Civil Engineering Reference
In-Depth Information
Assume
ε
=
0
.
00253
>ε
y
after a few trials
Equation [15
b
]
f
=
413 MPa (60 000 psi)
E
ps
(
ε
dec
+
ε
)
=
214 000(0
.
005
+
0
.
00253)
=
1611 MPa (233
.
6ksi)
E
ps
(
ε
dec
+
ε
s
)
1611
Equation [17
b
]
f
p
=
m
=
1
1862
4
1
E
ps
(
ε
dec
+
ε
s
)
f
pu
m
+
1611
1
1
4
+
=
1441 MPa (209 ksi)
738 000(0
.
0015)(11
.
73)
Then
ε
=−
0
.
0015
+
790(1441)
=−
0
.
0015
+
0
.
00404
5031(413)
+
=
0
.
00254
≈
0
.
00253 assumed. OK
Solve
ε
t
738 000(203)(0
.
0015)(11
.
73)
Equation [21]
ε
t
=−
0
.
0015
+
(3479)(284
f
t
+
0)
Assume yielding
=
Equation [16
a
]
f
t
413 MPa(60 000 psi)
Then
ε
t
=−
0
.
0015
+
0
.
00646
=
0
.
00496
>ε
y
OK
Check
ε
r
Equation [24]
ε
r
=
ε
+
ε
t
−
ε
d
=
0
.
00254
+
0
.
00496
+
0
.
0015
=
0
.
00900
=
0
.
00900 assumed. OK
Check t
d
.
.
+
.
738 000
3479
(0
0015)(0
00900
0
0015)
=
Equation [19]
t
d
(0
.
00254
+
0
.
0015)(0
.
00496
+
0
.
0015)
=
127
.
8mm(5
.
03 in
.
)
≈
127 mm assumed. OK
Calculate
α
r
,τ
t
,
T
,γ
t
and
θ
0
.
00496
+
0
.
0015
Equation [25] tan
2
α
r
=
0015
=
1
.
599
0
.
00254
+
0
.
66
◦
sin
α
r
=
51
.
α
r
=
0
.
784 cos
α
r
=
0
.
620
Equation [3]
τ
t
=
(11
.
73)(0
.
784)(0
.
620)
=
5
.
70 MPa (827 psi)
Equation [4]
T
=
5
.
70(2)(738 000)(127)
=
1068 kN m (9453 in
.
k)
Equation [7]
γ
t
=
2(0
.
00900
+
0
.
0015)(0
.
784)(0
.
620)
=
0
.
01021
3479
2(738 000)
0
10
−
3
Equation [8]
θ
=
.
01021
=
0
.
02406 rad
/
m(0
.
611
×
rad
/
in
.
)
