PROBLEM 11.33
The ship at A has just started to drill for oil on the ocean floor at a depth of 1500 m.
The steel drill pipe has an outer diameter of 200 mm and a uniform wall thickness
of 12 mm. Knowing that the top of the drill pipe rotates through two complete
revolutions before the drill bit at B starts to operate and using G = 77 GPa,
determine the maximum strain energy acquired by the drill pipe.
SOLUTION
ϕ = 2 rev = (2)(2π ) = 4π radians
1
L = 1500 m, co = do = 100 mm, ci = co − t = 88 mm
2
π 4 4 π
J=
co − ci = (1004 − 884 ) = 62.88 × 106 mm 4
2
2
(
)
= 62.88 × 10−6 m 4
GJ ϕ
L
ϕ=
TL
GJ
U=
T 2 L  GJ ϕ  L
GJ ϕ 2
=
=

2GJ  L  2GJ
2L
T=
2
=
(77 × 109 )(62.88 × 10 −6 )(4π )2
= 254.86 × 103 J = 254.9 kJ
(2)(1500)

PROBLEM 11.44
Collar D is released from rest in the position shown and is stopped by a
small plate attached at end C of the vertical rod ABC. Determine the mass
of the collar for which the maximum normal stress in portion BC is
125 MPa.
SOLUTION
σ m = 125 × 106 Pa
π 2
Portion BC:
ABC =
4
(9) = 63.617 mm 2 = 63.617 × 10−6 m 2
Pm = σ m ABC = 7952 N
Corresponding strain energy:
U BC =
π
(12)2 = 113.907 mm 2 = 113.907 × 10−6 m 2
4
P2 L
(7952) 2 (4)
= m AB =
= 10.574 J
2 E AB AAB (2)(105 × 109 )(113.907 × 10−6 )
AAB =
U AB
Pm2 LBC
(7952) 2 (2.5)
=
= 17.750 J
2 EBC ABC (2)(70 × 109 )(63.617 × 10−6 )
U m = U BC + U AB = 28.324 J
Corresponding elongation Δ m :
1
Pm Δ m = U m
2
2U m (2)(28.324)
Δm =
=
= 7.12 × 10−3 m
7952
Pm
Falling distance:
Work of weight = U m
h = 0.6 + 7.12 × 10−3 = 0.60712 m
Wh = mgh = U m
m=
Um
28.324
=
gh (9.81)(0.60712)
m = 4.76 kg 
PROBLEM 11.48
The steel beam AB is struck squarely at its midpoint C by a 45-kg
block moving horizontally with a speed v0 = 2 m/s. Using
E = 200 GPa, determine (a) the equivalent static load, (b) the
maximum normal stress in the beam, (c) the maximum deflection of
the midpoint C of the beam.
SOLUTION
From Appendix C, for W150 × 13.5,
I x = 6.83 × 106 mm 4 = 6.83 × 106 m 4
S x = 91.1 × 103 mm3 = 91.1 × 10−6 m3
Kinetic energy: T =
1
1
m v02 = (45)(2) 2 = 90 J
2
2
From Appendix D, Case 4:
ym =
PL3
,
48EI
M max =
PL
4
Principle of work and energy.
U =
(a)
1
P 2 L3
Pm ym = m = T
2
96EI
Equivalent static load.
Pm =
96 EIT
=
L3
(96)(200 × 109 )(6.83 × 10−6 )(90)
= 20.907 × 103 N
(3.0)3
Pm = 21.0 kN 
(b)
Maximum normal stress.
σm =
M max
P L
(20.907 × 103 )(3.0)
= m =
= 172.1 × 106 Pa
4S
S
(4)(91.1 × 10−6 )
σ m = 172.1 MPa 
(c)
Maximum deflection.
ym =
2U
(2)(90)
=
= 8.61 × 10−3 m
Pm
20.907 × 103
ym = 8.61 mm 
PROBLEM 11.55
A 72-kg diver jumps from a height of 0.5 m onto end C of a
d
diving
board having the uniform cross section shhown.
A
Assuming
that the diver’s legs remain rigid and using
E = 12 GPa, determine (a) the maximum deflection at pooint C,
( the maximum normal stress in the board, (c) the equivalent
(b)
s
static
load.
SOLUTION
W = 72 × 9.81 = 706.332 N
1
(0.4)(0.066)3 = 9.583 × 10−6 m 4
12
L = 2.9 m, a = 0.7 m
1
C = (0.066) = 0.0333 m
2
P L
M =− m x
a
a M2
P 2 L2 a
P 2 L2 a 2
x dx = m
U AB =
dx = m 2
0 2 EI
6 EI
EIa 0
2E
I=
Over portion AB:


M = − Pm v
L M2
P2
U BC =
dv = m
0 2 EI
2E
EI
Over portion BC:

1
Pm ym = U m
2
0
v3 dv =
Pm2 L3
6 EI
+ L)
6 EI
2U m Pm L2 ( a + L)
=
ym =
Pm
3EI
U = U AB + U BC =
Total

L
Pm2 L2 (a
3EI
(
(3)(12
× 109 )(9.583 × 10 −6 )
ym = 11394.8 ym
y
=
m
L2 ( a + L)
(2.9) 2 (2.9 + 0.7)
1
U m = Pm ym = 5697.44 ym2
2
Pm =
W (h + ym ) = (706..32)(0.5 + ym ) = 353.16 + 706.32 ym
Work of weight:
353.16 + 706.32 ym = 5697.4 ym2
Equating:
ym2 − 0.124 ym − 0.062 = 0
{
1
0.124 +
2
(a)
ym =
(c)
Pm = (11394.8)(0.3186) = 3630 N
}
0.1242 + (4)(0.062) =
ym = 318.6 mm
m 
Pm = 3.63 kN 
M m = −(3630)(2.9) = 10527 N ⋅ m
(b)
σm =
Mm c
I
=
(10527)(0.033)
= 36.25 M
MPa
9.583 × 10−6
σ m = 36.25 MPa
M 
1.109
PROBLEM 11
For the beam andd loading shown, and using Castigliano’s theorem, deteermine
(a) the horizontal deflection of point B, (b) the vertical deflection of point B.
SOLUTION
Add horizontal force Q at point B.
Use polar coordinate ϕ .
U=

π /2
0
M2
R ϕ
Rd
2EI
Bending moment:
ΣM J = 0: M − Pa
P − Qb = 0
M = Pa + Qb
= PR sin ϕ + QR (1 − cos ϕ )
∂M
∂M
= R (1 − cos ϕ )
= R sin ϕ
∂P
∂Q
Set Q = 0.
(a)
(b)

∂U
1
=
∂ Q EI

=
PR3
EI
(sin ϕ − sin ϕ cos ϕ )dϕ =
=
PR3 
π
1 2π 1 2 
+ sin 0 
 − cos + cos 0 − sin
EI 
2
2
2 2

=
PR3 
1

0 +1− + 0
EI 
2

δQ =

π /2
0
π /2
M
0
∂M
1
Rdϕ =
∂Q
EI

π /2
/
0
PR sin ϕ R (1 − cos ϕ ) Rdϕ
π /2
PR3
P
1
(− cos ϕ − sin 2 ϕ )
EI
2
0
δQ =
∂U
1
=
∂ P EI

=
PR 3
EI
sin 2 ϕ dϕ =
=
PR  1
PR3  1 π 1
1
1
1


sin
2
−
=
⋅ − ⋅ 0 − sin π + ⋅ sin 0 
ϕ
ϕ



EI  2
2
EI  2 2 2
2
2
0

=
PR3  π

 − 0 − 0 + 0
EI  4

δP =

π /2
0
π /2
0
M
∂M
1
Rdϕ =
∂P
EI
3
PR 3
EI

π /2
0

π /22
0
PR 3
→ 
2 EI
PR sin ϕ R sin ϕ Rdϕ
1
(1 − cos 2ϕ )dϕ
2
π /2
δP =
π PR3
4 EI
↓ 
PROBLEM 11.111
Determine the reaction at the roller support and draw the bending moment
diagram for the beam and loading shown.
SOLUTION
Remove the support at A and add the reaction RA as a load.
Bending moment:
M = RA x − M 0
Strain energy:
U=
1
2EI

L
M 2 dx
0
Deflection at point A is zero.
yA =
=
∂U
1
∂M
=
M
dx
∂RA EI
∂RA

1
EI
Reaction at A:
Bending moment:

L
0
( RA x − M 0 ) xdx =
1
EI

L3
L2 
− M 0  = 0
 RA
3
2 

RA =
3 M0
↑
2 L
3

M = M 0  x − 1 
2


PROBLEM 11.112
Determine the reaction at the roller support and draw the bending moment
D
diiagram for the beam and loading shown.
SOLUTION
R
Remove
support B and add reaction RB as a load.
U = U AC + U CB =
∂ U ∂ U AB
yB =
=
∂ RB
∂ RB
O
Over
AC:


L/ 2
0
M2
dv
d
2 EI
L
∂M 
L

= u + 
M = RB  u +  − Pu ,
2
2
∂
R



B
∂ U AB
1
=
EI
∂ RB
=
RB
EI

=
L/ 2 
0

R
= B
3EI
O
Over
CB:
M2
du +
0
2 EI
∂ U CB
+
=0
∂ RB
L/ 2

L
L

 RB  u +  − Pu   u +  duu
2
2
 

L/ 2 
2
L
P
 u +  dv −
EI
2

0

L/ 2
0
L

u  u +  du
2

2
3
 3 L
P 1  L 
L 1 L 
L −    −
   + ⋅   
2 2  2  
 2   EI  3  2 

3
7 RB L3 5 PL3
−
24 EI
48 EI
∂M
=v
∂ RB
M = RB v
∂ U CB
1
=
EI
∂ RB

L/ 2
0
3
( RB v)vdv =
RB  L 
1 RB L3
=
E
3EI  2 
24 EI
1  RB L3 5 PL3
 7
+
−
=0
yB = 

48 EI
 24 24  EI
M C = RB
L
2
M A = RB L − P
L  5 1
=
−
PL
2  16 2 
RB =
5
P↑ 
16
MC =
5
PL 
32
MA = −
3
PL 
16
MB = 0 