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� 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
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
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.
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
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
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 !!!
��������� 7-1 ���������� ����������� ���������������������������� ����!����"�����������#���������$���������� ������������"%������ �� 1,650 ��./�-�.
6 m
6 t/m(not including beam wt.)
breqd
F
M
c
IS ==
Design with the Flexure FormulaDesign with the Flexure Formula
����� ����! ��"������� 120 �.�./�.
M = (6.12)(6)2/8 = 27.54 t-m
Sreqd = 27.54(100)/1.65 = 1,669 cm3
��������������� W350××××106 (Sx= 1,670 cm3)
�����
������������� ��"%: 0��� = (2.5)(0.12)(2,400) = 720 ��./�.
����6������� � = 80 ��./�.
�������������8� = (2.5)(500) = 1,250 ��./�.
�����������9� ��"%������� = 2,050 ��./+.
M = 2.05(6.0)2/8 = 9.23 t-m
Sreqd = 9.23(100)/1.65 = 559 cm3
��������������� W350x41.4 (Sx = 641 cm3)
��������� 7-2 0��� ����" 12 -�.������D�� ����E�������� 2.5 �. ��F� �������"%���� 6.0 �. 0�������������������8� 500 ��./��. 8�0!8�6����� ����"%����0�%�������0��� �$����������#�� �GH�������0��� ����"��F������������#�� ����"���� 2,400 ��./���. ���������������"%������ 1,650 ��./��.
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
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
��������� 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����$� �:
��8������ �:
�����0� W300 ×××× 94 �23��������4�+5674
3010
2 2(1.5)
f
f
b
t= =
544 10.9
2,500= OK <
3030
1.0w
d
t= =
5,355107.1
2,500= OK <
Fb
b/2tf
Noncompact
Partially CompactCompact
0.60Fy
Major axis:0.66Fy
Minor axis:0.75Fy
yF
795
yF
544
Noncompact Section: flange and web not compact
Fb = 0.60Fy
0.79 0.000242
f
b y y
f
bF F F
t
= −
Partially Compact:
Lateral Buckling of BeamsLateral Buckling of Beams
Slender beam buckle sideways even under perfectly vertical loads.
Before After
Movement of cross-section
at midspan
Lateral Restraint of BeamLateral Restraint of Beam
To force the beam to buckle in a complicated shape (higher mode)
L
Buckling occurs
over whole length
Without restraints
L1 Buckling occurs
between restraints
With lateral restraints
L2
L3
yfy
f
cFAdF
bL
)/(
000,400,1or
636 ofsmaller =
AISCAISC Allowable Bending StressAllowable Bending Stress
cLL ≤ yb FF 66.0=
D��8�0!8�6���� ������F�������� ��
��8�+� 1: ��"%�"����� ���0E ��� Lb ≤ Lc
��8�+� 2: ��"%�"�����K�� ���0E ��� Lb ≤ Lc
��8�+� 3: ��"%�"����� ���0E ����K�� ���0E D�� Lb > Lc
��8�+� 1: ��"%�"����� ���0E ��� Lb ≤ Lc
��8�+� 2: ��"%�"�����K�� ���0E ��� Lb ≤ Lc
0.79 0.000242
f
b y y
f
bF F F
t
= −
��8�+� 3: ��"%�"����� ���0E ����K�� ���0E D�� Lb > Lc
��D������#���$����������-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%�����#�����������������
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
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
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����
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
Practical Use:
Lc < Lunbraced < Lu give Fb = 0.60 Fy
Allowable Bending StressAllowable Bending Stress
�������� .1 � � �������� �������������� W
Lunbraced
M
Lc
Mc
Mc-u
Lu
��������� 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
������� ������� 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/�������+�����+=+��/:������;�������=���
=+��/:������;�������=���=+��/:������;�������=���:: ( )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 ��������