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Deflection Control: Laboratory Exercise No. 3

This document provides details for Laboratory Exercise No. 3 on deflection control of a continuous beam that is part of a roof system. The beam has different dead and live loads applied across its spans. Students are asked to: 1. Calculate the immediate and total deflections at the midspan of span BC and the free end of cantilever span DE. 2. Determine if deflections are controlled at these locations based on a 4-year time dependent factor and weighted average moment of inertia. The solution shows calculations for the moment of inertia, cracking moment, and effective moment of inertia at various cross sections to evaluate deflections against limitations.

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Ma Khristina Cassandra Pacoma
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66 views13 pages

Deflection Control: Laboratory Exercise No. 3

This document provides details for Laboratory Exercise No. 3 on deflection control of a continuous beam that is part of a roof system. The beam has different dead and live loads applied across its spans. Students are asked to: 1. Calculate the immediate and total deflections at the midspan of span BC and the free end of cantilever span DE. 2. Determine if deflections are controlled at these locations based on a 4-year time dependent factor and weighted average moment of inertia. The solution shows calculations for the moment of inertia, cracking moment, and effective moment of inertia at various cross sections to evaluate deflections against limitations.

Uploaded by

Ma Khristina Cassandra Pacoma
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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CE 463/461L

CE-4G

LABORATORY EXERCISE NO. 3

DEFLECTION CONTROL
PROBLEM: A continuous beam as shown which is a part of a roof system supporting elements
likely to be damaged by large is to carry a dead load of 5kN/m on all span and a live load of 5
kN/m on all span BC and DE. Material properties used in the design are
'
f c =30 MPa , fy=414 MPa ,n=8. Reinforcement at critical sections are shown.

REQUIRED:

1. Determine the immediate deflection


iLL due to lie load and total deflection

iTOT = iLL + at the following locations.

a.) Midspan of Beam BC


b.) Free End of Cantilever Beam DE.

2. Are deflection controlled at the above locations? (Note: Use time dependent factor

corresponding to 4 years. Use weighted average for


I eff .)

CHECKED:
GROUP 2 Engr. Edwin L. Dela Vega
FERNANDEZ, Ma. Jan Nickole B. SCORE:
HERNANDEZ, Judy Ann B. `
PACOMA, Ma. Khristina Cassandra
CANLAS, Neil Justine
Use:
As=max bd

1
A s ' = max bd
2

Use:
As=0.667 max bd

A s ' =0.334 max bd

SOLUTION:

iLL
1. Determine the immediate deflection due to lie load and total deflection
iTOT = iLL +

a.) Midspan of Beam BC

a.1) Consider DL

W DL L2 5 ( 5 2 )
Mo= = =15.625 kN .m
8 8

13.94+6.57
Mave= =10.255
2

Mm=MoMave=15.62510.255=5.370 kN . m
Mo

K=1.20.2 ( Mm )=1.20.20 ( 15.625
5.370 )

K=0. 618

Ec=4700 fc '=4700 30
Ec=25742. 960 kN . m
fcr Ig
Mcr=
yt

where:

fcr=0.7 f c ' =0.7 30=3.834 Mpa

400


3
250
b h3
Ig= =
12

400
yt = =200 mm
2

3.834 ( 1333.333 x 106 )


= 200

Mcr=25 .560 kN . m

Compute Ie at Midspan (Iem)

As=0.667 max bd

'
A s =0.334 max bd

max =0.75 b + '


0.75 { [
1 600
m 600+fy
+
bd]}
As '

0.05(3028)
1=0.85 =0 . 836
7

414
m= =16 . 235
0.85(30)

max =0.75 { 0.836


[ 600
16.235 600+ 414
+ ]}
0.334 max ( 250 ) (350)
( 250 ) (350)

max 0.334 max =0.0229

max =0 . 034

As=0 . 667 ( 0 . 034 ) ( 250 ) (350 )=1984 . 325 mm2 A s ' =0 . 334 ( 0 . 034 ) ( 250 )( 350 )=993 . 650 mm2

Summation of areas about N.A

250 x ( x2 )+( 2 n1 ) A s ( x50 )=nAs(350x)


'

125 x2 + [ 2 ( 8 )1 ] ( 1022.875 ) ( x 50 )=8 ( 2042.688 ) (350x)

x=133 . 945 mm

x
d

0
x5

b x3
Icr= + nAs
3
133.945
350

50
133.945

3
250 ( 133.945 )
+ 8(1984.325)
3
6 4
Icr=9 43.705 x 10 m m

Mcr 25.560

= =4.760
Ma 5.370

Ma ]
( Mcr 3
1 Icr } Ig

Mcr 3
Ig+
Ma
I em={

1( 4.70 3 ] 1071.199 x 10 6 m m4 } Ig
4.760 3 (1333.333 x 106 )+
{

29342. 073 x 106 m m4 > Ig(Use Ig=1333 . 333 x 106 mm4 )

Compute Ie at supports (I e B , I e C )

As=max bd

1
A s ' = max bd
2
max =0.75 b + '

0.75 { [
1 600
m 600+fy
+
As '
bd ]}
1=0 . 836

m=16 . 235

1
( 250 ) (350)
max =0.75 {
0.836 600
[
16.235 600+ 414
+
2 max
]}
( 250 ) (350)

1
max max =0.0229 max =0 . 046
2

2
As=0 . 046 ( 250 ) ( 350 )=4025 mm

1
A s ' = ( 0 . 046 )( 250 )( 350 ) =2012. 5 mm2
2

Summation of areas about N.A

250 x ( x2 )+( 2 n1 ) A s ( x50 )=nAs( 350x)


'

4025
2
125 x + [ 2 ( 8 )1 ] ( 2012.500 ) ( x 50 )=8 ( ) (350x )

x=156 . 049 mm

x 156.049
d 350

0 50
x5 156.049

3
b x3 250 ( 156.049 )
Icr= + nAs + 8( 4025)
3 3
6 4
Icr=1867 . 434 x 10 m m
Mcr 25.560

= =1.834
MaB 13.94

] Icr } Ig
Mcr
1 ( MaB 3

Mcr 3
Ig +
MaB
I e B ={

1( 1.834 3 ] (1867.434 x 106 m m4) } Ig


1.834 3 (1333.333 x 106 )+
{
6 4 6 4
I e B =1427 . 308 x 10 m m > Ig(Use I e B=Ig=1333 .333 x 10 mm )

Mcr 25.560

= =3.890
MaB 6.57

1
Mcr
( Mac ]3
Icr } Ig

Mcr 3
Ig+
Mac
I e c ={

1( 3.890 3 ] (1867.434 x 106 mm 4 ) } Ig


3.890 3 (1333.333 x 10 6)+
{
6 4 6 4
I e c =2957 . 817 x 10 mm > Ig (Use I e c =Ig=1333 .333 x 10 mm )

Use weighted average


Ie=0.7 I e M + 0.15 ( I e B + I eC )

0.7 ( 1333.333 x 106 ) + 0.15 ( 1333.333 x 10 6+ 1333.333 x 106 )

Ie=1333 .333 x 10 6 m m4

2
5000

( 5.370 x 106 )

[ ][ ]
2
i DL =K
5 ML
48 EcIe
=0.618
5
48

[ ]
i DL =0 . 252 mm
a.2) Consider DL+LL

(W DL +W ) L2 (5+5) ( 5 2 )
Mo= = =31.250 kN . m
8 8

21.64+6.90
Mave= =14.270 kN . m
2

Mm=MoMave=31.25014.270=16.980 kN . m

Mo

K=1.20.2 ( Mm )=1.20.20 ( 31.250
16.980 )

K=0. 832

Mcr 25.560

= =1 . 505
Ma 16.980

Ma ]
( Mcr 3
1 Icr } Ig

Mcr 3
Ig+
Ma
I e M ={

1( 1.505 3 ] (1071.199 x 106 m m 4)} Ig


1.505 3 (1333.333 x 106 )+
{
6 4 6 4
I e M =1964 . 778 x 10 mm > Ig (Use I e M =Ig=1333. 333 x 10 m m )

I eB , I eC
Compute Ie at supports (

Mcr 25.560
= =1 . 181
M a B 21.640

1 ( MMcra ] Icr } Ig
B
3

Mcr 3
Ig+
M aB
I e B ={
1( 1.181 3 ] (1867.434 x 10 6 m m 4 )} Ig
1.181 3 (1333.333 x 106 )+
{

I e B =987 . 656 x 106 mm4 < Ig

Mcr 25.560
= =3 .704
M aC 6.900

1 ( MMcra ] Icr } Ig
C
3

Mcr 3
Ig+
M aC
I e C ={

1( 3.704 3 ] (1867.434 x 106 m m4) } Ig


3.704 3 (1333.333 x 106 )+
{

I e C =25274 . 221 x 106 mm4 > Ig (Use I e C =Ig=1333 .333 x 106 mm4 )

Use weighted average

Ie=0.7 I e M + 0.15 ( I e B + I eC ) =0.7 ( 1333.333 x 106 ) + 0.15 ( 987.656 x 106 +1333.333 x 106 )

Ie=1281. 481 x 10 6 m m4

5000 2

( 16.980 x 106 )

i DL+ =K [ ][ ]
5 M L2
48 EcIe
=0.832
5
48 [ ]

i DL+ =1 .115 mm

i = i DL+ i DL =1.1150.252

i =0 .863 mm

Compute Long Term Effect


= sus
51 21.4
=
Assume 100% of DL is sustained 41 x1.4

Therefore, x=1.85
sus = i DL =0 . 252 mm

=
( 1+501 )=1.85( 1+50 1( 0.011 ) )
' w h ere ; =1. 85

' As ' 993.650


=1. 194 = = =0.011
bd 250(350)

=1.194 (0.252)
=0 . 301

TOT = + iLL =0.301+0.863

TOT =1 . 164 mm

b.) Free End of Cantilever Beam DE


b.1. Consider the DL


K= ( 125 )
K=2 . 4

Ec=25742. 960 MPa

Mcr=25 .560 kN . m

Icr=1867 . 434 x 106 m m4

Compute Ie at Support D

Mcr 25.560
= =1 . 136
Ma 22.50
1 ( MMcra ] Icr } Ig
B
3

Mcr 3
Ig+
M aB
I e={

1( 1.136 3 ] (1867.434 x 106 mm4 ) } Ig


1.136 3 (1333.333 x 10 6)+
{

I e=1084 . 440 x 106 mm4 < Ig

i DL =K
5
[ ][ ]
48
M L2
EcIe

3000 2

( 22.50 x 106 )

2.4 [ ]
5
48

i DL =1 . 813mm

b.2. Consider DL+LL


K= ( 125 )
K=2 . 4
Mcr 25.560

= 0 .568
Ma 45.0

1 ( MMcra ] Icr } Ig
B
3

Mcr 3
Ig+
M aB
I e={
1( 0.568 3 ] ( 1867.434 x 106 mm 4 ) } Ig
0.568 3(1333.333 x 106)+
{
use Ie=Ig= 6 4
1333 .333 x 10 m m
I e=1769 . 560 x 106 mm 4 > Ig

2
3000

( 45 x 106 )

[ ][ ] [ ]
2
5 ML 5
i DL+ =K =2.4
48 EcIe 48

i DL+ =2 . 959mm

i = i DL+ i DL =2.9501.813

i =1. 137 mm

Compute Long Term Effect


= sus

51 21.4
=
Assume 100% of DL is sustained 41 x1.4

Therefore, x=1.85
sus = i DL =1 . 813mm

1 1
=
( 1+50 )
'
=1.85
(
1+50 ( 0.023 ) ) w h ere ; =1. 85

As ' 2012.5
=0 . 860 ' = = =0.023
bd 250(350)

=0.860 (1.813)
=1 . 559
TOT = + iLL =1.559+1.137

TOT =2 . 696 mm

2. Are the deflections controlled at the above locations?

For Midspan of Beam BC


L
max= ( Table 9.5 b)
480
5000
max= =10 . 417 mm
480

( i =0 .863 mm )< ( max=10 . 417 mm) ; DEFLECTION IS CONTROLLED


( TOT =1.164 mm)<
; DEFLECTION IS CONTROLLED

For Cantilever Beam DE


L
max= ( Table 9.5 b)
480
3000
max= =6 . 250 mm
480

( i =1. 137 mm ) < ( max=6 . 250 mm) ; DEFLECTION IS CONTROLLED


( TOT =2 .696 mm)<
; DEFLECTION IS CONTROLLED

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