fundamentals of engineering supplied-reference handbook - Ventech!

Pu
A's
As
Mu
A's
As
UNIFIED DESIGN PROVISIONS
A's
As
Internal Forces and Strains
d' Comp.strain
dt
Strain Conditions
0.003 0.003
c c
RESISTANCE FACTORS, φ
Tension-controlled sections ( εt ≥ 0.005 ): φ = 0.9
Compression-controlled sections ( εt ≤ 0.002 ):
Members with spiral reinforcement φ = 0.70
Members with tied reinforcement φ = 0.65
Transition sections ( 0.002 < εt < 0.005 ):
Members w/ spiral reinforcement φ = 0.57 + 67εt
Members w/ tied reinforcement φ = 0.48 + 83εt
Shear and torsion φ = 0.75
Bearing on concrete φ = 0.65
c
0.003
εt ≥ 0.005 0.005> εt >0.002 εt ≤ 0.002
Tension-
controlled
section:
c ≤ 0.375 dt
Cc
Ts
Cs'
Transition
section
Balanced Strain: εt = εy
d
dt
dt
c
Net tensile strain: εt
f y
εt = εy =
Es
Compression-
controlled
section:
c ≥ 0.6 dt
0.003
ε's
116
CIVIL ENGINEERING (continued)
BEAMS − FLEXURE: φMn ≥ Mu
For all beams
Net tensile strain: a = β1 c
ε t
0. 003(
dt −c
)
=
c
=
0.
003(
β1
dt
−a
)
a
Design moment strength: φMn
where: φ = 0.9 [εt ≥ 0.005]
φ = 0.48 + 83εt [0.004 ≤ εt < 0.005]
Reinforcement limits:
AS, max εt = 0.004 @ Mn
AS
, min
=
⎪
larger ⎨
⎪
⎩
⎧ ′
3
f b d
c w
f
y
Singly-reinforced beams
As,max =
As
f y
a =
0.
85 f ′ b
200b
d
w
or
f
y
As,min limits need not be applied if
As (provided ≥ 1.33 As (required)
0.85 fc'β1b⎛ 3dt
⎞
f
⎜
⎝ 7
⎟
⎠
c
y
a a
Mn = 0.85 fc' a b (d − ) = As fy (d − )
2
2
Doubly-reinforced beams
Compression steel yields if:
0.85β fd'b ′ ⎛ 1 87, 000 ⎞
c
As − As' ≥
f
⎜
87, 000 − f
⎟
⎝ ⎠
If compression steel yields:
As,max =
a
y y
0.85 f ′β b 3
c 1 ⎛ dt
⎞
⎜ + A′
s
f ⎝
⎟
7 ⎠
( As
− As′
) f y
=
0.
85 f ' b
c
y
⎡ ⎛ a ⎞
⎤
Mn = fy ⎢(
As − As′
) ⎜d
− ⎟ + As′
( d − d'
) ⎥
⎣ ⎝ 2 ⎠
⎦
If compression steel does not yield (four steps):
1. Solve for c:
c 2 ⎛ ( 87,
000 − 0.
85 fc
')
As
' − As
f y ⎞
+ ⎜
⎟
⎜
⎟
c
⎝ 0.
85 fc
'β1b
⎠
87,
000 As
'd
'
−
= 0
0.
85 f ' β b
c
1