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Floor Framing System Bending Stresses Compact Sections Lateral Support of Beams Timber and Steel Design Timber and Steel Design Lecture Lecture 7 7 - - Introduction to B E A M Mongkol JIRAVACHARADET S U R A N A R E E INSTITUTE OF ENGINEERING UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

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Page 1: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

� Floor Framing System

� Bending Stresses

� Compact Sections

� Lateral Support of Beams

Timber and Steel DesignTimber and Steel Design

Lecture Lecture 77 -- Introduction to

B E A M

Mongkol JIRAVACHARADET

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

Page 2: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

Columns

Stair

Stringer

Floor beam or Girder

Joist

Spandrel

Floor Framing SystemFloor Framing System

Layout of Beams and Columns

- Occupancy requirements

- Commonly used beam size

- Ceiling and services requirements

To transfer vertical loads on the floor to the beams and columns in a

most efficient and economical way

Page 3: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

Loading on BeamsLoading on Beams

Tributary area = Area for which the beam is supporting

One-way Floor System (m =S/L < 0.5)

C1

B1

B2

B3

SL

Floor load w kg/m2

Tributary area

wS kg/m

B1 Loading

Load from B1

B3 Loading

B1 = Secondary Beam

B3 = Primary Beam

If span of B3 is too large, more secondary beam may be used.

Page 4: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

Tributary area = 0.5SL sq.m

Load on beam = 0.5wSL kg/m

Floor load = w kg/sq.m

Precast Concrete Slab

C1B2

B3 S

L

Page 5: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

45o 45o

45o 45o

A

D C

B

Tributary

areaS

L

Span ratio m = S/L

Short span (BC):

Floor load = w kg/sq.m

Tributary area = S2/4 sq.m

Load on beam = wS/4 wS/3 kg/m

Long span (AB):

Floor load = w kg/sq.m

Tributary area = SL/2 - S2/4 = sq.m

Load on beam kg/m

−m

mS 2

4

2

−2

3

3

2mwS

Two-way Slab

Page 6: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

b

Mc Mf

I S= =

Fy

Fy

Fy

Fy

Fy

Fy

Fy

Fy

My = Yield Moment

Mp Mp

Plastic Hinge

Bending StressesBending Stresses

Mp = Plastic Moment

M

fb

fb

fb = Bending Stress

S = I/c = Section Modulus

My = SFy Mp = ZFy

Z = Plastic section modulus

Reserved Strength !!!

Page 7: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

��������� 7-1 ���������� ����������� ���������������������������� ����!����"�����������#���������$���������� ������������"%������ �� 1,650 ��./�-�.

6 m

6 t/m(not including beam wt.)

breqd

F

M

c

IS ==

Design with the Flexure FormulaDesign with the Flexure Formula

����� ����! ��"������� 120 �.�./�.

M = (6.12)(6)2/8 = 27.54 t-m

Sreqd = 27.54(100)/1.65 = 1,669 cm3

��������������� W350××××106 (Sx= 1,670 cm3)

Page 8: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

�����

������������� ��"%: 0��� = (2.5)(0.12)(2,400) = 720 ��./�.

����6������� � = 80 ��./�.

�������������8� = (2.5)(500) = 1,250 ��./�.

�����������9� ��"%������� = 2,050 ��./+.

M = 2.05(6.0)2/8 = 9.23 t-m

Sreqd = 9.23(100)/1.65 = 559 cm3

��������������� W350x41.4 (Sx = 641 cm3)

��������� 7-2 0��� ����" 12 -�.������D�� ����E�������� 2.5 �. ��F� �������"%���� 6.0 �. 0�������������������8� 500 ��./��. 8�0!8�6����� ����"%����0�%�������0��� �$����������#�� �GH�������0��� ����"��F������������#�� ����"���� 2,400 ��./���. ���������������"%������ 1,650 ��./��.

Page 9: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

Local Buckling of Compression FlangeLocal Buckling of Compression Flange

If the plate is "slender" it may buckle locally before Fy is reached.

C

T

C

T

Buckling of compression flange

The factor b/t is a measure of slenderness.

b

b

b

Page 10: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

Laterally supported compact section: Fb = 0.66Fy

Laterally supported noncompact section: Fb < 0.66Fy

Compact SectionCompact Section

��/��� 7.1 #"�8����#��������� �������� ������������ I� ����������

N.A.d/tw��� ����������8������

1/2 bf /tf�$�#��������$����� ����������������������

noncompactcompact

#"�8����������� b / t�������b / t��!�#���!������

544 / yF 795/ yF

5,355 / yF

Page 11: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

��������� 7-3 8���8���������� W300×94 ��F������ ���0E ����K�� ����� Fy

= 2,500 �.�./-�.2

����� ������� W300×94 (d = 30 -�. tw

= 10 �.�. bf= 30 -�. t

f= 15 �.�. S

x= 1,360 -�.3)

��8����$� �:

��8������ �:

�����0� W300 ×××× 94 �23��������4�+5674

3010

2 2(1.5)

f

f

b

t= =

544 10.9

2,500= OK <

3030

1.0w

d

t= =

5,355107.1

2,500= OK <

Page 12: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

Fb

b/2tf

Noncompact

Partially CompactCompact

0.60Fy

Major axis:0.66Fy

Minor axis:0.75Fy

yF

795

yF

544

Noncompact Section: flange and web not compact

Fb = 0.60Fy

0.79 0.000242

f

b y y

f

bF F F

t

= −

Partially Compact:

Page 13: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

Lateral Buckling of BeamsLateral Buckling of Beams

Slender beam buckle sideways even under perfectly vertical loads.

Before After

Movement of cross-section

at midspan

Page 14: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

Lateral Restraint of BeamLateral Restraint of Beam

To force the beam to buckle in a complicated shape (higher mode)

L

Buckling occurs

over whole length

Without restraints

L1 Buckling occurs

between restraints

With lateral restraints

L2

L3

yfy

f

cFAdF

bL

)/(

000,400,1or

636 ofsmaller =

Page 15: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

AISCAISC Allowable Bending StressAllowable Bending Stress

cLL ≤ yb FF 66.0=

D��8�0!8�6���� ������F�������� ��

��8�+� 1: ��"%�"����� ���0E ��� Lb ≤ Lc

��8�+� 2: ��"%�"�����K�� ���0E ��� Lb ≤ Lc

��8�+� 3: ��"%�"����� ���0E ����K�� ���0E D�� Lb > Lc

��8�+� 1: ��"%�"����� ���0E ��� Lb ≤ Lc

��8�+� 2: ��"%�"�����K�� ���0E ��� Lb ≤ Lc

0.79 0.000242

f

b y y

f

bF F F

t

= −

Page 16: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

��8�+� 3: ��"%�"����� ���0E ����K�� ���0E D�� Lb > Lc

��D������#���$����������-L%�8�������F�����!��!���������!��!��!�

������������"%�������" �K����!� 0.60Fy

������������"%������8������� ��"%�����#������D����������������!�

D��������� �������#���$�����������"%��������� �� L/rT ���%�

(d/2 - tf)/3

Neutral Axis

Compression

tf

d/2 - tfd/2

L = Lb = ����K�������������#��

rT

= ��M�"K8����%�#���$�������������0����"%��L%�����#�����������������

Page 17: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

Buckling EquationsBuckling Equations ���������"%������8�#L������������ ������� L/rT D��

3 47,173 10 3,585 10b b

y T y

C CL

F r F

× ×≤ ≤���%�

4��:������;5<<�������=��� ( )2

4

20.60

3 10,760 10

y T

b y y

b

F L rF F F

C

= − ≤

×

���%� 43,585 10 b

T y

CL

r F

×≥

4��:������;5<<����=��� ( )

4

2

1,195 100.60b

b y

T

CF F

L r

×= ≤

Torsion EquationsTorsion Equations843,600

0.60bb y

f

CF F

Ld A= ≤

���%� Cb ���� �6���� �8�9�#��D����I�"%�"����D���������#��

Cb �" ����������� 1.0 NL� 2.3 �����0�%� ��������8��� ��"%�9�%�K�� �� Cb = 1.0

Page 18: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

Fb

0.6Fy

× 37173 10 b

y

C

F

× 43585 10 b

y

C

F

L/rT

( ) = −

×

2

4

2

3 10760 10

y T

b y

b

F L rF F

C

( )×

=4

2

1195 10 bb

T

CF

L r

Fb

0.6Fy

Ld/Af1400000 /b yC F

=843600

/b

b

f

CF

Ld ATorsion EquationsTorsion Equations

Buckling EquationsBuckling Equations

Page 19: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

Lu = Maximum unbraced length beyond which Fb < 0.60Fy

Unbraced Unbraced Length Length LLuu

Fb

0.66Fy

0.60Fy

Lc

Lu

Buckling equation

Torsional equation

L

Lu = ��"%�����#�� 37,173 10 b

T

y

Cr

F

× 1,400,000

( / )

b

y f

C

F d A����

Page 20: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

cLL ≤ yb FF 66.0=

uc LLL ≤< yb FF 60.0=

uLL > formula) (use 60.0 yb FF <

Summary

Fb

0.66Fy

0.60Fy

Lc

Lu

Buckling equation

Torsional equation

L

Page 21: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

Practical Use:

Lc < Lunbraced < Lu give Fb = 0.60 Fy

Allowable Bending StressAllowable Bending Stress

�������� .1 � � �������� �������������� W

Lunbraced

M

Lc

Mc

Mc-u

Lu

Page 22: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

��������� 7-6 8�0!8�6������������"%������#�� ������ W300×94 ��F� �������"%������ � 3.5, 6.0 ��� 9.0 ��� K���"�����������#�� ����� F

y= 2,500 �.�./-�.2

����� ����� W300×94 (d = 30 -�. tw

= 10 �.�. bf= 30 -�. t

f= 15 �.�. I

y= 6,750 -�.3)

8��� .1 ����� W300×94: LLcc = 3.82 m= 3.82 m LLuu = 8.40 m= 8.40 m

L

Fb

Lc = 3.82 m

0.66Fy

Lu = 8.40 m

0.60FyFb

= 0.66Fy

= 0.66(2,500) = 1,650 �.�./-�.2

4��+��� LL = 3.5 m= 3.5 m < Lc= 3.82 m

Fb

= 0.60Fy

= 0.60(2,500) = 1,500 �.�./-�.2

4��+��� Lc= 3.82 m < LL = 6.0 m= 6.0 m < L

u= 8.40 m

4��+��� Lu= 8.40 m < LL = 9.0 m= 9.0 m

Page 23: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

������� ������� L/rT

= 900/8.26 = 109

4��+��� Lu= 8.40 m < LL = 9.0 m= 9.0 m

0!8�60����"%������"���������:

2

130 1.5 4.5 1.0

6

49.5 cm

f wA A+ = × + ×

=

N.A.

30 cm1.5 cm

1 cm

d = 30 cm

(30 - 2(1.5))/2

= 13.5 cm

13.5/3

= 4.5 cm

����6 Iy #����������" = 0.5 Iy #�����������

0.5 6,750/ 8.26 cm

49.5Tr I A

×= = =

3 47,173 10 (1.0) / 2,500 53.6 / 3,585 10 (1.0) / 2,500 119.7TL r× = ≤ ≤ × =

�����0�4����@�����5/�������+�����+=+��/:������;�������=��������0�4����@�����5/�������+�����+=+��/:������;�������=���

Page 24: Timber and Steel Design - Suranaree University of Technologyeng.sut.ac.th/ce/oldce/CourseOnline/430432/L07 Beam1.pdf · Timber and Steel Design Lecture 7 - Introduction to B E A M

=+��/:������;�������=���=+��/:������;�������=���:: ( )24

20.60

3 10,760 10

y T

b y y

b

F L rF F F

C

= − ≤

×

( )24

2,500 10922,500

3 10,760 10 (1.0)bF

= −

×

= 976.6 �.�./-�.2 < [0.60Fy = 1,500 �.�./-�.2] OK

�����5/�������+�����+=+��/��/<�������5/�������+�����+=+��/��/<��:: 843,6000.60b

b y

f

CF F

Ld A= ≤

843,600(1.0)

900 30 (30 1.5)bF =

× ×

= 1,406 �.�./-�.2 < [0.60Fy = 1,500 �.�./-�.2] OK

�����0������0� FFbb = 1,406= 1,406 ��..��././A+A+..22 ��������