Civil Engineering Reference
In-Depth Information
Interior support
C
v3
:= 1.15
Shear coefficient
l
n1
V =C
⋅
w
⋅
=⋅
2.5kip
Ultimate shear force
u3
v3
TFL
2
Shear forces (interior bay)
First support
C
v4
:= 1.15
Shear coefficient
l
n2
V =C
⋅
w
⋅
=⋅
3kip
Ultimate shear force
u4
v4
TFL
2
Midspan
C
v5
:= 0.15
Shear coefficient
l
n2
V =C
⋅
w
⋅
=
0.39 kip
⋅
Ultimate shear force
u5
v5
TFL
2
Second support
C
v6
:= 1.15
Shear coefficient
l
n2
V =C
⋅
w
⋅
=⋅
3kip
Ultimate shear force
u6
v6
TFL
2
6.6 STEP 4—DESIGN FRP PRIMARY REINFORCEMENT
Ty pe_of_Fiber :=
Bar_Size :=
#2
#3
#4
#5
#6
#7
#8
#9
#10
Glass
Carbon
Select the FRP reinforcement:
For the purpose of this design example,
it is assumed that GFRP bars of the same size are used everywhere in
the slab.
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