Civil Engineering Reference
In-Depth Information
Check t
d
(
A
o
p
o
−
ε
d
)(
ε
r
−
ε
d
)
Equation [19]
t
d
=
(
ε
−
ε
d
)(
ε
t
−
ε
d
)
793 000
3603
(0
.
0005)(0
.
00413
+
0
.
0005)
=
(0
.
00156
+
0
.
0005)(0
.
00207
+
0
.
0005)
=
96 mm
≈
96 mm assumed. OK
Calculate
α
r
,τ
t
,
T
,γ
t
and
θ
ε
t
−
ε
d
ε
−
ε
d
=
0
.
00207
+
0
.
0005
Equation [25] tan
2
α
r
=
0005
=
1
.
247
0
.
00156
+
0
.
16
◦
α
r
=
48
.
sin
α
r
=
0
.
745
cos
α
r
=
0
.
667
Equation [3]
τ
t
=
(
−
σ
d
)sin
α
r
cos
α
r
=
(13
.
51)(0
.
745)(0
.
667)
=
6
.
71MPa (973 psi)
Equation [4]
T
=
τ
t
(2
A
o
t
d
)
=
6
.
71(2)(793 000)(96)
=
1022 kN m (9040 in
.
k)
Equation[7]
γ
t
=
2(
ε
r
−
ε
d
)sin
α
r
cos
α
r
=
2(0
.
00413
+
0
.
0005)(0
.
745)(0
.
667)
=
0
.
00460
p
o
2
A
o
γ
t
=
3603
2(793 000)
0
Equation [8]
θ
=
.
00460
10
−
3
rad
=
0
.
01045 rad
/
m(0
.
265
×
/
in
.
)
2. Select
ε
d
=−
0
.
0015
ε
ds
=
2(
−
0
.
0015)
=−
0
.
0030
Assume
ε
r
=
0
.
00900 after several cycles of trial-and-error:
0
.
9
0
.
9
Equation [14]
ζ
=
√
1
ε
r
=
√
1
00900)
=
0
.
356
+
600
+
600(0
.
ε
ds
ζε
o
=
−
0
.
0030
002)
=
4
.
21
>
1 descending branch
0
.
356(
−
0
.
Equation [13
b
]
k
1
=
0
.
798 (from Table 7.1)
f
c
=
Equation [12]
σ
d
=
k
1
ζ
0
.
798(0
.
356)(
−
41
.
3)
=−
11
.
73 MPa (
−
1702 psi)
Assume
t
d
=
127 mm (5.00 in.) after two or three trials
1
2
(3987)(127)
(127)
2
738 000 mm
2
(1145 in
2
)
Equation [22]
A
o
=
975 000
−
+
=
.
Eq
.
[23]
p
o
=
p
c
−
4
t
d
=
3987
−
4(127)
=
3479 mm (137
.
0in
.
)
Solve
ε
738 000(0
.
0015)(11
.
73)
Equation [20]
ε
=
ε
d
+
(5031
f
+
790
f
p
)
