طه
TRANSCRIPT
1166
Mongkol JIRAVACHARADET
Reinforced Concrete DesignReinforced Concrete Design
S U R A N A R E E INSTITUTE OF ENGINEERING
UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING
One-Way Slabs
� Types of Slabs
� Load Paths & Flaming Concepts
� One-Way Slab
� One-Way Joist
� Slab on Grade
Types of Slab
One-way slab Two-way slabOne-way slab
Flat plate slab Flat slab Grid slab
Think we’ll need some additional framing members???
Load Path / Framing Possibilities
Ln = 4.4 m
Ln = 8.2 m
Ln = 3.2 m
Ln = 3.6 m
Plan
Framing Concepts
Let’s use a simple example
for our discussion…
Think about relating it to your
design project.
Column spacing 8 m c-c
Framing Concepts
We can first assume that
we’ll have major girders
running in one direction
in our one-way system
Framing Concepts
We can first assume that
we’ll have major girders
running in one direction
in our one-way system
If we span between girders
with our slab, then we have
a load path, but if the spans
are too long…
But we need to support the
load from these new beams,
so we will need additional
supporting members
Framing Concepts
We will need to shorten up
the span with additional beams
We again assume that we’ll
have major girders running in
one direction in our one-way
system.
Framing Concepts
Now let’s go back through with
a slightly different load path.
This time, let’s think about
shortening up the slab span by
running beams into our girders.
Our one-way slab will transfer our load to the beams.
Two Load Path Options
Framing Concepts - Considerations
For your structure:
Look for a “natural” load path
Assume walls are not there for structural support, but
consider that the may help you in construction (forming)
Identify which column lines are best suited to having
major framing members (i.e. girders)
ExampleExample
Condo Floor PlanCondo Floor Plan
t
SL
Simple supports
on two long
edges only
OneOne--Way Slab : Way Slab : ����ก���ก����� ���ก����ก���ก����� ���ก
S S
L
S S
���������� : ������ (L) > ����������������� (S)
การแอนตัวเกิดขึ้นบนดานสั้น
1.0 m
ก����ก� : ����������� ��ก������������
ก����ก�������������ก����ก�������������
L
S
����ก���� ���ก : �������
����ก���� ก������ : ������
1 m
t
�!"����#�
S
Sn
�!"�������
��������������������������������������������
������������������� 10-15 !�. �����#����ก�$� 2-3 !�.
��'����($���
��������������� ����������������
��������������� � !ก"#$���% �����
L / 20Ln
�)����������*����($��L / 24Ln
�)����������+��*���L / 28Ln
�����L / 10Ln
��!&ก�����'(��ก�ก�������!$ก���'!�����'!�� #�)*�� (��!&ก�����ก�����)
Spacing ≤≤≤≤ 3 t ≤≤≤≤ 45 cm
Main Steel (short direction):
As ≥ ∅ 6 mm
Max. Spacing ≤ 3 t ≤ 45 cm
Min. Spacing ≥ f main steel ≥ 4/3 max agg. ≥ 2.5 cm
RB24 (fy = 2,400 ksc) . . . . . . . . . . . . . . 0.0025
DB30 (fy = 3,000 ksc) . . . . . . . . . . . . . . 0.0020
DB40 (fy = 4,000 ksc) . . . . . . . . . . . . . . 0.0018
DB (fy > 4,000 ksc) . . . . . . . . . . . . . . . . 0.0018 4,0000.0014
yf
×≥
'����#��!&ก�����-�������������'����#��!&ก�����-�������������
�,���+����)-ก�+��� As �����.$���ก�$�.,���( Ag : As/Ag1 m
t
Ag = b ×××× t = 100 t
����.����� 9.1 ��ก/00��.���($�� S1 ������,01��,ก0��.#ก2'��� 300 ก.ก./�.2
1��,ก*���,+(#�����.��ก,0 50 ก.ก./�.2 ก1��(����/��.$����2�� f5c = 210 ก.ก./!�.2
/)� fy = 2,400 ก.ก./!�.2
2.7 m
S1
S1 S2
��3�� � 1) ��������������� �
+1���,0����������*����($�� tmin = L/24
�,����)(�������� fy = 2,400 ksc ��� 2,400
0.4 0.747,000
+ =
min
270t 0.74 8.3 cm
24= × =
�!��ก����������� t = 10 cm1��,ก��
WSD
= 0.1 x 2,400 = 240 kg/m2
1��,ก�,+(#��;�� = 50 kg/m2
1��,ก�� = 300 kg/m2
WSD1��,ก��� = 240 + 50 + 300 = 590 kg/m2
<����=.$�����,(��ก>��1���<(�2'����+,����+�.?�@2����� ก.10 (,�$
�. �#(����,0A��2:
�. ก)��'�����:
�. �#(����,0A���ก:
2M 590 2.7 / 9 477.9 kg-m− = × =
2M 590 2.7 /14 307.2 kg-m+ = × =
2M 590 2.7 / 24 179.2 kg-m− = × =
fs = 0.5x2,400 = 1,200 ksc fc = 0.45x210 = 94.5 ksc
134n 9
210= ≈
s
c
1 1k 0.293
f 1,200119 94.5n f
= = =++
×
c
1 1R f k j 94.5 0.293 0.902 12.49 ksc
2 2= = × × × =
j = 1 – 0.293/3 = 0.902
�1�������������=.$�2'�2ก����ก/00 :
WSD
d = 10 - 0.45 - 2 = 7.55 cm
����)�ก*������,(<(�+��#�����2'���)-ก RB9 �.�. �����#�� 2 !�.
<����=���.�<(���ก�$� :Mc = R b d
2 = 12.49 × 100 × 7.5522 = 71,196 kg-cm
= 712.0 kg-m > M .$���ก��.1� OK
�1�����������)-ก�+���.$�����ก�� : s
s
MA
f jd=
� ����.� M As ��!&ก�����
�#(����,0A��2 -477.9 5.85 RB9 @ 0.10
ก)��'����� 307.2 3.76 RB9 @ 0.16
�#(����,0A���ก -179.2 2.19 RB9 @ 0.29
RB9 : As = 0.636 cm2
Spacing = 0.636×100/As
��)-ก�+���ก,���� : As,min = 0.0025 × 100 × 10 = 2.5 cm2 USE RB9 @ 0.25 m
RB9 @ 0.25
��!&ก����������� : 2'���)-ก�+���ก,���� USE RB9 @ 0.25 m
0.7 �. 0.9 �.
2.7 �.
RB9 @ 0.25 �.
RB9 @ 0.16 �.
RB9 @ 0.25 �.RB9 @ 0.10 �.
10 !�.
WSD
����.����� 9.1 ��ก/00��.���($�� S1 ������,01��,ก0��.#ก2'��� 300 ก.ก./�.2
1��,ก*���,+(#�����.��ก,0 50 ก.ก./�.2 ก1��(����/��.$����2�� f5c = 210 ก.ก./!�.2
/)� fy = 2,400 ก.ก./!�.2
2.7 m
S1
S1 S2
��3�� � 1) ��������������� �
+1���,0����������*����($�� tmin = L/24
�,����)(�������� fy = 2,400 ksc ��� 2,400
0.4 0.747,000
+ =
min
270t 0.74 8.3 cm
24= × =
�!��ก����������� t = 10 cm1��,ก�� = 0.1 x 2,400 = 240 kg/m2
1��,ก�,+(#��;�� = 50 kg/m2
1��,ก�� = 300 kg/m2
SDM
1��,ก���),� wu = 1.4×(240+50) + 1.7×300 = 916 kg/m2
��������)-ก�+�����ก.$�+#( (����� ก.5):
SDM
<����=.$�����,(��ก>��1���<(�2'����+,����+�.?�@2����� ก.10 (,�$
�. �#(����,0A��2:
�. ก)��'�����:
�. �#(����,0A���ก:
2
uM 916 2.7 / 9 742.0 kg-m− = × =
2
uM 916 2.7 /14 477.0 kg-m+ = × =
2
uM 916 2.7 / 24 278.2 kg-m− = × =
ρmax = 0.0341
d = 10 - 0.45 - 2 = 7.55 cm
����)�ก*������,(<(�+��#�����2'���)-ก RB9 �.�. �����#�� 2 !�.
�1�����������)-ก�+���.$�����ก�� : un 2
MR
bd=φ
s c n
y c
A 0.85 f 2R1 1
bd f 0.85 f
′ρ = = − − ′
RB9 : As = 0.636 cm2
Spacing = 0.636×100/As
��)-ก�+���ก,���� : As,min = 0.0025 × 100 × 10 = 2.5 cm2 USE RB9 @ 0.25 m SDM
� ����.� Mu As ��!&ก�����
�#(����,0A��2 -742.0 4.75 RB9 @ 0.13
ก)��'����� 477.0 3.01 RB9 @ 0.21
�#(����,0A���ก -278.2 1.73 RB9 @ 0.36RB9 @ 0.25
��!&ก����������� : 2'���)-ก�+���ก,���� USE RB9 @ 0.25 m
0.7 �. 0.9 �.
2.7 �.
RB9 @ 0.25 �.
RB9 @ 0.16 �.
RB9 @ 0.25 �.RB9 @ 0.10 �.
10 !�.
Exterior span
Bottom bars
Top bars at
exterior beams
Top bars at
exterior beams
Interior span
Temperature bars
(a) Straight top and bottom bars
Exterior span
Bottom bars
Bent bar Bent bars
Interior span
Temperature bars
(b) Alternate straight and bent bars
ก���������!&ก-�ก���������!&ก-�������������������������
����������������������������������
RB9 @ 0.10 �.�. �+�����+�
RB9 @ 0.20 �+������KL
'����#��!&ก�������.�ก� (ก!��4.��, '!��4.��)
RB9 @ 0.10 m As = 6.36 cm2
3
1L
4
1L
8
1L
L1
Temp. steel
4.0 �.
.13 �.
1.0 �. 1.3 �.
[email protected] [email protected]
4.0 �.
.13 �.
1.0 �. 1.3 �.
����������������
[email protected] [email protected] ����������
�$�$�� ���!&ก������$�$�� ���!&ก�����
Floor Plan
�$�����$��������������������
1/14 1/16 1/16
1/24
1/12 1/12 1/12
����.����� 9.2 ������ก/00��.���($��+1���,01��,ก�� 500 ก.ก./�2 ก1��(����/��.$����2�� f’c = 210 ksc /)� fy = 2,400 ksc
SDM
AA
3 @ 12 m = 36 m
Section A-A
Ln = 3.7 m Ln = 3.7 m Ln = 3.7 m
0.4 + 2,400/7,000 = 0.74
1) Minimum depth :
Min. h = 0.74×370/24 = 11.4 cm
USE h = 12 cm
Slab weight = 0.12×2,400 = 288 kg/m2
Assume beam + super DL :
Service DL = 350 kg/m2
Service LL = 500 kg/m2
2) Factored Load :
wu = 1.4×350 + 1.7×500 = 1,340 kg/m2
3) Max. Moment :
Mu = 1,340 × 3.72 / 12 = 1,529 kg-m (Interior negative moment)
Max. reinforcement ratio (from Table ก.5) : ρρρρmax = 0.0341
USE RB9 with 2 cm covering: d = 12-2-0.45 = 9.55 cm
'
c n
'
y c
0.85 f 2R1 1 0.0082
f 0.85 f
ρ = − − =
2un 2 2
M 1,529 100R 18.63 kg/cm
bd 0.9 100 9.55
×= = =φ × ×
< ρρρρmax = 0.0341 OK
Required As = ρbd = 0.0082 × 100 × 9.55 = 7.83 cm2/m
RB9 : As = 0.636 cm2 → s = 0.636×100/7.83 = 8.12 cm
Select [email protected] : As = 0.636×100/8 = 7.95 cm2/m > Required As OK
Select [email protected] : As = 0.636×100/20 = 3.18 cm2/m
Temp. steel = 0.0025 × 100 × 12 = 3.00 < 7.95 cm2/m OK
����!$�����ก���������!&ก-�������������
RB9 @ 0.20 . �� ก���ก������
RB9 @ 0.08 .
�������������
RB9 @ 0.16 .
�������
RB9 @ 0.08 .
������������� + �������
12 ��.
0.95 . 1.25 .
3.7 .
0.55 . 0.95 .