Problem 15.46
The beam is subjected to a distrubuted load. For the
cross section at x = 0.6 m, determine the average shear
stress (a) at the neutral axis; (b) at y = 0.02 m.
Free Body Diagram:
Solution:
Summing the moments about point B to determine the reaction at point
A:
1
MB = 0 = −
(130, 000 N/m)(1.4 m) (0.133 m)+Ay (0.8 m) → Ay = 15, 130 N ↓
2
Cut the beam at x = 0.6 m and draw the FBD.
Summing the vertical forces to determine the shear force at x = 0.6 m:
1
Fy = 0 = −15, 130 N − (55, 714 N/m)(0.6 m) +V → V = 31, 844 N ↑
2
(a) Using Equation (15-18) to determine the average shear stress at the
neutral axis:
τ =
ANS:
3V
3(−31, 844 N
=
2A
2(0.04 m)(0.06 m)
τAVG = −19.9 MPa
(b) Using Equation (15-17) to determine the average stress at y =
0.02 m:
h 2 2
6(−31, 844 N)
0.06 m 2
6V
2
τAVG =
=
−
y
−
(0.02
m)
bh3
2
(0.04 m)(0.06 m)3
2
ANS:
τAVG = −11.06 MPa
Problem 15.47 Solve Problem 15.46 for the cross section at x = 1.0 m.
Free Body Diagram:
Solution:
Sum moments about point B to determine the reaction at point A:
1
MB = 0 = −
(130, 000 N/m)(1.4 m) (0.133 m)+Ay (0.8 m) → Ay = 15, 130 N ↓
2
Summing vertical forces to determine By :
1
Fy = 0 = −15, 130 N− (130, 000 N/m)(1.4 m) +By → By = 106, 130 N ↑
2
Cut the beam at x = 1.0 m and draw the FBD.
Summing the vertical forces to determine the shear force at x = 1.0 m:
1
Fy = 0 = −15, 130 N+106, 130 N − (92, 860 N/m)(1.0 m) −V → V = 44, 570 N ↓
2
(a) Using Equation (15-18) to determine the average shear stress at the
neutral axis:
τ =
ANS:
3V
3(−44, 570 N)
=
2A
2(0.04 m)(0.06 m)
τAVG = −27.86 MPa
(b) Using Equation (15-17) to determine the average stress at y =
0.02 m:
6V
h 2 2
6(−44, 570 N)
0.06 m 2
2
τAVG =
−
y
−
(0.02
m)
=
bh3
2
(0.04 m)(0.06 m)3
2
ANS:
τAVG = −15.48 MPa
Problem 15.48 At a particular axial position, the beam
whose cross section is shown is subjected to a shear force
V = 40 kN. What is the average shear stress at the
neutral axis (y = 0).
Solution:
The moment of inertia for the cross section is:
π
πRo4 πRi4
−
=
(0.08 m)4 − (0.05 m)4 = 2.726×10−5 m4
I=
4
4
4
Calculating y :
(0.08 m)2
4(0.08 m)
(0.05 m)2
4(0.05 m)
−π
π
2
3π
2
3π
y =
(0.08 m)2
(0.05 m)2
−π
π
2
2
Calculating Q:
Q = y A = (0.0421 m)(0.0061 m) = 0.000257 m3
The average shear stress at the neutral axis is:
τAVG =
ANS:
VQ
(40, 000 N)(2.57 × 10−4 m3 )
=
bI
2(0.08 m − 0.05 m)(2.726 × 10−5 m4 )
τAVG = 6.29 MPa
Problem 15.49
For the beam in Problem 15.48, what is the average shear
stress at y = 50 m?
Solution:
The moment of inertia for the cross section is:
π
πRo4 πRi4
−
=
(0.08 m)4 − (0.05 m)4 = 2.726×10−5 m4
I=
4
4
4
From the diagram, the angle α is:
0.05 m
α = cos−1
= 51.3◦
0.08 m
The thickness of the section as y = 50 mm is:
t = 2 (0.08 m) sin 51.3◦ ) = 0.125 m
Calculating Q using the derived equation:
3/2 3/2 2 (0.008)2 − (0.05)2
Q=
− (0.05)2 − (0.05)2
3
Q = 0.0001624 m3 = 1.624 × 10−4 m3
Calculating the shear stress at y = 50 mm:
τ =
ANS:
(40, 000 N)(1.623 × 10−4 m3 )
VQ
=
It
(2.73 × 10−5 m4 )(0.125 m)
τ = 1.902 MPa
Problem 12.3 The components of plane stress at a
point p of a material are σx = −8 ksi, σy = 6 ksi
and τxy = −6 ksi. If θ = 30◦ , what are the stresses σx ,
σy and τxy
at point p?
Solution:
Using Equation (12-7) to find σx :
σx =
σx
=
ANS:
σx +σy
σ −σ
+ x 2 y (cos 2θ) + τxy (sin 2θ)
2
−8 ksi+6 ksi
ksi
+ −8 ksi−6
(cos 60◦ ) + (−6
2
2
ksi) (sin 60◦ )
σx = −9.7 ksi
Using Equation (12-9):
σy =
σy
=
ANS:
σx +σy
σ −σ
− x 2 y (cos 2θ) − τxy (sin 2θ)
2
−8 ksi+6 ksi
ksi
− −8 ksi−6
(cos 60◦ ) − (−6
2
2
ksi) (sin 60◦ )
σy = 7.7 ksi
:
Using Equation (12-8) to find τxy
σx −σy
(sin 2θ) + τxy (cos 2θ)
2
(−8−6)
− 2
sin 60◦ + (−6) cos 60◦
=−
τxy
=
τxy
ANS:
= 3.062 ksi
τxy
Problem 12.4 During liftoff, strain gauges attached to
one of the Space Shuttle main engine nozzles determine
that the components of plane stress σx = 66.46 MPa,
σy = 82.54 MPa, and τxy
= 6.75 MPa at θ = 20◦ .
What are the stresses σx , σy and τxy at that point?
Solution:
Adding Equations (12-7) and (12-9):
σx + σy = σx + σy → σx + σy = 149 MPa
[1]
Using Equation [1] in Equation (12-9):
82.54 MPa =
149 MPa σx − σy
−
[cos (40◦ )] − τxy [sin (40◦ )]
2
2
8.04 MPa = −(σx − σy )(0.383) − (0.643)τxy
τxy = −12.5 MPa − (σx − σy )(0.596)
[2]
Substituting the given information into Equation (12-8):
6.75 MPa = −
σx − σ y
[sin (40◦ )] + τxy [cos (40◦ )]
2
6.75 MPa = −(0.321)(σx − σy ) + τxy (0.766)
τxy = 8.81 MPa + (0.419)(σx − σy )
[3]
Subtracting Equations [2] and [3]:
0 = −21.31 − 1.015(σx − σy ) → σx − σy = −20.99 MPa
Adding Equations [1] and [4]:
2σx = 128.01 MPa
ANS:
σx = 64 MPa
σy = 85 MPa
τxy = 0 MPa
[4]
Problem 12.13 The stress τxy = 14 MPa and the angle θ = 25◦ . Determine the components of stress on the
right element.
Solution:
Using Equation (12-7) to find σx :
σx =
σx
=
σx +σy
σ −σ
+ x2 y
2
8 MPa+(−6 MPa)
2
σy
=
+
8 MPa−(−6 MPa)
2
(cos 50◦ ) + (14 MPa) (sin 50◦ )
σx = 16.22 MPa
ANS:
σy =
(cos 2θ) + τxy (sin 2θ)
σx +σy
σ −σ
− x2 y
2
8 MPa+(−6 MPa)
2
(cos 2θ) − τxy (sin 2θ)
−
8 MPa−(−6 MPa)
2
(cos 50◦ ) − (14 MPa)(sin 50◦ )
σy = −14.22 MPa
ANS:
:
Using Equation (12-8) to find τxy
σx −σy
(sin 2θ) + τxy (cos 2θ)
2
8 MPa−(−6 MPa)
−
(sin 50◦ ) + (14
2
=−
τxy
=
τxy
MPa) (cos 50◦ )
= 3.64 MPa
τxy
ANS:
Problem 12.14 On the elements shown in Problem 12.13, the stresses τxy = 12 MPa, σx = 14 MPa,
σy = −12 MPa. Determine the stress τxy
and the angle
θ.
Solution:
Using Equation (12-7) to find θ:
σx +σy
σ −σ
+ x 2 y (cos 2θ) + τxy (sin 2θ)
2
8 MPa+(−6 MPa)
8 MPa−(−6 MPa)
+
MPa =
2
2
σx =
14
(cos 2θ) + (14 MPa) (sin 2θ)
A graphing calculator reveals two angles at which these conditions
exist. The angles are:
θ1 = 19.5◦ θ2 = 40.2◦
ANS:
:
Using Equation (12-8) to find the two corresponding values of τxy
σx −σy
(sin 2θ) + τxy (cos 2θ)
2
8 MPa−(−6 MPa)
−
(sin 2 ∗ 19.5◦ )
2
=−
τxy
=
τxy
+ (12 MPa) (cos(2 ∗ 19.5◦ )
ANS:
) = 4.9 MPa
(τxy
1
τxy
=−
8 MPa − (−6 MPa)
(sin 2 ∗ 40.2◦ )+(12 MPa) (cos(2 ∗ 40.2◦ )
2
ANS:
) = −4.9 MPa
(τxy
2
Problem 12.23 Determine the principal stresses and
the maximum in-plane shear stress and show them acting
on properly oriented elements. σx = 25 ksi, σy = 0 ksi,
and τxy = −25 ksi.
Solution:
Equation (12-15) is used to find the principal stresses.
ANS:
σ1,2 =
σx +σy
2
σ1,2 =
25 ksi+0
2
±
σ1 = 40.45 ksi
25 ksi−0
2
τMAX =
ANS:
2
+ τxy
2
+ (−25 ksi)2
σ2 = −15.45 ksi
Using Equation (12-19) to find τMAX :
τMAX =
2
σx −σy
2
±
2
σx −σy
2
25 ksi−0
2
2
+ τxy
2
+ (−25 ksi)2
τMAX = 27.95 MPa
The angles for the principal stresses and the maximum shear stress are:
tan 2θp =
tan 2θs =
ANS:
2τxy
2(−25 ksi)
=
= −2
σx − σy
25 ksi − o
σx − σy
25 ksi − 0
=
= −0.5
2τxy
2(−25 ksi)
θp = −31.7◦
(Equation 12-12)
(Equation 12-16)
θs = −13.3◦
Problem 12.24 Determine the principal stresses and
the maximum in-plane shear stress and show them acting
on properly oriented elements. σx = −8 ksi, σy = 6 ksi,
and τxy = −6 ksi.
Solution:
Equation (12-15) is used to find the principal stresses.
ANS:
σ1,2 =
σx +σy
2
σ1,2 =
−8+6
2
σ1 = 8.22 ksi
2
σ −σ
x
y
2
±
+ τxy
2
−8−6 2
±
+ (−6)2
2
σ2 = −10.22 ksi
Using Equation (12-18) to find τMAX :
τMAX =
τMAX =
ANS:
σx −σy
2
2
2
+ τxy
−8 ksi−6 ksi
2
2
+ (−6 ksi)2
τMAX = |9.22 MPa|
The angles for the principal stresses and the maximum shear stress are:
tan 2θp =
tan 2θs =
ANS:
2τxy
2(−6 ksi)
=
= 0.857
σx − σy
−8 ksi − 6 ksi
−σx − σy
(−8 − 6)
=−
= −1.1667
2τxy
2(−6)
θp = 20.3◦
θs = −24.7◦
Problem 12.25 Determine the principal stresses and
the maximum in-plane shear stress and show them acting
on properly oriented elements. σx = 240 MPa, σy =
−120 MPa, and τxy = 240 MPa.
Solution:
Equation (12-15) is used to find the principal stresses.
ANS:
σ1,2 =
σx +σy
2
σ1,2 =
240+(−120)
2
σx −σy
2
±
±
2
+ τxy
240−(−120) 2
2
+ (240)2
σ1 = 360 MPaσ2 = −240 MPa
Using Equation (12-19) to find τMAX :
τMAX =
τMAX =
ANS:
2
σx −σy
2
2
2
+ τxy
240−(−120)
2
2
+ (240)2
τMAX = 300 MPa
The angles for the principal stresses and the maximum shear stress are:
tan 2θp =
tan 2θs =
ANS:
2τxy
2(240)
=
= 1.333
σx − σy
240 − (−120)
σx − σy
240 MPa − (−120 MPa)
=
= 0.75
2τxy
2(240 MPa)
θp = 26.6◦ θs = 18.4◦
Problem 12.26 For the state of plane stress σx =
20 MPa, σy = 10 MPa, and τxy = 0, what is the
absolute maximum shear stress?
Solution:
Since τxy = 0, we can use σx and σy as σ1 and σ2 .
Using Equations (12-24) to find the absolute maximum shear stress:
σ1 − σ2 20 MPa − 10 MPa =
= 5 MPa
2
2
σ1 20 MPa =
= 10 MPa
2
2
σ2 10 MPa =
= 5 MPa
2
2
We see that the largest of these values is:
τMAX = 10 MPa
ANS:
Problem 12.55 A spherical pressure vessel has a 2.5-m
radius and a 5-mm wall thickness. It contains a gas with
pressure pi = 6 × 105 Pa and the outer wall is subjected
to atmospheric pressure p0 = 1 × 105 Pa. Determine
the maximum normal stress in the vessel wall.
Solution:
The projected area of half of the sphere is:
AP = πr 2 = π(2.5 m)2 = 19.6 m2
The net force on the hemisphere is:
F = (Pi − P0 )(AP ) = (6 × 105 N/m2 − 1 × 105 N/m2 )(19.6 m2 )
F = 9.8 × 106 N
The area over which the circumferential load is exerted is:
A = 2πrt = 2π(2.5 m)(0.005 m) = 0.0785 m2
The normal (circumferential) stress in the material is:
σ=
ANS:
9.8 × 106 N
F
=
A
0.0785 m2
σ = 125 MPa
Problem 12.56 A spherical pressure vessel has a 24in. radius and a 1/64-in. wall thickness. It contains a
gas with pressure pi = 200 psi and the outer wall is subjected to atmospheric pressure p0 = 14.7 psi. Determine
the maximum normal stress and the absolute maximum
shear stress at the vessel’s inner surface.
Solution:
The projected area of the cross-section is:
AP = πr 2 = π(24 in)2 = 1809.6 in2
The area of material which resists the pressure is:
A = 2πrt = 2π(24 in)(1/64 in) = 2.356 in2
Net force on the projected area of the hemisphere is:
F = (200 lb/in2 − 14.7 lb/in2 )(1809.6 in2 ) = 335, 319 lb
The normal stress in the material is:
σx = σy =
335, 319 lb
= 142.3 ksi
2.356 in2
The shear stress in the sphere is:
τxy = (Pi −PATM ) = (200 lb/in2 −14.7 lb/in2 ) = 185.3 lb/in2
Using Equations (12-26) to determine I1 , I2 , and I3 :
I1 = σx + σy + σz = 142.3 + 142.3 + 0 = 284.6
2 − τ 2 + τ 2 = (142.3)(142.3) + (142.3)(0) + (0)(142.3) − (0.185)2 − (0)2 − (0)2 = 20, 250
I2 = σx σy + σy σz + σz σx − τxy
yz
zx
2 − σ τ 2 − σ τ 2 + 2τ τ τ
2
2
I3 = σx σy σz − σx τyz
y xz
z xy
xy yz yz = (142.3)(142.3)(0) − (142.3)(0) − (142.3)(0) − (0)(0.185) + 2(0.185)(0)(0) = 0
Using Equation (12-29) to determine the absolute maximum shear
stress:
σ 3 −I1 σ 2 +I2 σ−I3 = 0 → σ 3 −284.6σ 2 +20, 250σ+0 = 0
The roots of Equation [1] are:
σ1 = 142.3
σ2 = 142.3
σ3 = 0
Absolute maximum shear stress on the sphere is:
σ1 − σ3 142.3 MPa − 0 MPa τMAX = =
2
2
ANS:
τMAX = 71.15 ksi
[1]
Problem 12.58 A cylindrical pressure vessel with
hemispherical ends has a 2.5-m radius and a 5-mm wall
thickness. It contains a gas with pressure pi = 6×105 Pa
and the outer wall is subjected to atmospheric pressure
p0 = 1 × 105 Pa. Determine the maximum normal
stress in the vessel wall. Compare your answer with the
answer to Problem 12.55.
Solution:
The area over which the axial load is distributed is:
AA = 2πrt = 2π(2.5 m)(0.005 m) = 0.0785 m2
The area over which the circumferential load is distributed is:
AC = 2rL = 2(2.5 m)(L) = (5L) m2
The projected area in the axial direction is:
(APROJ )A = πr 2 = π(2.5 m)2 = 19.63 m2
The projected area for a given section of the cylinder is:
(APROJ )C = 2tL = 2(0.005 m)(L) = (0.01L) m2
The force exerted in the axial direction is:
FA = (Pi −Po )(APROJ )A = (6×105 N/m2 −1×105 N/m2 )(19.63 m2 )
FA = 9.81 MN
The force exerted in the circumferential direction is:
FC = (Pi −Po )(APROJ )C = (6×105 N/m2 −1×105 N/m2 )(0.01L m2 )
FC = (2.5)L MN
The axial stress in the cylinder is:
σA =
FA
9.81 MN
=
= 125 MPa
AA
0.0785 m2
The circumferential stress in the cylinder is:
σC =
ANS:
FC
((2.5)L) MN
=
= 250 MPa
AC
(0.01L) m2
σMAX = 250 MPa
Problem 12.59 In Example 12-9, the wall thickness of
the cylindrical vessel is determined to be 8.04 mm. What
is the resulting maximum normal stress in the vessel
wall?
Solution:
Equations (12-33) and (12-34) demonstrate that the larger normal stress
will be in the circumferential direction.
The area over which the circumferential load for a given length of the
cylinder is distributed is:
AC = 2tL = 2(0.00804 m)(L) m2 = (0.01608L) m2
The force exerted in the circumferential direction upon a given length
of the cylinder is:
FC = (Pi −Po )(AC ) = (8×105 N/m2 −1×105 N/m2 )(0.01608L) m2 = (11.256×103 L) N
The circumferential stress in the cylinder is:
σh =
=
ANS:
(Pi −Po )(R)
t
(8×105 −1×105 )(2)
0.00804
σC = 174.1 MPa
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
σ2 =
RILEY, STURGES AND MORRIS
E
210 × 109
ε
νε
330 + 0.3125 ( 619 )  ×10−6
+
=
(
)
2
1
2
2 
1 −ν
1 − 0.3125
= 121.82 × 106 N/m 2 ≅ 121.8 MPa ............................................................................. Ans.
(b)
σa =
pr p ( 255 )
=
= ( 42.5 p ) MPa
2t
2 ( 3)
σh =
pr p ( 255 )
=
= ( 85.0 p ) MPa
t
( 3)
σ 2 = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ
= ( 42.5 p ) cos 2 ( 30° ) + ( 85.0 p ) sin 2 ( 30° ) + 0 = 121.82 MPa
p = 2.29 MPa ......................................................................................................................... Ans.
10-119 A 4-in. diameter shaft is subjected to both a torque of 30 kip⋅in. and an axial tensile load of 50 kip, as
shown in Fig. P10-119. Determine the principal stresses and the maximum shearing stress at point A on the
surface of the shaft.
SOLUTION
π d 2 π ( 4)
=
= 12.566 in 2
4
4
2
A=
π d 4 π ( 4)
J=
=
= 25.13 in 4
32
32
σx =
P
50
=
= 3.979 ksi
A 12.566
τ xy =
Tc 30 ( 2 )
=
= 2.388 ksi
25.13
J
4
2
σ p1, p 2
σ +σ y
 σ −σ y 
2
= x
±  x
 + τ xy
2
 2 
2
3.979 + 0
 3.979 − 0 
=
± 
+ 2.3882

2
2


σ p1 = 1.9895 + 3.1082 = +5.0977 ksi ≅ 5.10 ksi (T) .................................................... Ans.
σ p 2 = 1.9895 − 3.1082 = −1.1187 ksi ≅ 1.119 ksi (C)
................................................. Ans.
σ p 3 = σ z = 0 ksi ........................ τ max = τ p = 3.1082 ksi ≅ 3.11 ksi .............................. Ans.
2τ xy
2 ( 2.388 )
1
1
θ p = tan −1
= tan −1
= 25.10°
2
σ x −σ y 2
3.979 − 0
....................................................... Ans.
10-120 A hollow shaft with an outside diameter of 400 mm and an inside diameter of 300 mm is subjected to both
a torque of 350 kN⋅m and an axial tensile load of 1500 kN, as shown in Fig. P10-120. Determine the
principal stresses and the maximum shearing stress at a point on the outside surface of the shaft.
SOLUTION
A=
π ( 4002 − 3002 )
4
π ( 400 − 3004 )
= 54.98 × 103 mm 2 = 54.98 × 10−3 m 2
4
J=
32
= 1718.1× 106 mm 4 = 1718.1× 10−6 m 4
514
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
10-122 A 60-mm diameter shaft must transmit a torque of unknown magnitude while it is supporting an axial
tensile load of 150 kN. Determine the maximum allowable value for the torque if the tensile principal
stress at a point on the outside surface of the shaft must not exceed 125 MPa.
SOLUTION
π d 2 π ( 60 )
A=
=
= 2827 mm 2
4
4
π d 4 π ( 60 )
J=
=
= 1.2723 × 106 mm 4
32
32
2
σx =
4
P
150 × 103
=
= 53.06 ×106 N/m 2 = 53.06 MPa (T)
−6
A 2827 ×10
2
2
σx +σ y
 σ x −σ y 
53.06 + 0
 53.06 − 0 
2
+ 
+ 
+ τ xy2 = 125 MPa
σ p1 =
 + τ xy =

2
2
2


 2 
τ xy = 94.83 MPa
τ xy = Tc J = 94.83 MPa
6
−6
τ xy J ( 94.83 ×10 )(1.2723 ×10 )
T=
=
c
0.030
= 4.022 ×103 N ⋅ m ≅ 4.02 kN ⋅ m ................................................................................ Ans.
10-123 The T-section shown in Fig. P10-123 is used as a short post to support a compressive load P = 150 kip.
The load is applied on the centerline of the stem at a distance e = 2 in. from the centroid of the cross
section. Determine the normal stresses at points C and D on section AB.
SOLUTION
A = 2 ( 20 × 6 ) = 24 in 2
xC =
3( 2 × 6) + 7 ( 2 × 6)
= 5 in.
24
 2 ( 6 )3
  6 ( 2 )3

2
2
Iy = 
+ ( 2 × 6 )( 2 )  + 
+ ( 2 × 6 )( 2 )  = 136 in 4
 12
  12

P Mc
150 (150 × 2 )( 5)
+
=−
+
= +4.779 ksi ≅ 4.78 ksi (T) ....................... Ans.
A I
24
136
150 (150 × 2 )( 3)
P Mc
σD = − −
=−
−
= −12.868 ksi ≅ 12.87 ksi (C) .................. Ans.
24
136
A I
σC = −
10-124 The cross section of the straight vertical portion of the coil-loading hook shown in Fig. P10-124a is shown
in Fig. P10-124b. The horizontal distance from the line of action of the applied load to the inside face CD
of the cross section is 600 mm. Determine the maximum tensile and compressive stresses on section
CDEF for a 40-kN load.
SOLUTION
A = (100 × 50 ) + ( 30 × 80 ) + ( 60 × 20 ) = 8600 mm 2
xC =
25 (100 × 50 ) + 90 ( 30 × 80 ) + 140 ( 60 × 20 )
= 59.19 mm
8600
100 ( 50 )3

2
Iy = 
+ (100 × 50 )( 34.19 ) 
 12

516
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
σ p 2 = 22.07 − 36.91 = −14.84 MPa ≅ 14.84 MPa (C) ................................................. Ans.
σ p 3 = σ z = 0 MPa
.................... τ max
= τ p = 36.91 MPa ≅ 36.9 MPa ........................ Ans.
2τ xy
2 ( 29.58 )
1
1
θ p = tan −1
= tan −1
= 26.64°
2
44.14 − 0
σ x −σ y 2
....................................................... Ans.
10-127 The thin-walled cylindrical pressure vessel shown in Fig. P10-127 has an inside diameter of 24 in. and a
wall thickness of 1/2 in. The vessel is subjected to an internal pressure of 250 psi. In addition, a torque of
150 kip⋅ft is applied to the vessel through rigid plates on the ends of the vessel. Determine the maximum
normal and shearing stresses at a point on the outside surface of the vessel.
SOLUTION
J=
π ( 254 − 244 )
32
σx =σa =
σ y =σh =
τ xy =
= 5777.5 in 4
pr 250 (12 )
=
= 3000 psi = 3.000 ksi
2t
2 (1 2 )
pr 250 (12 )
=
= 6000 psi = 6.000 ksi
t
(1 2 )
Tc (150 × 12 )(12.5 )
=
= 3.894 ksi
5777.5
J
2
σ p1, p 2
2
σ +σ y
 σ −σ y 
3+ 6
3 − 6 
2
= x
±  x
+
=
±
+ 3.8942
τ
xy



2
2
 2 
 2 
σ p1 = 4.5 + 4.173 = +8.673 ksi ≅ 8.67 ksi (T)
............................................................... Ans.
σ p 2 = 4.5 − 4.173 = +0.327 ksi ≅ 0.327 ksi (T)
............................................................ Ans.
σ p 3 = σ z = 0 ksi ..................................................................................................................... Ans.
σ − σ min 8.673 − 0
τ max = max
=
= 4.337 ksi ≅ 4.34 ksi ................................................. Ans.
2
2
10-128 A steel shaft is loaded and supported as shown in Fig. P10-128. If the maximum shearing stress in the shaft
must not exceed 55 MPa and the maximum tensile stress in the shaft must not exceed 83 MPa, determine
the maximum torque T that can be applied to the shaft.
SOLUTION
A100
π d 2 π (100 )
=
=
= 7.854 × 103 mm 2
4
4
J100
π d 4 π (100 )
=
=
= 9.817 × 106 mm 4
32
32
A150
π d 2 π (150 )
=
=
= 17.671× 103 mm 2
4
4
J150
π d 4 π (150 )
=
=
= 49.70 × 106 mm 4
32
32
2
2
4
4
P
550 × 103
=
A 7.854 × 10 −3
= 70.03 × 106 N/m 2 = 70.03 MPa
P
550 ×103
=
A 17.671×10 −3
= 31.12 × 106 N/m 2 = 31.12 MPa
σ 100 =
τ100 =
σ 150 =
T ( 0.050 )
Tc
=
= 5093T N/m 2
−6
J 9.817 ×10
τ150 =
518
Tc 3T ( 0.075 )
=
= 4527T N/m 2
−6
J 49.70 ×10
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
M x = 9600 (1) = 9600 lb ⋅ in.
M y = 800 ( 24 ) = 19, 200 lb ⋅ in.
P 9600
=
= 400 psi (C)
24
A
σP =
σ Mx =
σMy =
M x c 9600 ( 2 )
=
= 600 psi (T & C)
Ix
32
M yc
Iy
=
19, 200 ( 3)
72
= 800 psi (T & C)
σ A = −400 + 600 + 800 = +1000 psi = 1000 psi (T)
σ B = −400 − 600 + 800 = −200 psi = 200 psi (C)
..................................................... Ans.
......................................................... Ans.
σ C = −400 − 600 − 800 = −1800 psi = 1800 psi (C) ..................................................... Ans.
σ D = −400 + 600 − 800 = −600 psi = 600 psi (C) ......................................................... Ans.
10-132 Determine the vertical normal stresses at points A, B, C, and D of the rectangular post shown in Fig. P10132. Neglect stress concentrations.
SOLUTION
A = ( 200 ×150 ) = 30 × 103 mm 2
Ix
( 200 )(150 )
=
3
= 56.25 ×106 mm 4
12
Iy =
(150 )( 200 )
12
3
= 100.0 ×106 mm 4
M x = ( 75 × 103 ) ( 0.075 ) = 5625 N ⋅ m
M y = ( 75 × 103 ) ( 0.050 ) = 3750 N ⋅ m
P 75 ×103
=
= 2.50 ×106 N/m 2 = 2.50 MPa (C)
A 30 ×10−3
M c 5625 ( 0.075 )
= x =
= 7.50 ×106 N/m 2 = 7.50 MPa (T & C)
Ix
56.25 ×10−6
σP =
σM
x
σMy =
M yc
Iy
=
3750 ( 0.100 )
100.0 × 10
−6
= 3.75 ×106 N/m 2 = 3.75 MPa (T & C)
σ A = −2.50 + 7.50 + 3.75 = +8.75 MPa = 8.75 MPa (T) ............................................. Ans.
σ B = −2.50 + 7.50 − 3.75 = +1.25 MPa = 1.25 MPa (T)
............................................. Ans.
σ C = −2.50 − 7.50 − 3.75 = −13.75 MPa = 13.75 MPa (C) ........................................ Ans.
σ D = −2.50 − 7.50 + 3.75 = −6.25 MPa = 6.25 MPa (C)
521
............................................ Ans.
STATICS AND MECHANICS OF MATERIALS, 2nd Edition
RILEY, STURGES AND MORRIS
10-134 The output from a strain gage located on the bottom surface of the hat-section shown in Fig. P10-134 will
be used to indicate the magnitude of the load P applied to the section. The hat-section is made of
aluminum alloy (E = 73 GPa and ν = 1 3 ) and is 24 mm wide. When the maximum load P = 500 N is
applied to the section, the strain gage should read ε = +1000 µ m/m . Plot a curve showing the
combinations of thickness t and height h that will satisfy the specification. Limit the range of h from 0 to
50 mm.
SOLUTION
Ix =
A = 24t mm 2
24t 3
= 2t 3 mm 4
12
500h ( t 2 ) (×10−6 )
P Mc
500
=
+
= ( 73 ×109 )(1000 ×10−6 ) N/m 2
σ P = Eε = +
−12
3
A I
24t ×10−6
2t (×10 )
1752t 2 − 500t = 3000h
gives
h (mm)
0
10
20
30
40
t (mm)
0.285
4.283
5.996
7.311
8.420
10-135 A solid shaft 4-in. in diameter is acted on by forces P and Q, as shown in Fig. P10-135. Determine the
principal stresses and the maximum shearing stress at point A on the surface of the shaft.
SOLUTION
π ( 4)
A=
= 12.566 in 2
4
2
π ( 4)
I=
= 12.566 in 4
64
4
π ( 4)
J=
= 25.133 in 4
32
4
At Point A:
ΣFx = 0 :
Vx − 2.25 = 0
ΣFy = 0 :
N − 18 = 0
ΣFz = 0 :
Vz = 0
ΣM x = 0 :
Mx = 0
ΣM y = 0 :
T − 2.25 ( 24 ) 0
ΣM z = 0 :
M z − 2.25 ( 36 ) = 0
Vx = 2.25 kip
N = 18 kip
T = 54 kip ⋅ in.
M z = 81 kip ⋅ in.
( 81)( 2 ) = 11.459 ksi (T)
N Mc
18
+
=−
+
A
I
12.566 12.566
Tc 54 ( 2 )
τ=
=
= 4.297 ksi
J 25.133
σ =−
523