Civil Engineering Reference
In-Depth Information
Solution:
If
ρ
f
is unknown and Example 5.8 does not suggest a failure mode, assume ductile
crushing failure.
60
4
10.19
12
Q
=
0.0221
×
×
=
0.2185
1
1518.66
0.9416
10.19
12
Q
=
+
0.2185
×
0.85
−
=
0.1833
2
2
×× ×
12
a
d
a
112.77
0.1833
0.298
=− −
×
=
f
=
3.58"
c
dc
c
=
4.21"
−
−
3
ε=
0.003
=
4.256
×
10
ductile transition
s
c
d
4.21"
10.19"
=
=
0.413
t
1
0.413
5
3
φ=
0.65
+
0.25
−
=
0.838
,,
2
ε
calculations, since
ϕ
has changed. Use Excel to repeat
the mechanical iterations.
Repeat the
Qc
a
d
s
ε
s
Q2
a/df
a
c
c/dt
Phi
0.183329
0.298446
3.581351
4.213354
0.004256
0.413479
0.837958
0.196885
0.325739
3.90887
4.598671
0.003648
0.451293
0.787298
0.209539
0.352253
4.227037
4.972985
0.003147
0.488026
0.745601
0.221244
0.377784
4.533405
5.333417
0.002732
0.523397
0.710982
0.232006
0.402217
4.826609
5.678364
0.002384
0.557249
0.681966
0.241867
0.425519
5.10623
6.007329
0.002089
0.589532
0.657399
0.250897
0.447718
5.372618
6.320727
0.001836
0.620287
0.650
It is evident that
ϕ
=
0.65 and the mode of failure is brittle crushing (compression
controlled).
87
d
d
87
4
10.19
12
ˆ
Q
=ρ
=
0.0221
×
×
=
0.4082
1
s
f
cf
M
fbd
d
d
1518.66
10.19
12
ˆ
u
cf
Q
=
φ
−Ψ−=
×× ×
Q
−
0.4082
×
0.85
−
=
0.2532
2
1
f
2
0.65
41612
2
f
3
2
a
d
a
d
(
)
−+ ×
20.208
0.4082
f
f
10.19
12
a
d
+
2.77
×
0.2532
+
0.208
×
0.4082
×
0.85
×
f
10.19
12
10.19
12
+
2.77
×
0.4082
×
0.85
×
×
0.85
−
=
0
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