Barber J.R. Intermediate Mechanics of Materials
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C Stress Concentration Factors 605
Figure C.4: Rectangular bar with a change in section through a fillet radius r, loaded
in bending;
σ
nom
=6M/hd
2
, where h is the plate thickness.
D
radius r
d
M
M
Figure C.5: Cylindrical bar with a change in section through a fillet radius r, loaded
in tension or compression;
σ
nom
=4F/
π
d
2
.
radius r
D
d
F
F
606 C Stress Concentration Factors
Figure C.6: Cylindrical bar with a change in section through a fillet radius r, loaded
in bending;
σ
nom
=32M/
π
d
3
.
radius r
M
M
D
d
Figure C.7: Cylindrical bar with a change in section through a fillet radius r, loaded
in torsion;
σ
nom
=16T /
π
d
3
.
radius r
D
d
T
T
D
Answers to Even Numbered Problems
Chapter 2
2.2:
τ
max
= 20.06 ksi,
θ
s
= 2.15
o
clockwise.
2.4:
σ
1
= 114 MPa,
σ
2
= −14 MPa,
θ
1
= 19.3
o
clockwise.
2.6:
τ
max
= 2.5 ksi.
2.8:
σ
1
= 303 MPa,
σ
2
= −8 MPa,
σ
3
= 145 MPa,
{l,m,n} = {−0.692, 0.715,0.095} .
2.10:
σ
1
= 3 ksi,
σ
2
= −6 ksi,
σ
3
= 3 ksi, {l,m,n} = {1/
√
3,1/
√
3,1/
√
3}.
2.12: U
0
= [
σ
2
1
+
σ
2
2
σ
2
3
−2
ν
(
σ
1
σ
2
+
σ
2
σ
3
+
σ
3
σ
1
)]/2E.
2.14:
σ
xx
+
σ
yy
+
σ
zz
< 0,
σ
xx
σ
yy
+
σ
yy
σ
zz
+
σ
zz
σ
xx
−
σ
2
xy
−
σ
2
yz
−
σ
2
zx
> 0,
σ
xx
σ
yy
σ
zz
−
σ
xx
σ
2
yz
−
σ
yy
σ
2
zx
−
σ
zz
σ
2
xy
+ 2
σ
xy
σ
yz
σ
zx
= 0.
2.16:
σ
1
= 330.4 MPa,
σ
2
= 89.6 MPa,
τ
max
= 130.2 MPa.
2.18:
σ
1
= 338 MPa,
σ
2
= 71 MPa,
σ
3
= 96 MPa,
τ
max
= 133.5 MPa.
2.20:
τ
= 193 MPa (Tresca) and 223 MPa (Von Mises).
2.24:
τ
E
=
q
(
σ
2
xx
+
σ
2
yy
+
σ
2
zz
−
σ
xx
σ
yy
−
σ
yy
σ
zz
−
σ
zz
σ
xx
+ 3
σ
2
xy
+ 3
σ
2
yz
+ 3
σ
2
zx
)/3
2.26: S.F. = 1.31.
2.28: M
Y
= 5 kNm (Tresca) and 4.33 kNm (Von Mises).
2.30: 451
o
.
2.34: 35 kN.
2.36: F =
π
D
3
S
t
/8L.
2.38: S = S
c
[sin(2
θ
) −
µ
−
µ
cos(2
θ
)]/2, maximum when
θ
=
1
2
tan
−1
(−1/
µ
).
2.40: 17.7 mm.
2.42:
µ
= 0.42.
2.44: 525 ∼675 MPa.
2.46: 0.53.
2.48: b = 33, S
0
= 405 MPa.
2.50:
σ
max
= 137 MPa.
2.52: S.F. = 1.29.
2.54: A machined shaft is sufficient.
2.56: S.F. = 2.0.
2.58: Probability of failure = 0.2 percent at A,C.
2.60: S.F. = 3.1.
608 D Answers to Even Numbered Problems
Chapter 3
3.2: U = 16T
2
L/
π
d
4
G.
3.4: k = k
1
k
2
/(k
1
+ k
2
).
3.6: a
3
[
σ
2
xx
+
σ
2
yy
+
σ
2
zz
−2
ν
(
σ
xx
σ
yy
+
σ
yy
σ
zz
+
σ
zz
σ
xx
)]/2E.
3.8: F[4 + 3 cos(2
α
) +
√
3sin(2
α
)]/k, u
max
/u
min
= 7.
3.10: u = 2.517FL
3
/Ec
4
.
3.12: u = 4.108FL/EA.
3.14:
θ
= 0 is stable if W < 2kL.
3.18: F = 2ka(1 −2 sin
θ
)cos
θ
sin
θ
1 +
a cos
θ
√
b
2
−a
2
sin
2
θ
.
3.20: u(L) = FL
3
/84EI.
3.22: C
i
= 2L
4
w
0
(−1)
i+1
/
π
4
EIi
4
.
3.24: u(L/4) = −2.6 ×10
−4
FL
3
/EI.
3.26: u
max
= M
0
L
2
/9
√
3EI.
3.28:
∆
AC
= 0.5FR
3
/
π
EI.
3.30: See 3.18 above.
3.32: u
B
= M
0
/kL.
3.34: The block rises by 4
ν
F
2
/
π
DE.
3.36:
K = k
1 −1 0 0
−1 3 −2 0
0 −2 5 −3
0 0 −3 7
; C =
1
k
25
12
13
12
7
12
1
4
13
12
13
12
7
12
1
4
7
12
7
12
7
12
1
4
1
4
1
4
1
4
1
4
.
3.38:
K =
44630 N/mm 11656 N/mm 875 N
11656 N/mm 11670 N/mm −3350 N
875 N −3350 N 1408246 N mm
.
3.40:
F
M
1
M
2
=
12EI
1
L
3
1
+
12EI
2
L
3
2
6EI
2
L
2
2
6EI
1
L
2
1
6EI
2
L
2
2
GK
1
L
1
+
4EI
2
L
2
0
6EI
1
L
2
1
0
GK
2
L
2
+
4EI
1
L
1
u
θ
1
θ
2
.
3.42:
α
= 22.5
o
or −67.5
o
.
3.44: W
max
= 4.4 lb.
3.46: u
A
= 7FL
3
/2EI.
3.48:
∆
BD
= 0.1366FR
3
/EI.
3.50: u
C
= M
0
L
2
/2EI.
3.52: R = 3F/2, u = 0.0169FL
3
/EI.
3.54:
u
B
=
π
FR
3
4EI
+
FR
3
GK
3
π
4
−2
.
3.56: 0.11
o
(horizontal) and 0.04
o
(vertical).
3.58: 0.37 ×10
−6
radians.
D Answers to Even Numbered Problems 609
Chapter 4
4.2:
σ
max
= 154 MPa.
4.4: Inside a diamond shaped region with corners at (0,±b/6),(±a/6,0).
4.6:
α
= −47.5
o
,
σ
max
= 58.3 MPa.
4.8:
σ
max
= 70 MPa.
4.10: u
max
y
= −3.98 mm, u
max
x
= 4.11 mm.
4.12: u
x
= 0.739 in.
4.14: I
x
= 560,000 mm
4
, I
y
= 290,000 mm
4
, I
xy
= 300,000 mm
4
.
4.16: I
x
= 22.6 ×10
6
mm
4
, I
y
= 3.84 ×10
6
mm
4
, I
xy
= −5.14 ×10
6
mm
4
.
4.18: I
x
= I
y
= 54.0a
4
, I
xy
= 1.01a
4
.
4.20: I
x
= 171,000 mm
4
, I
y
= 180,271 mm
4
, I
xy
= −65,250 mm
4
.
4.22: I
x
= 23a
3
t/16, I
y
= 37a
3
t/48, I
xy
= 7a
3
t/16.
4.24: I
1
= 315 ×10
6
mm
4
, I
2
= 101 ×10
6
mm
4
,
θ
1
= 58.3
o
clockwise from x.
4.26: I
1
= 754,000 mm
4
, I
2
= 96,000 mm
4
,
θ
1
= 32.9
o
clockwise from x.
4.28: I
1
= 35.04a
3
t, I
2
= 4.30a
3
t,
θ
1
= 32.8
o
anticlockwise from x.
4.30: 76.2 MPa, –67.8 MPa.
4.32: 0.6M
0
/a
2
t.
4.42: I
1
/I
2
< 11.5.
Chapter 5
5.2: 0.198
σ
0
bh
2
.
5.4: M
Y
= 15
π
S
Y
a
3
/8, M
P
= 28S
Y
a
3
/3, f = 1.584.
5.6: M = 9.86 kNm, R = 15 m.
5.8: M = 398 Nm.
5.10: M
P
= 934 Nm.
5.12: 0.828S
Y
a
2
t.
5.14: M
P
= 10.6 kNm, M = 9.7 kNm for second plastic zone.
5.16: M = 0.934 kNm.
5.18: M
P
= 1.66 kNm,
θ
= 0.0466
o
.
5.20:
θ
= (1 −4cot
2
α
)/2(1 + 4 cot
α
).
5.22: For −4 < tan
α
< 4, tan
α
= −cot
θ
+
√
cot
2
θ
+ 48,
M
P
=
p
6400 −(2300/9) tan
2
α
+ (25/9) tan
4
α
.
5.24:
σ
zz
= S
Y
(sgn(y) −8y/3
π
a).
5.26: R
u
= 1573 in.
5.28:
σ
zz
= −200 −4.883y ; −73.71 < y < 0
= −200 ; 0 < y < 16.29
= 200 −4.883y ; 16.29 < y < 36.29
5.30: F
P
= M
P
L/a(L −a).
5.32: d = 0.419L, w
0
= 22.8M
P
/L
2
.
5.34: F
P
= 2M
P
/L.
5.36: F =
π
D
3
S
t
/8L.
5.38: S = S
c
(sin(2
θ
) −
µ
−
µ
cos(2
θ
))/2, maximum when
θ
=
1
2
tan
−1
(−1/
µ
).
610 D Answers to Even Numbered Problems
5.40: 17.7 mm.
5.42:
µ
= 0.42.
5.44: 525 ∼675 MPa.
5.46: 0.53.
Chapter 6
6.2: 2.88 ksi.
6.4:
τ
(y) = 4V
y
(a
2
−y
2
)/5a
3
t (inclined segment),
τ
(y) = V
y
(7a
2
−4y
2
)/10a
3
t (vertical segment).
6.6:
τ
(
φ
) = 3V
y
cos
φ
/4at
0
.
6.8: 3.52 bolts per foot.
6.10: 25 ksi.
6.12: c = 5a/8.
6.14: c = 3
π
a/8.
6.16: c = 0.533a.
6.18: c = 2.03a.
6.20:
τ
max
= V
y
(4 −
ε
)/4
π
at
0
(1 −
ε
).
6.22:
τ
A
= 4V
y
/
π
at, d = 2a.
6.24:
τ
= V
y
(0.059 + 0.259 cos
θ
)/at (curved segment)
τ
= V
y
(0.06477x
2
−0.2652a
2
)/a
3
t (straight segments, x meas. from corner).
6.26:
τ
= 2T (1 + r)
2
/p
2
rt, K = p
3
r
2
t/4(1 + r)
4
.
6.28: 6.4a
3
t.
6.30:
τ
max
= T /2
π
a
2
t
0
(1 −
ε
),
θ
/L = T/2
π
Ga
3
t
0
√
1 −
ε
6.32: 67.5 MPa.
6.34: V
y
(2 −2
√
1 −
ε
2
+
ε
2
)/4
π
Ga
2
t
0
ε
√
1 −
ε
2
.
6.36:
θ
/L = 0.
6.38: c/a = −(2 −2
√
1 −
ε
2
+
ε
2
)/2
ε
.
6.40: The shear centre is at O.
6.42:
τ
max
= 0.13T /a
2
t, K = 6.4a
3
t.
6.44: K = 0.444at
3
0
.
6.46: 45.1
o
/m.
6.48:
τ
max
= 45.7 MPa,
θ
/L = 15.2
o
/m.
6.50:
τ
max
= 74 MPa,
θ
/L = 17.7
o
/m.
6.52:
θ
= 10.3
o
.
Chapter 7
7.2: u(0) = −6.17 mm,
θ
(0) = 0.88
o
.
7.4: F = −3564 N, |M|
max
= 1.02 kNm.
7.6: u(0) = 2.58 mm,
σ
max
= 103 MPa.
7.8: u
max
26% less, M
max
10% less
7.10:
σ
max
= 42.9 MPa, |F
s
|
max
= 1.367 N.
7.12: u(z) = −w
0
sin(az)/(a
4
EI + k), M(z) = w
0
a
2
EI sin(az)/(a
4
EI + k).
D Answers to Even Numbered Problems 611
7.14:
u(z) = −
F
0
4Lk
[2 − f
1
(
β
(z + L)) + f
1
(
β
(L −z))] ; −L < z < L
= −
F
0
4Lk
[ f
1
(
β
(z −L)) − f
1
(
β
(z + L))] ; z > L
M(z) = −
F
0
8
β
2
L
[ f
2
(
β
(z + L)) + f
2
(
β
(L −z))] ; −L < z < L
= −
F
0
8
β
2
L
[ f
2
(
β
(z + L)) − f
2
(
β
(z −L))] ; z > L .
7.16: M(z) = w
0
f
3
(|
β
z|)/8
β
3
.
7.18: M
max
= 0.14w
0
/
β
2
at
β
′
z = −0.699, where
β
is calculated for the segment
z > 0 and
β
′
for z < 0.
7.20: Yes.
7.22: –647 Nm at the center, 1303 Nm at the supports, 0.166 mm central deflection.
7.24: u(0) = −0.0103w
0
/k, M(L/2) = 0.0833w
0
/
β
2
.
Chapter 8
8.2: A selection of answers:
CC, R
1
= 300 mm, R
2
= 1025 mm,
σ
1
= 51 MPa,
σ
2
= −72 MPa.
DD, R
1
= ∞, R
2
= 924 mm,
σ
1
= 46 MPa,
σ
2
= 92 MPa.
8.4:
τ
max
= 20.4 ksi.
8.6:
σ
1
= −
ρ
gR
2
3t sin
2
φ
4
27
−cos
2
φ
+ cos
3
φ
.
σ
2
=
ρ
gR
2
3t sin
2
φ
−
50
27
+ 3cos
φ
+ cos
2
φ
−2cos
3
φ
.
8.8:
σ
1
=
pa(2b −a sin
φ
)
2t(b −asin
φ
)
;
σ
2
=
pa
2t
.
8.10:
σ
max
1
= −p
0
a/2t,
σ
max
2
= −p
0
a/t, |
σ
1
−
σ
2
|
max
= p
0
a/t.
8.12: t
min
= 244 mm.
8.14:
σ
1
= −
ρ
ga(5 −3 cos
φ
)
2(4 −3 cos
φ
)(1 + cos
φ
)
σ
2
= −
ρ
ga cos
φ
+
ρ
ga(5 −3 cos
φ
)
2(4 −3 cos
φ
)(1 + cos
φ
)
.
8.16: Include only internal pressure.
σ
max
= 425 kPa.
8.18: u
r
= −
ρ
s
gh
2
tan
α
[(2 −
ν
)tan
2
α
−
ν
]/2E.
8.20: u
r
= 0.405 mm at A
1
and 0.468 mm at A
2
.
8.22: At the equator, u
r
=
ρΩ
2
a
3
/E.
612 D Answers to Even Numbered Problems
Chapter 9
9.2: 19.6 ksi.
9.4:
σ
max
= 215 MPa.
9.6:
α
> F
0
p
3(1 −
ν
2
)/
π
f Et
2
.
9.8: R = 0.288a/
p
(1 −
ν
2
) (independent of p,t
0
).
9.10: M
z
(0) = p/16
β
3
b.
9.12: F
0
= p
0
a tan
α
/2, A = (18t/
β
) −(10attan
α
/(2 −
ν
)).
9.14: M
0
= E
θ
I
η
/a
2
.
9.16:
σ
max
= 22.86 MPa.
Chapter 10
10.2: 632 MPa.
10.4: −1875 psi; 0.86 ×10
−3
in.
10.6: S.F. = 7.9.
10.8: p
Y
= (5 −8
ν
)S
Y
/4(1 −2
ν
).
10.12: p = E
δ
(n
2
−1)/2n
2
a,
σ
max
= E
δ
(n
2
+ 1)/2n
2
a.
10.14: 9900 psi.
10.16: 0.090S
Y
.
10.18: p
Y
= 2S
Y
(
p
(b/a)−1) ;
σ
rr
= −2S
Y
(
p
(b/r)−1) ;
σ
θθ
= S
Y
(2−
p
(b/r)).
10.20: 11,543 rpm.
10.22: c
max
= 3.66 in.
10.24: p(r) = p
0
+
ρΩ
2
(r
2
−a
2
/2).
10.26:
σ
rr
= −S
Y
ln(r/a) +
S
Y
ln(n)
(1 −1/n
2
)
1 −
a
2
r
2
σ
θθ
= −S
Y
ln(r/a) +
S
Y
ln(n)
(1 −1/n
2
)
1 +
a
2
r
2
.
n
max
= 2.218 .
Chapter 11
11.2: 229 Nm.
11.4: 19.4 ksi.
11.8: 6.35 ksi.
11.10: 36.1 MPa (compressive), 27.0 MPa (tensile).
11.12:
σ
max
θθ
=
M
a
3
(1 −n ln(1 + 1/n))
n +
1
2
ln(1 + 1/n) −1
σ
max
rr
=
M(exp[n ln(1 + 1/n) −1] −1)
a
3
n +
1
2
ln(1 + 1/n) −1
.
11.14: 10.7 ksi.
D Answers to Even Numbered Problems 613
Chapter 12
12.2:
λ
L = 3.44.
12.4: kL
3
/3.
12.6: M
max
= 0.428w
0
L
2
.
12.8: M
max
= (F/
λ
)tanh(
λ
L).
12.10: 20.2EI/L
2
.
12.12: k = 3.382EI/L.
12.14: P
0
= 6.48 ×10
6
N, l
0
= 3.58 m,
12.16: P
0
= 165 lbs, l
0
= 0.17 in.
12.18: 438 rpm.
12.20: 753 rpm.
12.22: 66 rpm.
12.24: The approximation is almost indistinguishable from the exact result.
12.26:
P
0
=
3EI
L
2
+
3kL
4
.
12.28: 5.93 m.
12.30:
P
0
=
Ec
4
4L
2
.
12.32: P
0
= kL/2.
12.34: 4
π
2
EI/L
2
.
12.36:
π
2
EI/L
2
.
12.38:
π
2
EI/L
2
.
APPENDIX A
A.2:
{C
1
,C
2
,C
3
} =
−
1
24
,
5
24
,
23
24
.
A.4: C
1
= 0.91941, C
2
= 2.49937.
A.6: 2.592 mm.
A.8: −p
0
L/3.
A.10: P
0
= 16.47EI/L
2
.
A.12: u
2
= −FL
3
/3EI,
θ
2
= −FL
2
/2EI.
A.14:
F
i
= −
F(
∆
−a)
2
(
∆
−2a)
∆
3
; F
i+1
= −
Fa
2
(3
∆
−2a)
∆
3
;
M
i
= −
Fa(
∆
−a)
2
∆
2
; M
i+1
=
F(
∆
−a)a
2
∆
2
;u
∗
(z) = −
Fz
2
(27L −10z)
81EI
.
A.16: u
2
= −w
0
L
4
/128EI.
A.18: 5w
0
L/4.