Strength of Materials by R.K.Bansal_text

’

1074 STRENGTH OF MATERIALS OBJECTIVE TYPE QUESTIONS 1075

A m45. rod having cross-sectional area 100 x 10”6 2 is subjected to a tensile load. Based on = 50 + (-10) 2—r50-(-10)f +4°?
Mthe Tresca failure criterion, if the uniaxial yield stress of the material is 200 Pa, the 2 J

failure load is

(a) 10 kN (6) 20 kN (c) 100 kN (d) 200 kN. = 20 ± 2 + 40 2

V30

46. Wire diameter, mean coil diameter and number of turns of a closely-coiled steel spring = 20 ± 50,

D N Aare d, and respectively and stiffness of the spring is k. second spring is made of = 20 + 50 = 70 MPa

the same steel but with wire diameter, mean coil diameter and number of turns as 2d, and o = 20 - 50 = - 30 MPa = 30 MPa. (compressive) Ans.
2
N2D and 2 respectively. The stiffness of the new spring is
— —...

2.
(a) k (6)26 (0 46 id) 8k. p = _. ratio = - Lateral strain
Poisson s
: r;

47. If the diameter of a long column is reduced by 20%, the percentage of reduction in Longitudinal strain

Euler’s buckling load is The lateral strain is opposite to longitudinal strain. This means if longitudinal strain is
tensile, then lateral strain is compressive. Hence p is negative. For most of the material u lies
(o) 4 (6) 36 (e) 49 (d) 59. between - 0.25 to - 0.40.

48. With one fixed end and other free end, a column of length L buckles at load Pr Another
column of same length and same cross-section fixed at both ends buckles at load P2 .

Then P I P is 4. Bending stress, a - -y x y
21 b

(all (6)2 (c) 4 (d) 16. M ^, , d
where I = n dl4
49. The principal stresses av a2 and a at a point respectively are 80 MPa, 30 MPa and “ jr x
3 2

- 40 MPa. The maximum shear stress is M Md 32

(a) 25 MPa (6) 35 MPa (c) 55 MPa (d) 60 MPa. ^=

50. The Poisson’s ratio of a material which has Young’s modulus of 120 GPa and shear

modulus of 50 GPa, is T=—xds xz
ii
(a) 0.1 (6)0.2 (c) 0.3 (d) 0.4.

25.4. ANSWERS WITH EXPLANATIONS ml 3

1. (d) 2.(o) 3.(o) 4. (a) 5. (6) 6. (a) Th, 6n 321W 16T =rM T
7. (a) 8.(6) 10. (d) 11. (c) 12. (c)
13. (6) 14. (6) 9. (c) 16. (d) 17. (c) 18. (6) Due . ° r AnS'
19. (d) 20. (c) 15. (c) 22. (d) 23. (d) 24. (6)
25. (6) 26. (6) 21. (c) 28. (6) 29. (d) 30. (c) 5. to bending moment (m), the bending stress will be produced in the shaft. This
31. (6) 32. (c) 27. (c) 34. (d) 35. (6) 36. (a)
37. (d) 38. id) 33. (6) 40. (e) 41. (c) 42. (d) bending stress (oj,) is given by
43. (a) 44. (c) 39. (6) 46. (a) 47. id) 48. id)
49. (d) 50. (6). 45. (6) Mrt — 32

' Due to torque (71, the shear stress will be produced. This shear stress (t) is given by

16T
X ~ nd 3
The principal stresses due to bending and shear stresses are

a and a
r 2-

M1 32 fr f 16T 2
l
2 * Kd 3 =

For finding equivalent torque (7 when the shaft is subjected to bending moment and

e)

torque, we should determine the maximum shear produced by the principal stresses.

22

—

1077

OBJECTIVE TYPE QUESTIONS

10.

Temperature stress = a x (At) + E

Max. shear stress due to principal stresses is given by Where a = Co-efficient of linear expression,

16 T At = Temperature rise, and

+ (“33
—^ YF—wihere 16M
_ q l ~ °2 - =- + (fieiwY E = Modulus of elasticity.
:s- , tt I
°i Hence temperature stress depends upon all the three and id) is the answer.
12. The energy absorbed by a part subject to dynamic force is given by
lewf fierf M16 flSM'f

and a, - U = q2 x volume
2E
~Jm 2 + t 2
when o and E are constant then
Equivalent torque is U <x volume

vT^ xd riJl .q Hence (c) is the answer.
:
13. The criterion of failure according to maximum shear stress theory is
.

J(

«w *UU Jm=n [.16 x r2 + ; Qi ~ q 2 = ± i'A when stresses a and o are opposite
xaxrf3“ x 22 the principal l 2
3 '

= Jm 2 +T2 * Ans. i e one is tensile then other is compressive

o x AB X cos 9 14. But if both are tensile (or compressive), then will not represent the maximum

-A shear stress. It will represent the stress less than maximum shear stress. But will repre-
sent the maximum shear stress. Hence criterion of failure is
7' \Xg a £. ABx x. <? 1
4 >

l— I

Xo L Ans.

77T1

' Plane inclined at 22
angle 0 with x*axis Buckling load for column with different end condition are .

^BoE ...both ends hinged
_ b2j
-
L2

Force on AS -ox Area (Thickness = unity) _ k2eI ..o.onnee end is fixed (or clamped) other is hinged
Force on = a x AB x 1
.

4B2

AC — (ox AB) x cos 0 _ 2jt2 £7 ...oonnee end is fixed (clamped) other is hinged
x = Shear stress on the plane AC
8 L2

Shear force o x AB x cos 6 _ a x sin 0 x cos 0 _ 4x"EI ...both ends fixed (or clamped)

Area AC x 1 L2

= | |x 2 sin 0 cos 9 = x sin 29. Ans. When both ends are clamped, the buckling load is maximum. wire mean
terms of dia_ of id),
The spring stiffness (k) for a close-coiled helical spring in
in the bar at a distance y from the support 15 radius of coil (fi), no. of turns (n) and modulus of rigidity (c) is given by
_ Weight at the lower end + Weight of bar
8. The tensile force lengt _y

foi a W Cd 4

W w= + (L - y). Ans. Tk = = 3

9 . The normal stress on the inclined plane in case of biaxial stress system is given y 64fi xn

When the dia. of wire id) and material of coil is same, then

|(ox -o )cos 29 C[. is constant]
y .
o = |(o + 0 .)+ For the same material,
0 x }
3
fi x n

K-oa45.= |(ox + o,)+ 5 y) cos 90” k x R 3 x n = constant

R R/e x z x n - k x 3 x n2
L l 2
x 2

= 2 (a* + °y )- Ans ‘

^

1078 STRENGTH OF MATERIALS

\ /n
k — — in—/ -d f
x -j X =k x. i-
l
1 1 „

75 mm, n = 8
L
=©!)'
[ d = 60 mm, n2 - 10
2

16.

1.56 k,. Ans.

.-. Answer is (c).

Here d=lm = 100 cm

p = 100 N/cm2
Max. permissible tensile stress = 2 kN/cm2 = 2000 N/cm 2

For thin cylinder, the maximum stress is circumferential (or hoop) stress.

Hence here a = 2000 N/cm2
c

Let t = thickness.

17. o =^ (Here p and o should have same unit.
c c
Then pxd

Then d and t will have the same unit]

100 x 100 = 2.5 cm = 25 mm. A,ns.

2 x 2000

Original volume, V" = “Z12 x L

Change in volume, dV will be obtained by taking differential as

dV = ~D2 xdL + ~ Ld(2D2)

44

= y4 D2 X dl + vL X 2D d(D)

4

d.V.. -D 2 xdL + -Lx2Dxd(D)

Volumetric strain, 44

?£> 2 xL

dL 2d(D)
+
D
L

—r~ = Longitudinal strain = e jt(D + c/H) - jlD _ c/H
L/ 2
i(D) HjtD

= Circumferential strain

—= e + 2ej = 2ej + e = Change in volume per unit volume
2 2

18. The part which is cheapest in overall cost and can be easily replaced when there is some
damage, is made the weakest part. The key in comparison to pulley and shaft can be

replaced easily.

19. The shear stress distribution in a beam of rectangular cross-section is parabolic and
having maximum value at the neutral axis. Hence the answer is (d).

Fig. 15

2

STRENGTH OF MATERIALS OBJECTIVE TYPE QUESTIONS 1081

M Md d_
T(‘Vmax =
bd 3 * 2

_^ _6M_ 6M .. 6 = d being square)
= s= (

~ bd 2 dxd 2 d 3

The bending stress when beam is placed as shown in Fig. 15(6) 2nd position is given by

—(''q'“V)*max ss £ T* x JV* where y* = Distance of top layer from (a) 1 st position (6) 2nd position

OCN.A. i.e.i distance Fig. 16
= half of diagonal of the square = OC
EHere load P, span of beam L and is same for both positions,

- d x sin 45° = 5«y or 6x7 = constant

I* = M.O.I. of 2nd position about N.A. ^

or 6i x /j = 6g x /2 or \* ~ A X
J2

M12 bh 3 where d Now 4 x 63
1_ = 72
d3 *J2 =2x b = BD = 2x -j= and
12
12M
d3 xJ 12 ‘ V2 6 x 43
2 " 12
h ~~7z = 32

—S = Sj x = 2.25 6 r Ans.
2

J2d x j 1 d 4 Hence answer is (d).

12 * 2 12 23. Couple acting on beam = P x C anti-clockwise moment at any point should be zero
6

Hence the answer is (c). Fig. 17

M Md 32 M ma = o
Y —°6= X
*y ~ 3 or Rb x 2L + P x C = 0
k x a4 2 ltd

64 R - PxC

2L
T=~ x d3 x t ——or (-ve sign shows that reaction R, is acting downwards)
16 t

There is no load on the beam

T - nd3 Px_C

«A + RB = 0 or «a = -HB “ 2L

—a 6 3211/ ^ 3 2M M -(vv • T)’ The reaction PA is acting upwards.
red "•*
-
,
x ad 3 16P T
. F RA PxC
%
22. For a simply supported beam carrying a point load at the centre the deflection (6) is Shear force at x, = - . Ans.

given by Resolving forces along the inclined plane and normal to the plane, we get

F Wcos a = [iPN + sin a ...( 1 )

OBJECTIVE TYPE QUESTIONS 1083

30. To find the correct answer proceed as given below

Ductility is determined by tension test (3)

Toughness is determined by Impact test (1)

Endurance limit is determined by Fatigue Test (2)

Resistance to penetration is determined by Hardness test (4)

Hence the correct code is which contains 3, 1, 2, 4.

Hence the correct code is (c).

31. E = 21 x 106 N/cm2 C = 8 x 106 N/cm2
,
E = 2C(1 + p)
E
= ~1
2C

21xl0 6 , = 21 . = 1.315 - = 0.315. Ans.

2 x 8 x 10 b 1 - 1 1

16

P = 2jtNT, where —T= xd3 xx — x xx
16

P 1= 2nN T N= *± X (2c? 1)3 x T
1l 22

P2 — 2jtN’ 21 = -x8xd 2 ,
1
22

= 8xT
1

= 2JIAIJ xTjX4 I

= PjX4 = 90x4 = 360 kW. Ans.

33. According to maximum shear stress theory for design purpose, we have the equation

(a, - a ) = a. where a = Permissible stress in simple tension
2 t

or 60 - (- 60) = a or 60 + 60 = a
(t

— —or
cr, = 120 and safety factor = = - = 3. Ans.

1 a 120
f

W34. Here each rivet is subjected to direct stress due to load and bending stress due to
Wbending moment. Bending moment is equal to x e where e = eccentricity.

The value of e is maximum for rivets P and S. The direct stress is same for all the rivets,
Bending stress is maximum when e is maximum. Eccentricity is maximum for rivets P

and S. Hence rivets P and S are having maximum bending stress. Hence rivets P and S

are most loaded. Ans.

35. The stiffness of a helical compression spring is given by,

Cd 4
Rk, =
35 x n
64

where C = Modulus of rigidity

d = dia. of wire

n = no. of turns

R - mean radius of coil.

.:

STRENGTH OF MATERIALS OBJECTIVE TYPE QUESTIONS 1085

When spring is cut into two equal halves, only no. of turns will be effected it will become R = 2R
half. Other value such as C, d and R will be same. x2

From (i) and iii) 3R = 120 N R = 40 N. Ans.
and 2 ?

= 2 x k. Ans. Ry = 120 - 40 = 80 N. Ans.
64.^ x n
64 x R 3 x C = 100 G Pa = 100 x 109 N/m2

36. For close-coiled helical spring, p = 0.25

X= 16 WR E = 2C (1 + p)

nd 3i" . = 2 x 100 x 109 (1 + 0.25)

While deriving this equation, the effect of curvature of spring and stress concentration = 200 x 1.25 x 10 9 Ans.
effect are neglected. Hence the correct expression for shear stress will be
= 250 x 109 N/m2 = 250 GPa.
16 WR
39. Steel rod Aluminium rod
where K - Wahl’s correction factor
fj = 2m tj = lm

Ay = 1 cm2 A = 2 cm2
z
Ey = 200 GPa
Ey = 100 GPa

_ 45-1 0.615 The rigid beam will be horizontal if
~
4S - 4 + S Extension of steel rod = Extension of aluminium rod

... —S„ = spring index =- D mean dia. of coil bLy = Si

where = 2

P Lx Ly

jj. _
d dia. of wire 2x 2

37. Free body diagram through C —Aj XEy A-2 X Ery

R + R = 120 N p =p x1 -
l 2 A E1 2
or 2 2 100 2 2
2 2 Ly

or 2 Py - P or P = 2P r Ans.
2 2

or load on aluminium rod = 2 times the load on steel rod-

P = 1 20 N kB 40. Radial displacement = u.

Initial radius =r

=r+u
.... — —Final radius
T_angential, strain = Circumferential strain = Final circumference - Initial

-

Initial circumference

^2 —= 2it(r + u) - 2xr = 2itu = u Ans.
.
^
— —2nr 2 nr r

41. Stiffness =- = ...(c)
Deflection Deflection under P

Fig. 19 Let us find deflection under load P. This can be done by conjugate Beam Method. In this

Extension in AC - Compression in CB “method, the beam carries the load corresponding to actual load. The deflection at

El

For AC: bLy _ Oj a, R, any section will be equal to B. M. at that section due the load carried by conjugate

Y X Ll = AE X Ll beam.

For CB : bL2 O 2 Refer to Fig. 20 (6).

As L5Z/ = 6 —Load for conjugate beam is
l
2

Ry X Ly 1?2 X B.M. at A = 0, hence value of —7 at A = 0

AE AE till
—M — —B.M.
or Ry x Ly = R x L But Ly= L and L = 2L „ at C„ = P x 2L PL
2 2 %
C = P 2L, hence val, ue of -
RyxL=R x2L at x

2 ^

i

STRENGTH OF MATERIALS OBJECTIVE TYPE QUESTIONS 1087

1086

Pxl„

for

EJ
M ^B.M. at B = PxL, hence value of
at S for AB = Bn/C, - PxL PL
&*
-

(a) Actual beam

(6) Conjugate beam

Fig. 20

ADeflection (5) at = B.M. at A* due to load carried by conjugate beam

^(\ PL r A 2L (PL ier (1 PL 5L

Ul 2 EI El J 1 2 2EL J3
J3

„PU_( 1 + n 5 'l PI? 18 1.5 x PL?
+ j"
EI l 3 EI X EI
12 12

—Stiffness = P_ EI _ (200 x 10 9 ) x (375 x 10 6

)

:

o PL1.5 x 3 1.5 x 3 1.5 x 3
Z/ (0.5}
'j

EI

—= 200 x 10 9 x 375 x 10'4 x 10 3 = 4, x IO 10 N/m. Ans.

1.5 x 125

42. The maximum stress induced in a thin cylinder is hoop stress (o ). It is given by
c

pxif

°c= 21

The maximum hoop stress produced in spherical vessel = p xd

.

Max, stress in cylindrical vessel _ 21
Max. stress in spherical vessel p * d

At

Fig. 21

Strain energy stored in spring 1,

T

STRENGTH OF MATERIALS

Cd 4 x 16 —Kv T
=
64 xf— Wx 8 x 2 Cd*_ k, kvU- A
.2) -
64 xN 1 ' Ans.
DY(
{k ^]' — xN64 Subject Index

V V2J

47. Euler’s buckling load,

—P„ —* 2 El re For circular column when diameter is reduced by 20%, A Chimneys, 458
= a— . I/, =
ddr4 Analysis of bars' of varying — wind pressure, 458
where
If 64 sections, 14 Clapeyron’s equation, 653
Columns and struts, 808
then dia. of new column = 0.8 d — of composite sections, 30
— long, 808
—New moment of inertia, I* = —x (0.8 d)4 = x d4 x (0.8)4 Analysis of a compound weld, 905 — beams, 858
Area moment method, 546 — -with eccentric load, 849
Initial buckling load, — —E2 Assumptions mode in the theory of simple bend- — with initial curvature, 853

k it ing, 293 Combined direct and bending, 377

P = Lt2 t * 64 dr4 B Composite shaft, 706

New buckling load, ^ —p* = 2 y-, X d4 X 4 Bar of uniform strength, 51 Composite sections, 30
Combined bending and torsion, 710
X 0.8 — composite sections, 30 Complementary shear stresses, 73

j} 64 — varying sections, 14 Compressive stress, 3

— —reduction in load = P-—P* x 100 Beams, continuous, 652 — strain, 3

—— —E dif it ,4 -x—Eix -djr ,4 rt — fixed, 613 Conjugate beam method, 578
x 5 x 0.8 — deflection of beams, 511 Continuous beams, 652
L2 64 — propped, 609
L2 64 — columns, 858 Coupling, 695
Critical load, 810
. = - (x - 0.4096) x 100 = 59%. Ans. Bending, 292 Curved bars, 957

48. For a column of one end fixed and other free, the bucking load is — plane, 293 — Rectangular, 965
n _ re 2 E/ — moment, 235 — Triangular, 966
— moment and shear force, 235 — Circular, 969
For a second column of same length and same cross-sectional area when both ends are — stresses in beams, 292 — Trapezoidal, 968
— of curved bars, 957
fixed, the buckling load is Cylinders, thin, 740
Bulk modulus, 70
„ 4n 2 El Buckling load, 810 — thick, 781
Butt joint, 871
P> ~ L2 D
— weld, 901
USm] Dam, 409
c Deflection of beams, 511
L ) = 4 x 4 = 16. Ans.
_k Cantilever, 239 — of cantilever, 554
•• ~ /• o \
— propped, 597 Design of columns, 820
49. omax = | [C! - a3 ] = |[ 80 - (- 40)] = 60 MPa. Ans. — truss, 478
— thick cylindrical shell, 781
E C50. Centre of gravity, 170 — riveted joint, 892
= 2 (1 + ft) Circumferential stress, 741
Centroid, 170 Diamond riveting, 872
. JL _ x - 12P_ _ l = 1.2 - 1.0 = 0.2. Ans. Chain riveting, 872 Direct and bending stress combined, 377
2C 2x50 Disc with uniform strength, 936
^•'

1089

2

1090 STRENGTH OF MATERIALS

E INDEX 1091

Earth pressure, 443 Inclined loads, 279 N Simple bending, 294
Eccentric load, 378 Lsection, 356
Eccentricity, limit of, 398 Neutral axis, 295 Section modulus, 300
Elasticity, 5 J Shafts, torsion of, 672
Elastic constants, 59 — layer, 295 Shafts, compound and composite, 706
Elastic limit, 5 Johnson’s formula, 847
Elastic modulii, 6 Joints : riveted, 871 o Shear force diagram, 235
Elongation due to self weight, 50
Efficiency of a joint, 746, 880 — welded, 900 Oblique, 85 — for cantilevers, 239
Energy of destortion, 1033 Overhanging beams, 236
Effective length, 818 K —for a simply supported beam, 252
Euler’s theory, 809 Shear strain energy, 1033
Kernal, 401
— limitations of, 820 Shear modulus, 6
— of hollow circular section, 401
F — of hollow rectangular section, 402 P
Spherical shells, 771
Factor of safety, 6, 848 L
Failure of a column, 808 Pitch of rivets, 871 Springs : Laminated, 721
Failure of riveted joints, 876 Lap joint, 871 Poisson’s ratio, 60
Fixed beams, 613 — helical, 724
Formula— Johnson’s, 847 — weld, 902
Polar moment of inertia, 687 Straight line formula, 847
— I.S.C., 848 Lateral strain, 59
— straight line, 847 Leaf springs, 721 Polar Modulus, 688 Strain, 2
Long columns, 808
Frames, 465 Longitudinal strain, 59 Power transmitted by shafts, 677 — types, 2
Flanged coupling, 695 — compressive, 2
— stress,742 Principal planes, 85 — shear, 2
G — tensile, 2
M — Strain, 85
Gordon’s formula, 838 — Stress, 85
Gradual loading, 143 Macaulay’s method, 531
Graphical method, 498 Principle of complementary shear stresses, 73 Strain energy due to torsion, 713
Maximum principal stress theory, 1005 Proof resilience, 143 Strain energy, 143
— principal stresses, 85 — strain theory, 1006
Propped cantilevers and beams, 578 Stress, 1
Guest’s theory, 1010 — shear stress theory, 1010
Gyration, radius, 195 — strain energy theory, 1014 R — in a curved bar, 957
— shear strain energy theory, 1018 — compressive, 3
H Rankine’s formula, 835 — shear, 4
Middle third rule, 398 Rankine theory, 516, 1006 — principal, 86
Haigh’s theory, 1014 Middle quarter rule, 400 Relationship between modulus of elasticity, 79 — tensile, 2
Helical springs, 724 Mises-Henky theory, 1018 — thermal, 42
Hooke’s Law, 6 Method of joints, 467 — S.F. and B.M. 287
Hoop stress, 741 .Stress due to rotation
— sections, 488 Resilience, 143
1 stresses in a hook, 978
Modulus, bulk, 70 — proof, 143
Impact loading, 143 — in a simple chain link, 993
— of elasticity, 6 Resistance of moment, 295 — in a curved plate, 957
— load, 152 — of rigidity, 6 Riveted joints, 871 — in a circular ring, 987

Mohr's circle, 128 — failure, 876 Strength of a shaft, 688
Mohr’s theorem, 548 — strength, 879
— efficiency, 880 — rivetted joint, 879
Moment area method, 546 — design, 892
Moment of inertia, 194 Struts, 808
Rotational stress, 919
— resistance, 295 — with lateral load, 858
— in a thin cylinder, 773, 919
— in a thin disc, 919 T
— in a solid cylinder, 944
— Hollow cylinder, 940 Thermal stresses, 42

s Tensile, 2

— strain, 2
— stress,

Safety factor, 6 Thick cylinder, 781
Thick spherical shells, 800
Saint-Venent theory, 1006

i
[

r 092 STRENGTH OF MATERIALS

Thick sphere, 781 V
' hin cylinder, 740
Value of h- for curved
:
— rectangular bar, 965
— spherical shells, 770 — triangular bar, 966
— circular bar, 969
( heorem of — trapezoidal bar, 968

— parallel axis, 196 Volumetric strain, 62
— perpendicular axis, 195 V-bjutt joint, 901
Vibrations, 805
,
w
Theories of failure, 1005
Theory of Walls retaining, 443
Water and earth pressure, 409
— maximum principal stress, 1005 Welded joints, 900
— maximum strain, 1006
—{ maximum shear stress, 1010 — butt weld joint, 901
— maximum strain energy, 1014 — fillet weld joint, 901
—i . maximum shear strain energy, 1018 — compound weld joint, 905

" Torque by a Wire wound pipes, 765
Winkler-Bach Formula, 964
— solid shaft, 674 Wind pressure on chemneys, 461
—• hollow shaft, 676
Y
Torsion of shafts, 672
... .’orsion of tapering shafts, 716 Yield points, 7
Yield strength, 1005
Torsional rigidity, 688 Young’s modulus, 6, 59
’ypes of riveted joints, 871
z
— welded joints, 901
— beam, 235 Zig-Zag riveted joint, 872
— load, 236

J

Ultimate strength, 8
Uniform strength, 51

Unwin’s formula, 892
( Asymmetrical bending, 312

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Author's Profile

‘-

"J He Dcgro

Dr. R.K. Bansal graduated in 1966. obtained his Master's

,

with 'HONOURS’ from I.I.T., Delhi and Pli.D. in 1931 from Universih

He joined Delhi College of Engineering, as a Lecturer and became

Professor in 1 984. He was selected for the post of Professor in 1 99!

held the posts of Head of Much. Engineering (Faculty of Fei

University of Delhi). Dean (P.G.) Studies and Dean (U.6.) Stud

College of Engineering, Delhi. During his'teaching career of about

years, he guided a large number of research students and got publisl

papers. Dr. Bansal is the author of many books, , which are fol

textbooks. 1

s:

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