = bending moment at C for conjugate beam
= EI
WlllEI
WlllEI
Wl7685
321.
2.
324.
2.
32
2
=+ Deflection at B = bending moment at B for conjugate beam
-
22
= EI
WlEI
WlllEI
WlllEI
WlllEI
Wl38423
768)3259(
3.
21.
2.
265
21.
2.
3243.
2.
32
22
=++=++
Chapter-8
8.12
A B
15 kN
6m3m
20 kN
6m
8 kN/m
Fixing moment at each end for uniformly distributed load, 12
2wlM = (Example 8.2) Fixing moment for point load,
22
lbWaMb = and 2
2
lWabM a = . (Example 8.3)
By combination,
6.20712158
156920
1512315 2
2
2
2
2
=++= XXXXXM a kN.m
4.20012158
156920
1512315 2
2
2
2
2
=++= XXXXXMb kN.m Maximum bending moment = 207.6 kN.m
Section modulus = 66
10977.1105
104.200 XX = mm3 or 0.00198 m3 8.13
A B
5 m200 kN
4 m
80 kN/m
C
Fixing moment at each end for uniformly distributed load, 12
2wlM = (Example 8.2) Fixing moment for point load,
22
lbWaMb = and 2
2
lWabM a = . (Example 8.3)
By combination,
5.34212
9809
45200 22
2
=+= XXXM a kN.m (hogging)
-
23
1.29312
9809
45200 22
2
=+= XXXMb kN.m (hogging) To find reactions, take moments about B,
kNRorX
XXR aa 7.3411.2935.3422980
420092
==+ 3.1787.341200980 == XRb kN
8.15
W
l/2 l/2
A B
C
MaMc =
(Ma+ Mb)/2
Mb
Wl/4
As the slope at A is equal to slope at B =0, net area of the moment diagram must be zero, i.e.
bab
bba
a
baa
EIlWl
EIlWl
EIlMMM
EIlMMM
2.
4.
21
2.
4.
21
2.
22/)(
2.
22/)( +=+++++
or bab
baa
ba IWl
IWl
IlMM
IlMM
22)3()3(
22
+=+++
+=
++
+
babab
baa II
WlII
MII
M 112
3113
++=
+++
ba
ba
ba
baba II
IIWlIIIIMM
3233 (i)
Deflection of A relative to B is zero, so net moments of areas about B must be zero, i.e.
+
+=+
++
++
32.
22.
4.
21
31.
22.
2.
4.
21
6.
21
2.
2)(
4.
2.
2)(
2.
31
221.
2.
2)(
43.
2l
EIlWlll
EIlWll
EIlMM
lEIlMMll
EIlMMl
EIlM
bab
ab
b
ba
a
ab
a
a
-
24
bab
ab
b
ba
a
ab
a
a
IWl
IWl
IlMM
IlMM
IlMM
IlM
482448)(
16)(
12)(
83 332222 +=++++
+=
+++
+
babbab
bbaaa II
WlIII
MIIII
M211
6121
41
31
121
41
31
23
++=+
++ba
ba
ba
baba II
IIWlIIIIMM
7)2
27)(2 (ii)
From (i) and (ii),
+++=
baba
baaa IIII
IIIWlM14
)3(2 22
and
+++=
baba
babb IIII
IIIWlM14
)3(2 22
8.20
A B C
6 kN/m
5 m4 m
10 kN/m
I 2I
12
31.2525.5
5.625
18.375
30.1
19.9
8.22
AB C
Dw
l2ll The load diagram is shown in Fig. Applying the three moment equation for uniformly distributed loads to the spans AB and BC,
( )
+=+++
12
33
26
4)2(
4222
llEIlwwllXMllMlM accba
In this equation, ,0=aM and 0; == cbc MM (From symmetry of the beam, B and C have same load, so level of C is the
same as of B)
-
25
The above equation reduces to 22
43
329
lEIwlM ab
= (i)
Also, cbda RRRR == ; (From symmetry of the beam) Thus, badcba RRRRRRW 22 +=+++= or bba RwlRwlR =
= 2
24
and ba R=
(i) becomes, 22
4.3
329
lREIwlM bb
= (ii) Bending moment at B,
=
+=+=
82820
2
2 WRWlWRWlwllRMM bbaab (hogging)
or
=8
3WRlM bb (iii)
From (ii) and (iii)
2
22
4.3
329
812
lREIwlwlRl bb
=
+=+=+==
EIl
lwlEIl
wl
lEI
wl
Rorl
REIwlR bbb 3457
8348
57
43132
57
4.3
3257
3
3
3
4
3
3
++=
+=
+=
=
EIlEIlwl
EIllwl
EIllwllwXRWR ba
34487
8345716
8
3457
824
2
3
3
3
3
3
3
8.28
-
26
A B6 m 4 m
60kN/mC
+
+
_
180
8032.7
A
197.3
Maximum bending moment for the spans AB and BC treating as simply supported beams,
For span AB, 2708
6608
22
max === XwlM kN.m
For span BC, 1208
4608
22
max === XwlM kN.m For fixing moments, assume the continuous beam ABC to be made up of fixed beams
AB and BC.
For span AB: Fixing moments at A, 18012
66012
22
=== XwlM a kN.m
Fixing moments at B, 18012
66012
22
=== XwlMb kN.m
For span BC: Fixing moments at B, 8012
46012
22
=== XwlMb kN.m
Fixing moments at C, 8012
46012
22
=== XwlMc kN.m In span AB, moments at A and B due to sinking of support B by 8 mm,
2
6lEIM = (Example 8.7, -ve being counter-clockwise)
or 626
103.1090006
8104000002056 XXXXXM == N.mm or -109.3 kN.m Due to sinking of support B by 8 mm, moments at B and C (Refer example 8.7),
2
6lEIM = (-ve being counter-clockwise)
or 626
102460004
8104000002056 XXXXXM == N.mm or 246 kN.m Distribution factors at B:
-
27
Stiffness factor for AB, 3/26
4 EIEIsba == (as the beam is fixed at A)
and for BC, 4
34
3 EIEIsbc == (as the beam is simply supported at C)
Distribution factor for AB, 178
4/33/23/2 =+=+= EIEI
EIss
skbcab
abba
Distribution factor for BC, 179
4/33/24/3 =+=+= EIEI
EIss
skbcab
bcbc
Distribution factors 16/31 15/31 Fixed end moments Moments due to sinking
-180 180 -109.3 -109.3
-80 80 246 246
Initial moments Release C Carry over
-289.3 70.7
166 326 -326 -163
Net moments Distribute Carry over
-289.3 70.7 -38 -19
3 0 -35.7
Final moments -197.3 32.7 -32.7 0
Chapter-9 9.16 M =WR (1-cos )
[ ] [ ] dEI
RWdRWREIEI
dsMU .cos1.)cos1(2
122 0
232
0
22 ===
Vertical deflection = [ ] ( ) dEIWRd
EIWR
Wu .cos2cos12.cos12
0
23
0
23 ++==
0
3
0
3
sin242sin
22.cos2
22cos112
+++=
+++= EIWRdEIWR
68.3)12/612(000200
150533002
23
333
===
+++=XX
XXXEIWR
EIWR mm
9.8
-
28
20 kN
20 kN
100 m
m
80mm
P Q
d = 80 mm ; R = 50+40 = 90 mm ; A = 5.502680
42 =X mm2
5.4399080.
1281
1680...
12816 242
2
422 =+=++=
Rddp mm2
As bending moment tends to increase the curvature, it is positive. Resultant stress = Direct stress +Bending stress Stress at outside face (P), y = 290 - 202=88 mm
409040
5.43990.
5.502620000.1
2
2
2
2
2
+=+=
+++= yR
ypR
AW
yRy
pR
ARWR
AW
o 56.22= MPa (tensile) Stress at inside face (Q)
yRy
pR
AW
yRy
pR
ARWR
AW
i =
+= ..1 2
2
2
2
67.584090
4082.105 == X MPa (compressive) Chapter-10
10.12
TNP60
2= , If P is constant, T
N 1
But 16
3dT = , If stress is same,
3dT ; 31dN or 3/11
Nd , 64.4
505000 3/1 =
=
b
a
dd
10.16
aa
aa
ss
ss
JGlT
JGlT ==
Angle of twist per unit length,
-
29
aa
a
ss
s
JGT
JGT = or 444 60
327.0)60(2
32XXGTX
dXXGT
aa = d =50.5 mm
Also, asast ll
=
7.0 , ast GrGr
=
7.0
As r is same, 4.17.027.0 === XGG
a
st
a
st
As maximum limits are 65 MPa for alloy and 80 MPa for steel, the maximum stress will reach in steel first. 10.18 Maximum shear stress occurs at smallest diameter.
MPaorXX 5.99.16801010
36 ==
( ) ( ) radXX
rrrrrl
GT 0171.0
401
601
101200.
8200031010
'1
'1
'.
3 336
33 =
=
+=
= 0.98o
Chapter-11
11.13
Under axial torque, 3704.012000200
1210020000646444 === XXXX
EdTDn rad
Increase in the number of turns = 059.023704.0 =
Bending stress, 9.11712200003232
33 == XX
dT
MPa Torsional stiffness (torque /rad) = 54
3704.020 = N.m/rad
11.17
4
38Gd
nWD= or 441.220)5.28(8
5.2000808 3
4
3
4
1 ==== XXXX
nDGdWs N/mm
'302.115)'8(8'00080
8 34
3
4
2 dXdXXd
nDGdWs ====
21
21
sssss += or mmdordXdord
dX 136.2''302.1441.2'693.1173.3'302.1441.2'302.1441.23.1 ==++=
The maximum load will be that taken by the spring with smaller diameter.
-
30
11.18 E =2G(1+) =2G(1+0.3)=2.6G
For a closed-coiled spring, 438
GdnWD
c =
and for an open-coiled spring,
+=
EGdnWD
o
22
4
2 sin2coscos.
8
Now, 02.0=o
io
or 02.0
sin2coscos.
8
8
122
4
2
4
2
=
+
EGd
nWDGd
nWD
975.0sin2cos6.2
cos6.2025.0sin2cos
cos1 2222 =+=+
GGGor
GEE
or 975.02cos6.0
cos6.2975.0)cos1(2cos6.2
cos6.2222 =+=+
or
02cos6.2cos585.0975.02cos6.0
cos6.2 22 =+=+ or
On solving, o2.17= 11.20
'RE
RE
IM
y== (Eq. 11.14)
At proof load, R is infinite and thus E/R = 0
RE
y= or
R000208
2/10630 = or R = 1650 mm or 1.65 m
Let We be the equivalent static load which produces the same maximum stress and deflection as the impact load.
also, 223
nbtlWe=
As the maximum stress is one half of the proof stress, 21080101400.
23
2630
XXXWe=
or 12000=eW N 2.74
000208108010140012000.
83
83
3
3
3
3
===XXX
XEnbt
lWe mm
Chapter-12
-
31
12.16 Assuming the deflected form under the action of the crippling load be
=
=400
cos12
cos1 xalxay and thus,
400sin
400xa
dxdy =
dxx
adxxadxdxdyl .
2200
cos1
400.
400sin
400.
180
0
2180
0
22
0
2
=
=
22180
0
2
17006.02004002
1200
sin2004002
1 aXaxxa =
=
=
Also, 400
cos400
cos1)( xPaxPaPayaPM =
==
+= 200100
2
2100
01
222
0
2400
cos400
cos
22dx
I
x
dxI
x
EaPdx
EIMl
12.18
( ) 95100445664
44 ==I mm4; ( ) 30044564
22 ==A mm2
317300
951002 ===AIk mm2
Euler load, 1261122200
951000002072
2
2
2
=== XXlEIPe N
Actual load for failure = 126112 X 0.75 = 94584 N
77.49.77sec36.1sec95100000207
945842
2200sec2
sec2
sec ===== oXEI
Pll
+=
+= 77.4317
28130094584
2sec..1 2max X
eXlkye
AP c
MPa
or ( )e421.0136.100290 += or 1 + 0.421e = 2.89 or e = 4.49 mm 12.21 w = 0.8 kN/m = 0.8 N/mm;
( ) 400004064
4 ==I mm4; ( ) 400404
2 ==A mm2
968.140000000200
000362
320022
===
XEIPll rad (=109.7o)
-
32
( )17.109sec00036
400000002008.012
sec =
= oXXlP
wEIM 66 10215.2)1967.2(105585.0 XXX == N.mm
7.3527.2840000
2010215.2400
00036. 6max ===
XXIyM
AP c
- 381.4 MPa (compressive) and -324 MPa compressive
Chapter-13
13.13
Internal volume of cylinder = 62 107.70314008004
XXX = mm3 180
1028005.4
2===
XX
tpd
c MPa (tensile) and 904 == tpd
l MPa (tensile) Increase in volume, [ ]).(2).().2( lcclcl E
VVv +=+=
[ ] )360450(2050000
1500800)4/()90180(2)18090(1012452
3 =+= XXXXEVX
340360450)360450(205000
107541012506
3 == orXX 306.0=
13.16
Internal volume of shell = 767.15.16
3 =X m3 = 6101767X mm3 For tube:
Volumetric strain = 3 X hoop strain = ( ) ( ) == 14313
tEpd
E (13.9)
Decrease in volume, ( ) ( ) tEVVtEXXV
tEpd /15753.01
41500231
43 ===
For water: Increase in volume, V
KpVv w .. ==
Thus change volume of water VKp
tEV += 1575
or 63 1017672000
2000200
1575104000 XXt
X
+= or t=6.23mm
13.20
-
33
Maximum circumferential stress in a tube subjected to internal pressure p is at the inner
surface and is pdddd
io
ioc .22
22
+= or pc 57.3150200
15020022
22
=+=
Thus 4457.3 =p or p =12.32 MPa Longitudinal stress =
)(.
22
2
io
i
rrpr
= pXp 2857.1
7510075
22
2
=
)150200(
4
2200002875.122
=X
p or p =12.43 MPa
Maximum pressure =12.43 MPa 13.22
The maximum hoop stress is at d=di, iio
ioc pdd
dd .2222
+=
or 35606055 22
22
Xdd
o
o
+= or )60(
553560 2222 += oo dd
or )60(7601111 2222 += oo dXd or 8.84=od mm Maximum shear stress,
)( 222
io
oi
dddp= or )60(
3550 222
= oo
dd
or 222 7601010 oo dXd = or 5.109=od mm Thickness = 75.24
2605.109 = mm
13.25 (i) Increase of inner diameter of the collar
[ ]E
pdEpdpp
Edd
Ed rc 425.23.0125.2.
150250150250. 22
22
=+=
++=+=
or 000200150425.21.0 pX= or p = 55 MPa
13.27
)(2)2(
33
33
io
ioi
ddddp
+= or 30
)80(280275 33
33
Xd
Xd
o
o
+=
or )80(5802 3333 =+ oo dXd or 33 8074 Xdo = or do = 96.4 mm t = 16.4/2=8.2 mm
-
34
Chapter-14 14.12 Let the shrinkage pressure between the disc and the shaft at stand still be p.
09.0=iR m and 4.0=oR m Shrink allowance = 08.0 mm
At stand still, the hollow disc acts similar to a thick cylinder subjected to internal pressure (Refer section 13.9).
Shrinkage allowance (initial difference in radii))(
222
2
io
oi
RRERpR= (Eq. 13.31)
or )90400(000206
40090208.0 222
=XpX or p = 86.9 MPa
Maximum hoop stress is at the inner radius and is given by Eq. 13.7.
2.969.86107.1107.1.09.04.009.04.0. 22
22
22
22
===+=
+= XpppRRRR
iio
io MPa
(ii) When the disc rotates and the shrink fit loosens, radial pressure is zero and thus
the radial stress is also zero.
[ ]222 )3()1(4 oi
RR ++= (Eq. 14.17)
14.13
1006030002 == X
At stand still Let the shrinkage pressure between the disc and the shaft at stand still be p.
At stand still, the hollow disc acts similar to a thick cylinder subjected to internal pressure only and thus the results of the same may be used. - For the outer disc, hoop stress at the inner radius (50 mm),
08.0=iR m and 24.0=oR m ppp
RRRR
iio
io 25.1.08.024.008.024.0. 22
22
22
22
=+=
+= N/mm2
Hoop strain = ss Ep
Epp 55.13.025.1 =+
14.15 Let t be the thickness of the disc at radius r. Now as 2/..
22
. reAt = At r = 0, AeA == 2/0.. 22.180 (i)
-
35
At r = 0.2 m, 2/2.0..22
. = eAt (ii) Dividing (ii) and (i),
2/)04.0(.2/2.0..
222
.8.180
== eA
eAXt
where 1234.0101202
04.0603000275002/04.0 6
22 =
=
XXXXXX
1598 1234.0 == et mm
Chapter-15
15.6
62
3
2
3
108.61)3.01(12
15000200)1(12
XXEtC ===
=
=
2300
108.61324.0
232
22
6
222
2 xXX
xxRC
wxy
Deflections at different cross-sections are tabulated below:
x (mm) 0 60 120 180 240 300 y (mm) 0 0.064 0.24 0.48 0.72 0.82
0.9
0.30.6
r (mm)0 12060 240180 300
Figure shows the profile of the deflected plate.
[ ] [ ]222223 22 )3()1(.83)3()1()1(12.16.1 )2/( xRtwxREtwEtx ++=++= [ ] [ ]2222 3.3117000000667.0)3.03(300)3.01(154.0.83 xxX =++=
[ ] [ ]222223 22 )13()1(83)13()3()1(12.16.1 )2/( xRtwxREtwEtz ++=++= [ ] [ ]2222 9.1000117000667.0)13.03(300)3.01(158 4.03 xxXXX =++=
15.8
-
36
62
3
2
3
1026.2)28.01(12
5000200)1(12
XXEtC ===
++= )1(log
11log2
21
8
2
xRC
Pxy
++= )1(log
28.0128.01100log2
21
1026.282000
6
2
xXXx
[ ]xxX log773.101021.35 26 =
Chapter-16
16.9
60
80
10
60
y'
AN
10
A'N' y10
(a) (b)
.45
(mm)
Let N-A and N-A be the elastic and plastic neutral axes respectively.
=++=
1070106045107051060
XXXXXXy 26.54 mm
I = 23
23
)54.2645(701012
7010)554.26(106012
1060 +++ XXXXXX =807.76X103 mm4 5000 278383 285833 238540 Yield will start at the bottom edge,
Moment of resistance at first yield, My= yyyX
yI 15110.
)54.2680(1076.807.
3
== In fully plastic state, the neutral axis divides the total area of the beam into two
equal parts.
16.12
STRESS
30
72
12
24
(mm)
280 MPa
Let Wy be the load at first yield at the mid-span,
-
37
322
10725828067230.
6XXXbhM yy === N.mm or 7258 N.m
Thus, 72584
2.1 =XWy or Wy = 24.2 X103 N or 24.2 kN
For elastoplastic state, moment of resistance,
=
34.
22 ahbM y
=
324
47230280
22
X =9274 X 103 N.mm or 9274 N.m
or 92744
2.1 =XWy or Wy = 30.9 X103 N or 30.9 kN For this required depth of yield, let the yield occur beyond a cross-section at
distance x m from one end of the beam. Then this cross-section will be having the first yield and thus the moment of resistance will be equal to the first yield at the midspan i.e. 7258X103 N.mm.
16.13
120 mm
200mm15mm
15 mm
y
50mm
yyy
150
.
yyy
150)15(
y
150-y
(a) (b) Solution As the load increases and the yield spreads upto 50 mm from the lower edge of web, assume that the top of the flange is still in the elastic state. Let the neutral axis be at a distance y from the top edge. The stress distribution in the beam section will be as shown in Fig. b.
Stress at the top of flange, t = ..150 yy
y
Stress at the bottom of flange, b = ..150
15yy
y
Total tensile stress force 15)15.(2
)15120.(2
++= yX bbt
-
38
+=
+++=
+
+=
yyy
yyyyy
yy
yy
yy
y
y
y
yy
1505.1181215755.7
1505.16872255.713500900900
)15(150
155.7150
15150
900
2
2
Compressive force = )5.71875(1550)150(15.2
yXXyX yyy =+
16.14
10 mm
100 mm
80mm
AN
10 mm
y
y(a) (b)
y
y
Solution Let the neutral axis be at a distance y from the top edge. The yield starts at the bottom of the web and will cover some portion when the top of the flange will just yield. The stress distribution in the beam section is shown in Fig. b. Let the stress at the bottom of the flange be when the top of flange just yields. Then yy
y .15=
Positive force = yyX yy 545050010)10.(
2)10100.(
2++=++
Negative force = yyyy yXyy 800.1510)280(10.
2+=+
Chapter-17
17.8
A B
C D
E F
6 m
3 m
3 m
50 kN
AB
C D
E F
6 m
3 m
3 m
50 kN
50 kN 50 kN
AB
C D
E F 50 kN
50 kN
50 kN 50 kN
-
39
To find the reaction at support B, take moments about A,
6506 XXRb = or 50=bR kN (upwards); 50=aR kN (downwards) 50=ahR kN
oABC 6.26)6/3(tan == Joint A: kNFAB 50= (T) )(50 TFAC = kN, Joint B Considering FH = 0, 06.26cos50 =+ oBCF or kNFBC 9.55= (C) Considering FV = 0, 06.26sin9.5550 =+ BDo F or 25=BDF kN (compressive) 17.9
A B
C D
E F
2 m
2 m
2 m
50 kN
2 m
100 kN
H
G
A
B
C D
E F
2 m
2 m
2 m
50 kN
2 m
100 kN
H
G100 kN
50 kN
F EG
F CE
E
G100F GD
F GF =o
FH= 0D
FV= 0D
50 kN
100 kNG100
F GD
F GF =o
FV= 0
To find the reaction at support B, take moments about A, 45041004 XXXRb = or 50=bR kN (upwards); 0=avR
100=ahR kN Joint E: kNFCE 50= (C) 100=EGF kN (C) by judgement Joint F: 0== FDFG FF (C) by judgement Joint G Considering forces along GD,
FGD = 0, 045cos =+ oGEGD FF or 045cos100 =+ oGDF or kNFGD 7.70= (C) Considering FV = 0, 045cos45cos = oGCoGD FF or 7.70=GCF (T)
-
40
17.11
200 kN
C
D
E
A
F
B
4 m 4 m
2 m
2 m
200 kN
C
D
E
A
F
B
200
447.2
400
200
400
400
400
0
0
447.1200
0 Taking moments about F,
82004 XXFAH = or 400=AHF kN (inwards); 400=FHF kN (outwards) Joint D: 0=EDF as there is no other horizontal member to balance the force. 200=CDF kN (C) tan-1(4/2)= CBE or oCBE 43.63= Joint C: 20043.63cos =oBCF or 2.447=BCF kN (C)
CEo F=43.63sin2.447 or 400=ECF kN (T)
Joint B: BEo F=43.63cos2.447 or 200=BEF kN (T)
ABo F=43.63sin2.447 or 400=ABF kN (C)
Joint A: 0=AEF 0=AVF Joint F: 200=FVF 40043.63sin =oEFF or 1.447=EFF kN (T) 17.12
DCBA
EFG 40 kN
80 kN40 kN
60o 60o
DCBA
EFG 40 kN
80 kN40 kN
40 kN
Taking moments about F, 464.34068024012 XXXXFDV ++= or 21.58=DVF kN; 40=DHF kN
8.612.58120 ==AVF kN Joint D: 2.6721.5860sin == EDoED ForXF kN (C) 6.3360cos2.67 == oCDF kN (T)