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This material is protected under all copyright laws as they currently

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*12–4. Determine the equations of the elastic curve using P

the x1 and x2 coordinates. EI is constant.

a
L
x1 x2
1
d y1
EI = M1(x)
dx 21

d2y1
M1(x) = 0; EI = 0
dx12

dy1
EI = C1 (1)
dx1

EI y1 = C1x1 + C2 (2)

M2(x) = Px2 - P(L - a)

d2y2
EI = Px2 - P(L - a)
dx22

dy2 P
EI = x22 - P(L - a)x2 + C3 (3)
dx2 2

P 3 P(L - a)x22
EI y2 = x2 - + C3x2 + C4 (4)
6 2

Boundary Conditions:

dy2
At x2 = 0, = 0
dx2

From Eq. (3), 0 = C3

At x2 = 0, y2 = 0

0 = C4

Continuity Condition:
dy1 dy2
At x1 = a, x2 = L - a; = -
dx1 dx2

From Eqs. (1) and (3),

P(L - a)2 P(L - a)2

C1 = - C - P(L - a)2S; C1 =
2 2

At x1 = a, x2 = L - a, y1 = y2

1133
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12–4. Continued

From Eqs. (2) and (4),

P(L - a)2 P(L - a)3 P(L - a)3

a b a + C2 = -
2 6 2

Pa(L - a)2 P(L - a)3

C2 = - -
2 3

From Eq. (2),

P
v1 = [3(L - a)2x1 - 3a(L - a)2 - 2(L - a)3] Ans.
6EI

From Eq. (4),

P
v2 = [x 3 - 3(L - a)x22] Ans.
6EI 2

1134
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12–7. The beam is made of two rods and is subjected to P

the concentrated load P. Determine the maximum
deflection of the beam if the moments of inertia of the rods B
are IAB and IBC , and the modulus of elasticity is E. A C

l
L

d2y
EI = M(x)
dx2

M1(x) = - Px1

d2y1
EIBC = - Px1
dx1 2

dy1 Px21
EIBC = - + C1 (1)
dx1 2

Px31
EIBC y1 = - + C1x1 + C2 (2)
6

M2(x) = - Px2

d2y2
EIAB = - Px2
dx2 2

dy2 P
EIAB = - x2 2 + C3 (3)
dx2 2

P 3
EIAB y2 = - x + C3x2 + C4 (4)
6 2
Boundary Conditions:
dy2
At x2 = L, = 0
dx2

PL2 PL2
0 = - + C3; C3 =
2 2
At x2 = L, y = 0

PL3 PL3 PL3

0 = - + + C4; C4 = -
6 2 3
Continuity Conditions:
dy1 dy2
At x1 = x2 = l, =
dx1 dx2

1139
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12–7. Continued

From Eqs. (1) and (3),

PI 2 PI 2 PL2
c- + C1 d = c- d
1 1
+
EIBC 2 EIAB 2 2

Pl2 PL2 Pl2

c- d +
IBC
C1 = +
IAB 2 2 2
At x1 = x2 = l, y1 = y2
From Eqs. (2) and (4),

Pl3 Pl2 PL2 Pl2

e- + c a- b + d l + C2 f
1 IBC
+
EIBC 6 IAB 2 2 2

Pl3 PL2l PL3

c- d
1
= + -
EIAB 6 2 3

IBC Pl3 IBC PL3 Pl3

C2 = - -
IAB 3 IAB 3 3
Therefore,

Px1 3 Pl2 PL2 Pl2

e- + c a- b + d x1
1 IBC
y1 = +
EIBC 6 IAB 2 2 2

IBC Pl3 IBC PL3 Pl3

+ - - f
IAB 3 IAB 3 3

At x1 = 0, y1 |x = 0 = ymax

IBC Pl3 IBC PL3 Pl3

e f = e l3 - L3 - a bl f
I P IAB 3
ymax = - -
EIBC IAB 3 IAB 3 3 3EIAB IBC

e a1 - b l - L3 f
P IAB 3
= Ans.
3EIAB IBC

Ans:

e a1 - bl - L3 f
P IAB 3
vmax =
3EIAB IBC

1140
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*12–20. Determine the equations of the elastic curve 40 kN

8 kip
using the x1 and x2 coordinates, and specify the slope at A
and the deflection at C. EI is constant.
A B C

30
20 kNm
kipft
x11 x22
6 mft
20 3m
10 ft

Referring to the FBDs of the beam’s cut segments shown in Fig. b, and c,

a + ©Mo = 0; M(x11) ++25x

5x1 ==00 M(x1) ==(–25x kip· #m
(-5x11))kN ft
And

a + ©Mo = 0; -M(x2) - 40x

8x2 2 -–30
20==0 0 M(x
M(x ==
2)2) ( -8x
(–40x - 20)
2 –2 30) · m # ft
kNkip

d2v
EI = M(x)
dx2
For coordinate x1,

d2v1
EI = (–25x kN #· ft
(- 5x12)) kip m
dx21

EI
dv1
= aa–- xx2121 ++ CC11bbkN
5
25 # ft22
kip· m (1)
dx1 22

EI v1 = a– b2kN
a - xx1 3 ++CC1x1x1 1+ +C2C · m# 3ft3
b kip
5
25
(2)
6
For coordinate x2,

d2v2
EI = (–40x
(-8x22 –- 30) kip· #m
20)kN ft
dx2 2

A -4x222 –-30x B kip

· m# ft2
dv2
EI = (–20x 20x2 2+ +
C3C
) 3kN (3)
dx2

EI v2 = a–
a - xx223 –-1510x x32x+2 C+4bCkN · m3# ft3
4 b kip
4
20
x222 2+ +C3C (4)
3
At x1 = 0, v1 = 0. Then, Eq (2) gives

A(0
0 B) +
25
5 33
EI(0) = –- + CC11(0)
(0) ++CC
22 C2 = 0
66
Also, at x1 = 20 ft, v1 = 0. Then, Eq (2) gives
6 m,

A(6
20 )B ++CC m2# ft2
525 33
EI(0) = -
– 1(6) +0+ 0
1 (20) C1 == 150
333.33
kN ·kip
66
Also, at x2 = 10 ft, v2 = 0. Then, Eq. (4) gives
3 m,

A(3
10 )B –-15(3
10 2A )10+2 BC3+(3)C3+(10)
420 33
EI(0) = –- C4 + C4
33
3C3 3++ C44 =
10C = 315
2333.33 (5)

1157
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*12–20. Continued

dv1 dv2
At x1 = 20
6 mft and x2 = 10 ft,
3 m, = - . Then Eq. (1) and (3) gives
dx1 dx2

A(6
20 )B ++150 = - C2)-4
– A30(3)
102 B + + C3 D
525 22
-
– = –[–20(3
333.33 - C20(10)
3]
22

C3 = 1266.67 m2 # ft2
570 kN ·kip
Substitute the value of C3 into Eq (5),

C4 = –1395
-10333.33 m3 # ft3
kN · kip
Substitute the value of C1 into Eq. (1),

a - xx112 ++150b
a– m2 # ft2
dv1 1 5
25
= 333.33b
kN · kip
dx1 EI 26
At A, x1 = 0. Thus,

dv1 333 kN ·# ft
150 kip m22
uA = 2 = uA Ans.
dx1 x1 = 0 EI
EI

Substitute the values of C1 and C2 into Eq. (2) and C3 and C4 into Eq (4),

aa–- xx113 ++ 150x

333 1xb1kN · m# 3ft3
b kip
1 25
5
v1 = Ans.
EI 6

3 # 3
aa–- xx223 –-15x
1 4
20
v2 = 10x 2 2
+ 1267x
2 2 + 570x 2 - 10333b
2 – 1395b kN · mkip ft Ans.
EI 3
At C, x2 = 0. Thus

10 333kN
1395 # ft3 3 1395
kip· m 10 333 · m3# ft
kNkip
3
vC = v2 冷x2 = 0 = - = T T Ans.
EI
EI EIEI

40 kN

30 kN · m
6m 3m

25 kN 65 kN

40 kN

25 kN
30 kN · m

1158
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12–83. The W 250 * 22 cantilevered beam is m ade of A-36 45 kN/m

steel and is subjected to the loading shown. Determine the
slope and displacement at its end B.
B
Here, A
1.8 m 1.8 m
= a b (1.8)
1 72.9
uB = 冷uB>A冷
3 EI

43.74 k N # m2
= -
EI

For W 250 * 22 I = 28.8 (10) 6 mm4, and for A36 steel E = 203 MPa. Thus

43.74(1000)(1000)
uB = -
203(28.8) 106

= - 0.007481 rad Ans.

yB = 冷 tB>A冷 = c (1.8) + 1.8 d c a b(1.8) d

3 1 72.9
4 3 EI

136.40 kN # m3
=
EI

136.40 A 10 9 B
=
203 * 28.8 A 10 B 6

= 23.33 mm. T Ans.

Ans:
uB = - 0.007481 rad, vB = 23.33 mmT

1230
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*12–88. Determine the slope at A and the deflection at 3 kN 3 kN

point C of the simply supported beam. The modulus of
elasticity of the wood is E = 10 GPa. 100 mm

A C B 200 mm

1.5 m 1.5 m 3m

Elastic Curves. The two concentrated forces P are applied separately on the beam
and the resulting elastic curves are shown in Fig. a.
Method of Superposition. Using the table in the appendix, the required slopes and
deflections are

Pab(L + b) 3(1.5)(4.5)(6 + 4.5) 5.90625 kN # m2

(uA)1 = = =
6EIL 6EI(6) EI

A L2 - b2 - x2 B = A 6 - 4.52 - 1.52 B
Pbx 3(4.5)(1.5) 2
(¢ C)1 =
6EIL 6EI(6)

7.594 kN # m3
= T
EI

PL2 3 A 62 B 6.75 kN # m2
(uA)2 = = =
16EL 16EI EI

A 3L2 - 4x2 B = a 3(6)2 - 4(1.5)2 b =

Px 3(1.5) 9.281
(¢ C)2 =
48EI 48EI EI
Then the slope at A and deflection at C are

uA = (uA)1 + (uA)2

5.90625 6.75
= +
EI EI

12.65625 kN # m2 12.6525 A 103 B

= = = 0.0190 rad Ans.
10 A 109 B c (0.1) A 0.23 B d
EI 1
12
and

¢ C = (¢ C)1 + (¢ C)2

7.594 9.281 16.88 A 103 B

= + = = 0.0253 m = 25.3 mm Ans.
10 A 10 B c (0.1) A 0.2 3 B d
EI EI 9 1
12

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13–19. Determine the maximum force P that can be 0.9 m

applied to the handle so that the A992 steel control rod AB 0.6 m
does not buckle. The rod has a diameter of 30 mm. It is pin
connected at its ends. A

0.9 m

B
a + ©MC = 0; FAB (0.6) - P(0.9) = 0

2
P = F (1)
3 AB

Bucking Load for Rod AB:

p
I = (15.625 4) = 46813.37 mm4
4

A = p(15.625 2) = 767 mm2

p2EI
Pcr =
(KL)2

p22(105)(46813.37)
FAB = Pcr = = 114 kN
[1.0(900)] 2

From Eq. (1)

2
P = (114) = 76 kN Ans.
3

Check:

Pcr 114000
scr = = = 148.6 MPa 6 sY OK
A 767

Therefore, Euler’s formula is valid.

Ans:
P = 76 kN

1317
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*13–24. An L-2 tool steel link in a forging machine is pin P P

connected to the forks at its ends as shown. Determine the
maximum load P it can carry without buckling. Use a factor
of safety with respect to buckling of F.S. = 1.75. Note from
the figure on the left that the ends are pinned for buckling,
whereas from the figure on the right the ends are fixed.
1.5 in.
36 mm 0.5 in.
12 mm

24 in.
600 mm

Section Properties:

A 1.5(0.5)=432
= 36(12) mmin
0.750 2 2

(0.5) A 1.5 B 46656(10

11
Ixx =
I (12)(36 )
3 3
= 0.140625
6
in4 4
) mm
12
12 P P

(1.5) A 0.5 B 5184

11
IIyy = (36)(12 )
3 3
= 0.015625
mm4 in4
12
12

Critical Buckling Load: With respect to the x - x axis, K = 1 (column with both
ends pinned). Applying Euler’s formula,

p2EI
Pcr =
(KL)2

p2(29.0)(10 3
(200)(103)(46656)
)(0.140625)
=
[1(24)]2 2
[1(600)]

= 255820 N  255.8 kN
69.88 kip

With respect to the y - y axis, K = 0.5 (column with both ends fixed).

p2EI
Pcr =
(KL)2
2
(29.0)(1033)(5184)
p2(200)(10 )(0.015625)
= 2 2
[0.5(24)]
[0.5(600)]

= 113698 N  113.7
31.06 kip kN (Control!)
(Controls!)

Critical Stress: Euler’s formula is only valid if scr 6 sg.

Pcr 113698
31.06
scr = = =41.41
263.2ksi
MPa6 s
<gs=g 
102 ksiMPa
703 O.K.
A 0.75
432

Factor of Safety:

Pcr
F.S =
P

31.06
113698
1.75 =
PP

P = 17.7
64970kip
N = 64.97 kN Ans.

1322