L17 Composite Construction - Suranaree University of...
Transcript of L17 Composite Construction - Suranaree University of...
Lecture Lecture 1717 Composite ConstructionComposite Construction
� Effective Flange Width
� Nonencased Composite Sections
� Shear Transfer
� Partially Composite Beams
Timber and Steel DesignTimber and Steel Design
Mongkol JIRAVACHARADET
S U R A N A R E E INSTITUTE OF ENGINEERING
UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING
Shear connectors
Reinforced
concrete slab
Steel
stringer
Shear connectors
Reinforced
concrete slab
Steel
stringerPlaster
on lath
Shear transfer made by bond and friction
along top of W section and by the shearing
strength of the concrete along the dotted lines
Composite sectionsComprise a steel beam and a concrete slab,joined with shear connectors to achieve composite action between the two elements.
Composite sections
Using formed steel deck
Ribs
Reinforced
concrete slab
Formed steel deck
Ribs
Reinforced
concrete slab
Formed steel deck
Advantages of Composite Construction
- Increasing beam’s strength
- Less steel required
- Greater stiffness
- Smaller deflections
- Greater overload capacity
- Smaller floor depth
- Lower cost
Design Issues
1. Flexural strength of section - Complete shear connection
Failure: Yielding of steel beam in tension
or Crushing of concrete slab in compression
2. Flexural strength of section - Partial shear connection
Failure occurs in shear connection
3. Longitudinal shear failure within the slab
Failure plane develops within the slab
4. Shear strength of section
Design for shear as a plain steel beam (ignore the concrete)
5. Deflection must be computed in 2 stages
5.1 Deflection of steel beam due to dead load of wet concrete
5.2 Deflection of composite section due to live load
Effective Flange Widths, be
t = slab thickness
be = effective width of flange
1. 1/8 of beam span
2. 1/2 of beam distance
3. Beam centerline to edge of slab
be is the minimum of:
Nonencased Composite Sections
Modular ratio: n = Es/Ec
Es = 2.1 x 106 kg/cm2
′= 1.54,270c c cE w f
For wc = 1.45-2.48 ton/m3 (usually wc = 2.4 ton/m
3)
Before concrete hardens, bending stress in steel beam from dead
load of wet concrete and self-weight of beam must not exceed the
allowable bending stress of beam
Ds b
s
Mf F
S= ≤
where MD = Dead load moment and Ss = Section modulus of steel beam
Flexural Strength of Composite Section
After concrete hardens
Transformed section (concrete to steel)
be
t
be / n
t
0.9D Ls y
s trbot
M Mf F
S S= + ≤Stress in steel:
0.45Lc c
trtop
Mf f
nS′= ≤Stress in concrete:
��������� 16-1 ������������ ������������������������� ���������� AISC % ����&�����'�������& ����� ���()�������'���� *�+��'�������,���� ���&������-������ '.�� *��'�����* ��',
LL = 500 ��./�.2, ,�����6����,���� = 80 ��./�.2
,������),����'��� 10 9�. = 240 ��./�.2
f’c = 210 ��./9�.2, fc = 0.45 f’c = 94.5 ��./9�.2, n = 9, ��*<� A36
3 @ 2.5 m
= 7.5 m
8 m
10 cm concrete slab
W400x107 (A = 136.0 cm2,
d = 390 mm,bf = 300 mm,
Ix = 38,700 cm4, Sx = 1,980 cm
3)
����� �����%����?:
������������������������������������ (75% ������*��D'. 28 ��)
�), = (2.5)(240) = 600 ��./����
�� = 107 ��./����
���,�����D�,��� = 707 ��./����
MD = 0.707(8)2/8 = 5.66 ��-����
������������������������������������
6����,���� = (2.5)(80) = 200 ��./����
LL = (2.5)(500) = 1250 ��./����
���,�����D�,��� = 1450 ��./����
ML = 1.45(8)2/8 = 11.6 ��-����
��� ��!��"��#���$����"%�:
b = 2(1/8)(800) = 200 9�. (����� )
b = 2(1/2)(250) = 250 9�.
200/9 = 22.2 cm
W400x107 (A = 136.0 cm2,
Ix = 38,700 cm4, Sx = 1,980 cm
3)
10 cm
Neutral axis39 cmyb
��)# ���������!���*+,��"�����:
A = 136 + (10)(22.2) = 358 9�.2
yb = (136x19.5 + 10x22.2x44)/358
= 34.7 9�.
Itr = 38,700+136(34.7-19.5)2+(1/12)(22.2)(10)3+10(22.2)(44-34.7)2
= 91,172 9�.4
Str bot = 91,172/34.7 = 2,627 9�.3
Str top = 91,172/(49-34.7) = 6,376 9�.3
5.66(1,000)(100)286 ksc < [0.66 1,650 ksc]
1,980
Ds y
s
Mf F
S= = = = OK
���-#�����������+��*�.������!���*:
������'� �<����
�*������'� �<����
11.6(1,000)(100)286
2,627
728 ksc < [0.9 2,250 ksc]
D Ls
s trbot
y
M Mf
S S
F
= + = +
= = OK
11.6(1,000)(100)
9 6,376
20.2 ksc < [0.45 94.5 ksc]
Lc
trtop
c
Mf
nS
f
= =×
′= = OK
Shear Connectors
Weld Stud connectors
Weld Channel connectors
Weld Spiral connectors
Horizontal Shear Transfer
C
T
C
T
Allowable Horizantal Shear Load
For one connector (q), ton
CONNECTOR210 245 280
f’c , ksc
≥12 x 50 mm hooked or hooked end 2.27 2.45 2.63
16 x 62.5 mm hooked or hooked end 3.57 3.84 4.11
19 x 75 mm hooked or hooked end 5.13 5.58 5.94
22 x 87.5 mm hooked or hooked end 6.96 7.5 8.04
Channel C75 x 6.92 0.78w 0.85w 0.91w
Channel C100 x 9.36 0.83w 0.91w 0.96w
Channel C125 x 13.4 0.90w 0.96w 1.02w
w = length of channel, cm
Design of Shear Connectors
Neutral axis in slab
Neutral axis in beam
Total horizantal force
below plane between beam
and slab = As Fy
0.85f’c
Fy
Total horizantal force
above plane between beam
and slab = 0.85 Ac f’c
0.85f’c
Fy
Fy
2
ys
h
FAV =
2
85.0 cc
h
AfV
′=
N1 = Number of connectors = Vh/q
q = Strength of one connector, ton
��������� 16-2 ����� ������� ��� &6��% ������*<� A36 *��������
AISC ������������� +��'�������,���� ��+�I���()��&� *���-������ '.��
������������� �������,��������D&���D'. *�,��������D&��� ��������* ��',:
LL = 500 ��./�.2, ,������� � = 50 ��./�.2
,�����6����, = 75 ��./�.2, ,���������'� = 2,400 ��./�.3
f’c = 210 ��./9�.2, fc = 94.5 ��./9�.2, n = 9
A
A
9 m
3 @ 3 m
= 9 m
10 cm concrete slab
2 cm plasterd ceiling
on metal lath
Section A-A
����� �����%����? *� ���()�:
��������������)����#�!��
�), = (0.10)(2,400)(3.0) = 720 ��./����
���&��,�������(W400x66) = 66 ��./����
,�����D�,��� = 786 ��./����
MD = (0.786)(9)2/8 = 7.96 ��-����
������������������������������
�� � = 3(50) = 150 ��./����
6����, = 3(75) = 225 ��./����
LL = 3(500) = 1500 ��./����
,�����D�,��� = 1875 ��./����
ML = (1.875)(9)2/8 = 18.98 ��-����
%����?���D'.�& Mmax = MD + ML=7.96+18.98 = 26.94 ��-����
���+�0��1,!��!���* W400x66 (As = 84.12 ; .2, d = 400 .,
tw = 8 ., tf = 13 ., Is = 23,700 ; .4, Ss = 1,190 ; .3)
���()����D'.�& Vmax =(9/2)(0.786+1.875) = 11.97 ��
��� ��!��"��#���$����F0��:
b = (1/4)(900) = 225 9�. (����� )
b = 300 9�.
G *H��#��!���*��!�����:
Str ������� Mmax= (26.94)(100)/(0.66 x 2.5) = 1,633 9�.3
���&�����'����K ��,� ��������������L���� ����
Ss ������� MD = (7.96)(100)/(0.66 x 2.5) = 482 9�.3
��)# ���������!���*��#*�$# :
A = 84.12 + (10)(225/9) = 334 9�.2
yb = (84.12x20 + 10x25x45)/334 = 38.7 9�.
Itr = 23,700+84.12(38.7-20)2+(1/12)(25)(10)3
+ 10(25)(45-38.7)2 = 65,122 9�.4
Str bot = 65,122/38.7 = 1,683 9�.3
Str top = 65,122/(50-34.7) = 5,763 9�.3
W400x66
225/9 = 25 cm
10 cm
yb = 38.7 cm
�����������������
fs2 = fs1 + ML/Strbot = 669 + 18.98(1,000)(100)/1,683
= 1,797 ��./9�.2 < 0.9Fy = 2,250 ��./9�.2 OK
fc = ML/Strtop = 18.98(1,000)(100)/(9x5,763)
= 36.6 ��./9�.2 < fc = 94.5 ��./9�.2 OK
�����������������
fs1 = MD/Ss = 7.96(1,000)(100)/1,190
= 669 ��./9�.2 < 0.66Fy = 1,650 ��./9�.2 OK
���-#�����������+��*�.������!���*:
���6�N��?�*���*����*'�����D'.�& = 2.5 tf = 2.5(1.3)
= 3.25 9�. > 1.9 9�. OK
4
6
5 7.86 900
384 (2.1 10 )(23,700)DL
×∆ =
×
4
6
5 18.75 900
384 (2.1 10 )(65,122)LL
×∆ =
×
�����)�����������:
������'� �<����
= 1.35 9�. < [900/360 = 2.5 9�.] OK
�*������'� �<����
= 1.17 9�. < [900/360 = 2.5 9�.] OK
�&����?�K ��� ���()���������O��������� &6����<�D'.:
���1,!#���+���� 19 . ��� 7.5 ; .
1,!#���+���� 19 . ��� 7.5 ; . 41 ��� �������+��L��� 20 ���1������*!�� �����.����������,��������!���* W400x66
0.85 0.85(0.21)(225 10)201 ton
2 2
c ch
f AV
′ ×= = =
84.12 2.5105 ton
2 2
s y
h
A FV
×= = =
���+M0�� ���#�*1�������:
Control
��������D'. 16-1 �� q = 5.13 ��/�& ��
�����& ��D'.������� = N = Vh /q = 105/5.13
= 20.47 ���� �*� ������& %����?������D'.�&
20 studs 20 studs
Partially Composite Beams
When allowable moment more then the requirement no need for shear
connectors of full composite action.
So we reduce the number of shear connectors to save the money.
( )heff s tr s
h
VI I I I
V
′= + −
Effective Moment of Inertia:
2
reqd s
h h
tr s
S SV V
S S
− ′ =
−
0.25h hV qN V′ = ≥
Reduced Shear Force:
Live load deflection:
partially composite fully composite trLL LL
eff
I
I
∆ = × ∆
��������� 16-3 ��� ���& ���������� 16-2 �������'���*����'�����������%����?���D'.�& D'.�����D�� *���������� �������,��������D&������ ���
����� Seff = Str D'.������� = 1,633 9�.3 ����������D'. 16-2
2 21633 1190
105 84.8 ton1683 1190
reqd s
h h
tr s
S SV V
S S
− − ′ = = = − −
0.25 Vh = 0.25(105) = 26.3 �� < 84.8 �� OK
�����& ��D'.������� = N = Vh/q = 84.8/5.13 = 16.53 ���
� �*� ������& %����?������D'.�&
1,!#���+���� 19 . ��� 7.5 ; . -����� 34 ���
�����)�����������:
V’h = (5.13)(17) = 87.21 ��
4
( )
87.2123700 (65122 23700)
105
61,450 cm
heff s tr s
h
VI I I I
V
′= + −
= + −
=
∆LL = (65,122/61,450)(1.17)
= 1.24 ; . < [900/360 = 2.5 9�.] OK
Composite Beams with Formed Steel Deck
1. Rib height max. = 7.5 cm
2. Avg. width of concrete rib min. = 5 cm
3. Shear connector dia. max. = 19 mm
4. Concrete slab above steel deck min. = 5 cm
5. ribs perpendicular to beam neglect lower concrete rib
slab
rib
Stud dia. not greater then 19 mm4 cm or more
5 cm or more
7.5 cm or more
Reduced factor for q
0.851.0 1.0sr
r rr
Hw
h hN
− ≤
Parallel ribs:
Perpendicular ribs:
0.6 1.0 1.0sr
r r
Hw
h h
− ≤
��������� 16-4 D��9,���������D'. 16-2 % ���� ����*<� 6�K,���(����,�(���������*<�) �������������D���� 5 9�. *��),����'��� 5 9�. ���&�� wr �D���� 6 9�. �� � �����
Stud 19mm x 9cm
t = 5 cm
hr = 5 cm9 cm
6 cm 9 cm6 cm
15 cm rib spacing
����� ��������������)����#�!��
,������), *�������������D'. 16-2 = 786 ��./����
���&��,����� ����*<� = 30(3) = 90 ��./����
���,�����D�,��� = 876 ��./����
MD = (0.876)(9)2/8 = 8.87 ��-����
������������������������������
ML = 18.89 ��-���� ����������D'. 16-2
%����?���D'.�& Mmax= MD + ML= 8.87+18.98 = 27.76 ��-����
���()����D'.�& Vmax = (9/2)(0.876+1.875) = 12.38 ��
��������������DP�6*����), b = 225 9�. ���)�������� 16-2
G *H��#��!���*��!�����:
Str ������� Mmax= (27.76)(100)/(0.66�2.5) = 1,682 9�.3
Ss ������� MD = (8.87)(100)/(0.66� 2.5) = 538 9�.3
���+�0��1,!��!���* W400x94.3 (As = 120.1 ; .2, d = 386 .,
tw = 9 ., tf = 14 ., Is = 33,700 ; .4, Ss = 1,740 ; .3)
��)# ���������!���*��#*�$# :
A = 120.1 + (5)(225/9) = 245.1 9�.2
yb = (120.1x19.3+5x25x41.1)/245.1 = 30.4 9�.
Itr = 33,700+120.1(30.4-19.3)2+(1/12)(25)(5)3
+ 5(25)(41.1-30.4)2 = 63,069 9�.4
Str bot = 63,069/30.4 = 2,075 9�.3
Str top = 63,069/(43.6-30.4) = 4,778 9�.3
���������� �� *���� �����������+ �
�&����?�K ���()�����*����*'�� 19 ��. ��� 9 9�.
���6�N��?�*���*����*'�����D'.�& = 2.5 tf = 2.5(1.3)
= 3.25 9�. > 1.9 9�. OK
21682 1740
105 3.15 ton < 0.25(105) 26.3 ton2075 1740
hV−
′ = = = −
���()����D'.�& � �� Vh = 105 �� ����������D'. 16-2
��������D'. 16-1 �� q = 5.13 ��/�& ��
�����������* ���*���*����*'�� Nr = 1, Hs = 9 9�. *���� hr = 5 9�. wr = 6 9�.
0.85 6 91
5 51
−
������* �� = = 0.816
q D'.* �� *�� = 0.816(5.13) = 4.19 ��
Nreqd = 105/4.19 = 25.1 (1,! 50 #���+����)
1,!#���+���� 25 ����������*!���.���������
��� ���*����*'����������O��������� &6�������:
N = 26.3/4.24 = 6.2 1,!#���+���� 6 ����������*!���.���������