aect480 lecture 13
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AECT480-Lecture 13.pdfTRANSCRIPT
Lecture 13 - Page 1 of 7
Lecture 13 – Serviceability Serviceability refers to the structural “performance” of the finished building under service loads.
• Beam deflection • Lateral drift • Vibration
We will be focusing our discussion on beam deflection. The ACI 318-02 Code dictates that the deflections be checked on the basis of effective moment of inertia, Ie, under service loads. Before we can determine the value of the effective moment of inertia, we must first have an understanding of the gross moment of inertia, Ig, and the cracked moment of inertia, Icr. Gross Moment of Inertia Ig:
The gross moment of inertia is not appropriate for reinforced concrete beams because the concrete under the neutral axis is in tension and is ineffective. Since tension is carried by the steel rebar, the beam becomes composite and therefore must be analyzed as such (See AECT 360 lecture notes). The calculated value of gross moment of inertia is higher than what is actually present.
h
b
Ig = Gross moment of Inertia
= 12
3bh
Lecture 13 - Page 2 of 7
Cracked Moment of Inertia Icr:
The cracked moment of inertia takes into consideration the composite action between the concrete and steel rebar. This assumes that the concrete in the tension zone is totally ineffective, which is overly conservative. However, the cracked moment of inertia is far closer to predicting the actual moment of inertia of a reinforced concrete beam than the gross moment of inertia.
Icr = 23
)(3
ydnAbys −+
Where: n = Modular ratio
= cconc
steel
fPSI
EE
'000,57000,000,29
=
As = Area of steel rebar in tension, in2
y = b
nAbdnA
ss ⎥
⎦
⎤⎢⎣
⎡−+ 121
nAs
N.A.
y
d
b
Lecture 13 - Page 3 of 7
Effective Moment of Inertia, Ie:
The effective moment of inertia is typically used to determine the section property of the member at a specific point along the moment diagram. In most cases, the effective moment of inertia is used to determine the actual deflection of the member when comparing to Code-dictated maximums.
Ie = gcra
crg
a
cr IIMM
IMM
≤⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−+⎟⎟
⎠
⎞⎜⎜⎝
⎛33
1
where: Mcr = moment that would initially crack the section
= t
gr
yIf
fr = modulus of rupture for the concrete = cf '5.7 yt = dist. from N.A. of uncracked cross-section to extreme tension fiber
= 2h
Ma = maximum unfactored moment at specific location along the moment diagram Ig = gross moment of inertia Icr = cracked moment of inertia
Lecture 13 - Page 4 of 7
Example 1 GIVEN: A simply-supported rectangular beam is shown below. The loads indicated are SERVICE loads. Use concrete f’c = 4000 PSI and grade 60 bars. REQUIRED:
1) Determine the gross moment of inertia Ig of the beam. 2) Determine the cracked moment of inertia Icr of the beam. 3) Determine the maximum allowable mid-span deflection of the beam
assuming ∆allow = 360L .
4) Determine the actual mid-span deflection of the beam using Ie.
d = h – conc. cover – stirrup bar dia. – ½(main tension bar dia.) = 20” – ¾” – ⅜” – ½(8/8”) = 18.375”
∆
25’-0”
Wservice = 1500 PLF
¾” concrete cover
20”
12”
Section A-A
2 - #4 hanger bars
#3 stirrup bars @ 9” o.c.
3 - #8 main bars
Lecture 13 - Page 5 of 7
Step 1 – Determine gross moment of inertia Ig:
Ig = 12
3bh
= 12
)20)("12( 3
Ig =8000 in4
Step 2 – Determine cracked moment of inertia Icr:
Icr = 23
)(3
ydnAbys −+
Where: n = Modular ratio
= PSI
PSIfPSI
EE
cconc
steel
4000000,57000,000,29
'000,57000,000,29
==
= 8.04 As = Area of steel rebar in tension, in2 = 3 bars(0.79 in2 per #8 bar) = 2.37 in2
y = b
nAbdnA
ss ⎥
⎦
⎤⎢⎣
⎡−+ 121
= "12
1)37.2)(04.8()"375.18)("12(21)37.2)(04.8( 2
2⎥⎦
⎤⎢⎣
⎡−+
inin
= 6.2”
Lecture 13 - Page 6 of 7
Icr = 23
)(3
ydnAbys −+
= 223
)"2.6"375.18)(37.2(04.83
)"2.6)("12(−+ in
Icr = 3778 in4
Step 3 - Determine the maximum allowable mid-span deflection of the beam assuming ∆allow = L/360:
∆allow = 360L
= 360
/"12)"0'25( ft−
∆allow = 0.83”
Step 4 - Determine the effective moment of inertia Ie:
fr = modulus of rupture for the concrete = cf '5.7
= PSI40005.7 = 474.3 PSI
Mcr = moment that would initially crack the section
= t
gr
yIf
=
2"20
)8000)(3.474( PSIPSI
= 379,473 Lb-In = 379.4 KIP-In = 31.6 KIP-FT
Lecture 13 - Page 7 of 7
Ma = maximum unfactored moment at specific location along the moment diagram
= 8
2wL
= 8
)"0'25)(5.1( 2−KLF
Ma = 117.2 KIP-FT
Ie = ⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−+⎟⎟
⎠
⎞⎜⎜⎝
⎛cr
a
crg
a
cr IMM
IMM
33
1
=
gIinftkftkin
ftkftk
≤⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−−
−+⎟⎟⎠
⎞⎜⎜⎝
⎛−− )3778(
2.1176.311)8000(
2.1176.31 4
34
3
Ie = 3861 in4
Step 5 – Determine actual mid-span deflection using Ie:
∆act = econc IE
wL384
5 4
= )3861)(400057000(384
)/"12"0'25(12
15005
4
4
inPSI
ftxPLF−⎟
⎠⎞
⎜⎝⎛
∆act = 0.95”
Since ∆act = 0.95” > ∆allow = 0.83” → member is NOT acceptable