Civil Engineering Reference
In-Depth Information
Table 5.1
Trial-and-error method for solution of angle
α
r
Left-hand side
of equation
α
r
cot
α
r
cot
3
α
r
cot
4
α
r
30
◦
1.7321
5.1961
9.0000
4.6668
35
◦
1.4281
2.9129
4.1599
0.9170
37.5
◦
1.3032
2.2134
2.8845
0.0113
37.6
◦
−
1.2985
2.1895
2.8431
0.0170
37.55
◦
−
1.3009
2.2014
2.8638
0.0028
37.54
◦
1.3013
2.2038
2.8678
−
0.0001
5.3.4.2 Solution
α
r
will be solved by Equation (5.84). When
ρ
=
ρ
t
=
ρ
First of all, the angle
, Equation (5.84)
becomes
σ
t
τ
t
σ
τ
t
n
) cot
4
cot
3
(1
+
ρ
α
r
+
α
r
−
cot
α
r
−
(1
+
ρ
n
)
=
0
The given property
ρ
n
and the external stress ratios
σ
/τ
t
and
σ
t
/τ
t
in this equation are:
ρ
n
=
0
.
0103(8
.
06)
=
0
.
0833
σ
τ
t
=
2
.
13
577 and
σ
t
τ
t
=
−
2
.
13
69
=
0
.
69
=−
0
.
577
3
.
3
.
0833 cot
4
577 cot
3
Therefore, 1
0.
This equation can be easily solved by a trial-and-error procedure using Table 5.1.
For
.
α
r
−
0
.
α
r
−
0
.
577 cot
α
r
−
1
.
0833
=
37.54
◦
,tan
α
r
=
α
r
=
0.7684, cot
α
r
=
1.3013, and sin
α
r
cos
α
r
=
0.4831.
The stresses are calculated as follows:
−
τ
t
−
.
3
69
.
σ
d
=
α
r
=
4831
=−
.
Equation (5
69)
7
64 MPa
α
r
cos
.
sin
0
Equation (5
.
67)
ρ
f
=
σ
+
τ
t
tan
α
r
=
2
.
13
+
3
.
69(0
.
7684)
=
4
.
97 MPa
4
.
97
f
=
0103
=
482 MPa
>
413 MPa (yield)
0
.
Equation (5
.
68)
ρ
t
f
t
=
σ
t
+
τ
t
cot
α
r
=−
2
.
13
+
3
.
69(1
.
3013)
=
2
.
67 MPa
2
.
67
f
t
=
0103
=
259 MPa
<
413 MPa (yield)
0
.
It can be seen that the longitudinal steel stress has exceeded the yield point while the
transverse steel stress is still in the elastic range.
First yield of steel
The first yield of the longitudinal steel should occur when
f
=
f
y
=
413 MPa. It will occur
=
when the applied stresses are reduced by a factor of 413/482
0.857, i.e.
σ
=
0
.
857 (2
.
13)
=
1
.
82 MPa
σ
t
=
0
.
857 (
−
2
.
13)
=−
1
.
82 MPa
τ
t
=
0
.
857 (3
.
69)
=
3
.
16 MPa
