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1 1 6 6 Mongkol JIRAVACHARADET Reinforced Concrete Design Reinforced 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 slab One-way slab Flat plate slab Flat slab Grid slab

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Page 1: طه

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

Page 2: طه

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

Page 3: طه

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…

Page 4: طه

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.

Page 5: طه

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)

Page 6: طه

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

Page 7: طه

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

Page 8: طه

��!&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

Page 9: طه

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

Page 10: طه

��!&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

Page 11: طه

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 !�.

Page 12: طه

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

Page 13: طه

3

1L

4

1L

8

1L

L1

Temp. steel

4.0 �.

.13 �.

1.0 �. 1.3 �.

[email protected]

[email protected] [email protected]

4.0 �.

.13 �.

1.0 �. 1.3 �.

[email protected]

����������������

[email protected] [email protected] ����������

�$�$�� ���!&ก������$�$�� ���!&ก�����

Floor Plan

�$�����$��������������������

1/14 1/16 1/16

1/24

1/12 1/12 1/12

Page 14: طه

����.����� 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

Page 15: طه

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 .