ENGR0135 - Statics and Mechanics of Materials 1 (2161)
Homework #11
Solution Set
1. (a) The torques transmitted by cross sections in intervals AB, BC, CD, and DE are
TAB = 25 kip·ft ,
TBC = −75 kip·ft ,
TCD = −35 kip·ft ,
TDE = 45 kip·ft ,
(b) The torque diagram for the shaft is
T (kip · ft)
45
25
A
B
C
D
E
−35
−75
2. The radius of the shaft is c = 4 in and the polar second moment for its cross section is
1
J = π(4 in)4 = 402.1239 in4
2
(a) From Problem 1, it is seen that the maximum absolute value of torque in the
shaft is Tmax = 75 kip · ft = 900 kip · in. It follows that the maximum shearing
stress in the shaft is
Tmax c
(900 kip · in)(4 in)
τmax =
=
= 8.95 ksi
J
(402.1239 in4 )
(b) The rotation of a section at D with respect to a section at B is
θD/B = θC/B + θD/C
TBC LBC
TCD LCD
=
+
GBC JBC GCD JCD
(TBC + TCD )L
=
GJ
(−900 − 420 kip · in)(24 in)
=
(12, 000 kip/in2 )(402.1239 in4 )
= −6.57 × 10−3 rad
(c) The rotation of a section at E with respect to a section at A is
(TAB + TBC + TCD + TDE )L
GJ
(300 − 900 − 420 + 540 kip · in)(24 in)
=
(12, 000 kip/in2 )(402.1239 in4 )
= −2.39 × 10−3 rad
θE/A =
3. Let’s determine the torque corresponding to the allowable shearing stress first. The
polar second moments for the two segments of the shaft are
1
JAB = π(1.25 in)4 = 3.8350 in4 ,
2
1
JBC = π(0.875 in)4 = 0.92077 in4
2
The maximum shearing stress in shaft AB will equal the allowable shearing stress at
the following torque:
T =
τall JAB
(8000 psi)(3.8350 in4 )
=
= 24, 544 lb · in
cAB
(1.25 in)
(A)
The maximum shearing stress in shaft BC will equal the allowable shearing stress at
the following torque:
T =
τall JBC
(8000 psi)(0.92077 in4 )
=
= 8418 lb · in
cBC
(0.875 in)
(B)
To determine the torque corresponding to the allowable angle of twist, note first that
TAB LAB
TBC LBC
T LAB LBC
θC/A =
+
=
+
GAB JAB GBC JBC
G JAB
JBC
It follows that
−1
LAB LBC
+
T = θall G
JAB
JBC
−1
48 in
36 in
6
= (0.04)(4 × 10 psi)
+
3.8350 in4 0.92077 in4
= 3100 lb · in
(C)
The maximum permissible torque is the smaller of those in (A)–(C):
Tall = 3100 lb · in
4. The polar second moment for the shaft is
J=
π
π 4
(ro − ri4 ) = [(0.06 m)4 − (0.035 m)4 ] = 1.8000 × 10−5 m4
2
2
The maximum compressive stress is equal to the maximum shearing stress (on an
inclined plane with angle α = 45◦ ). So,
σmax = τc =
Tc
(8000 N · m)(0.06 m)
=
= 26.7 MPa
J
(1.8000 × 10−5 m4 )
5. The weight of the block is W = (100 kg)(9.81 m/s2 ) = 981 N and its free-body diagram
is shown below.
y
F
W
35◦
Fx = F cos 35◦ − 0.3N = 0
Fy = N − W − F sin 35◦ = 0
x
0.3N
N
Solving the two equilibrium equations for F and N gives
F = 454.8130 N ,
N = 1241.8700 N
(a) The work done on the block by the force F is
Uforce = (F cos 35◦ )(20 m) = 7451 N · m
(b) Gravity does no work on the block, since its direction is perpendicular to the
direction of motion:
Ugravity = 0
(c) The normal force, N , exerted on the block by the floor does no work on the block,
since its direction is perpendicular to the direction of motion. Thus, the work
done on the block by the floor is just the work done by the friction, which is
negative since it is in the opposite direction of the motion:
Ufloor = Ufriction = −(0.3N )(20 m) = −7451 N · m
6. The power transmitted by the shaft at the rated load is
lb · ft
lb · ft/min
= 6.6000 × 108
P = (20, 000 hp) 33, 000
hp
min
The angular velocity of the shaft is
rev rad
1
ω = 60
2π
= 376.99
min
rev
min
It follows that the torque in the shaft at the rated load is
T =
P
= 1.7507 × 106 lb · ft = 2.1008 × 107 lb · in
ω
The polar second moment for the shaft is
1
1
J = πc4 = π(15 in)4 = 7.9522 × 104 in4
2
2
(a) The maximum shearing stress in the shaft at the rated load is
τc =
(2.1008 × 107 lb · in)(15 in)
Tc
=
= 3963 psi
J
(7.9522 × 104 in4 )
(b) The magnitude of the angle of twist in the shaft at the rated load is
θ=
(2.1008 × 107 lb · in)(240 in)
TL
=
= 5.28 × 10−3 rad
GJ
(12 × 106 lb/in2 )(7.9522 × 104 in4 )
7. The angles of twist in a length L for the inner and outer tubes are
θi =
Ti L
,
Gi Ji
θo =
To L
Go Jo
where Ti and To are the torques carried by the inner and outer tubes, respectively, Gi
and Go are their moduli of rigidity, and their polar second moments are
π
[(0.0625 m)4 − (0.05 m)4 ] = 1.4151 × 10−5 m4
2
π
Jo = [(0.09 m)4 − (0.0625 m)4 ] = 7.9091 × 10−5 m4
2
Ji =
The two tubes are constrained to have the same angle of twist, so
Gi
Ji
80
1.4151
θi = θo
=⇒
Ti =
To =
To = 0.22021To
Go
Jo
65
7.9091
On the other hand, the sum of the torques carries by each tube must equal the total
torque of 12 kN · m applied to the composite shaft, so
Ti + To = 12 kN · m
=⇒
Ti = 2.1656 kN · m ,
To = 9.8344 kN · m
(a) The maximum shearing stress in the steel is
τi =
(2165.6 N · m)(0.0625 m)
Ti ci
=
= 9.5647 × 106 Pa = 9.56 MPa
Ji
(1.4151 × 10−5 m4 )
The maximum shearing stress in the Monel is
τo =
To co
(9834.4 N · m)(0.09 m)
=
= 1.1191 × 107 Pa = 11.2 MPa
Jo
(7.9091 × 10−5 m4 )
(b) The angle of twist in a 2 m length of the composite shaft is
θ = θi = θo =
(9834.4 N · m)(2 m)
= 3.83 × 10−3 rad
9
2
−5
4
(65 × 10 N/m )(7.9091 × 10 m )
8. The angles of twist for the two shafts are
θAB =
TAB LAB
,
GAB JAB
θBC =
TBC LBC
GBC JBC
where TAB and TBC are the torques carries by the two shafts, LAB and LBC are their
lengths, GAB and GBC are their moduli of rigidity, and their polar second moments
are
π
(0.05 m)4 = 9.8175 × 10−6 m4
2
π
= (0.04 m)4 = 4.0212 × 10−6 m4
2
JAB =
JBC
The two shafts are constrained to have the same angle of twist, so
θAB = θBC
=⇒
TAB =
GAB JAB LBC
(39)(9.8175)(600)
TBC =
TBC = 1.0986TBC
GBC JBC LAB
(65)(4.0212)(800)
On the other hand, the sum of the torques carries by each shaft must equal the torque
of 20 kN · m applied to the disk, so
TAB + TBC = 20 kN · m
=⇒
TAB = 10.4698 kN · m ,
TBC = 9.5302 kN · m
(a) The maximum shearing stress in shaft AB is
τAB =
TAB cAB
(10469.8 N · m)(0.05 m)
=
= 5.3322 × 107 Pa = 53.3 MPa
JAB
(9.8175 × 10−6 m4 )
The maximum shearing stress in shaft BC is
τBC =
TBC cBC
(9530.2 N · m)(0.04 m)
= 9.4800 × 107 Pa = 94.8 MPa
=
−6
4
JBC
(4.0212 × 10 m )
(b) The angle of rotation of the disk is
θ = θAB = θBC =
(9530.2 N · m)(0.6 m)
= 2.19 × 10−2 rad
9
2
−6
4
(65 × 10 N/m )(4.0212 × 10 m )