L16 Base Plate · Axial Load Small Moment Large Moment Timber and Steel Design Column Base Plate. ....
Transcript of L16 Base Plate · Axial Load Small Moment Large Moment Timber and Steel Design Column Base Plate. ....
� Axial Load
� Small Moment
� Large Moment
TimberTimber andand SteelSteel DesignDesign
Column Base PlateColumn Base Plate
������ ����..����..���� � ��������� ��������S U R A N A R E E INSTITUTE OF ENGINEERING
UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING
Visit www.tumcivil.com/DRMK/ for apps:
weld
Anchor bolts
(a) (b)
Concrete
footing
ColumnColumn BaseBase PlatesPlates
A1 = ���������� �����ก�������
= B x N
A2 = ���������ก��������� �����ก
Transfer bearing to concrete footing
B
N
A1
A2
��� A1 = A2,
��� A1 ���ก� � A2,
Allowable Bearing Pressure, Allowable Bearing Pressure, FFpp
A1 A2Fp = �� ����� ก�������!������ก���
f’ c = ก"��##$%�&�#����ก���
0.35p cF f ′=
2
1
0.35 0.7p c c
AF f f
A′ ′= ≤
Fp
A2/A101 2 3 4
0.35f’c
0.7f’c2
12
1
0.35 c
PA
A f
= ′
2
1 1
0.35 c
APf
A A′=
MinimumMinimum ThicknessThickness ofof ColumnColumn BaseBase PlatePlate
P
fp
t
B or N
1 cm
1 cm
bf
B
d N
m
0.95d
m
n n0.80bf
��$#� fp ����� �������� = P / ( B × N )
�*+��,��%-��������ก��� 1 /!.���� �&�*1�� 2$�!�� �3$�� ���� 0.80bf ��& 0.95d
2!�!��8$#$���� �&�*1��:2
2 2p
p
f nnM f n= = ��&
2
2 2p
p
f mmM f m= =
1 cm
t2!$:�#������#$��� ��ก��� 1 /!.��� t ��
3 21/ (1)( ) /( / 2) / 6
12S I c t t t= = =
�� ����$#$!�� ��� �ก# M/S ��!�� �;! �ก*��� ����$#$����!��� Fb +&;$�� �
( )2 2
2 2
/ 2 3
/ 6p p
b
f m f mMF
S t t= = =
23 p
b
f mt
F=
�< ��$���ก#� ���ก�*1��,23 p
b
f nt
F=
AISC ก"���$�� ����$#$����!������� �� Fb
= 0.75Fy:
2 p
y
ft m
F= 2 p
y
ft n
F=��&
Critical Bending DimensionsCritical Bending Dimensions
B
N
bf
d
m
0.95d
m
n n0.80bf
b
B
b N
m
0.95b
m
n n0.95b D
B
N
m
0.80D
m
n n0.80D
DesignDesign ofof ColumnColumn BaseBase PlatesPlates
1 cm
1 cm
bf
B
d N
m
0.95d
m
n n0.80bf ก��ก� �� �����ก+&%�&��#$����C$
�!���&�& N ��& B �:ก���ก+��"���� m = n
�D��&�< ����+&�ก*$�3���!��
1N A= + ∆
�!�� A1 = ���������� ������������ก��
∆ = 0.5(0.95d – 0.80bf)
�!���"���,���!��� N ;$����� ���!ก��� B �����ก��+&�"���,;$�+�ก B = A1/N
��������� 6-4 +ก� �� ���# �������"�$�������ก A36 �"���# ��� W300×94.0 (d
= 30 /!., bf= 30 /!.) ��&��"���#ก 160 �#� �����ก��ก��������# !����$ 2.5 × 2.5
�!�� = 210 กก.//!.2
����� ��������� ��!����$���กก� ������ก��� ������� A1 +&�<�� ����!�กก� ��
( )
22
21 ' 2
2
1 1 16075.8 cm
0.35 250 0.35 0.210c
PA
A f
= = =
21 '
1601,088 cm
0.7 0.7(0.210)c
PA
f= = = ������
��������������:
0.5(0.95 0.80 ) 0.5(0.95(30) 0.80(30)) 2.25 cmfd b∆ = − = − =
1 1088 2.25 35.2 cmN A= + ∆ = + = (� ! 35 $�.)
1 108831.1 cm
35
AB
N= = = (� ! 32 $�.)
��������(����ก���ก��:
2160(1000)143 kg/cm
(32)(35)p
Pf
B N= = =
×
�����*������� m �+, n :
0.95 35 0.95(30)3.25 cm
2 2
N dm
− −= = =
0.80 32 0.80(30)4.00 cm
2 2fB b
n− −
= = =
����+�ก�� ��;! �� ��C!��ก����#��!$
2' 22
1
2500.35 0.35(210) 549 kg/cm
(32)(35)P c
AF f
A= = =
' 20.7 147 kg/cmP cF f= = ������
fp
< Fp
OK
�����*����/���������������:
1432 2(4.0) 1.9 cm
2,500p
y
ft n
F= = = (� ! 2 $�.)
� !����0/+1ก PL 2 ×××× 32 ×××× 35 $�. �
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Base Plate 1 : Axial Load
Moment Resisting Column BaseMoment Resisting Column Base
MomentMoment--Resisting Column BasesResisting Column Bases
PM
ColumnFilletWeld
NutWasher
Anchor bolt
PM
Column
Angle welded orbolted to columnin shop
Anchor bolt
PM
Anchor bolt
Weld
2.5 cm
Bearing PressureBearing Pressure
B
d
N
bf
fp,min
fp,max
P
M
Pressure from axial load: fpa = P / A
Pressure from moment: fpb = M / S
where A = B x N and S = B x N2 / 6
0NB
M6NB
Pf
2min,p ≥−=
p2max,p FNB
M6NB
Pf ≤+=
Small Moment without Uplift (Full Plate Bearing)
fp1
fp2
n
Base Plate Design
1 2
2 21 2
1 2 12 3 2 3
3 6
p p
p p
n nM f n f n
f n f n
= +
= +
Moments at critical sections:
��������� �����������ก���� 1 ��. � t ���
32
1(1)( )
12/ 2 6
tI tS
C t= = =
�� ��� AISC ก!��"� Fb = 0.75Fy
P
fp
M
n
bf
B
d N
m
m
n n0.80bf
0.95d
n/3 n/3
2 21 2Another direction, 3 6
p pf m f mM = +
2
6b
M MF
S t= =
2
6 6 80.75 y
b y
M M MF t
F Ft= → = =
�������������ก���� 50 x 80 "�.(�����กก����������$)
&������'() 14-6 ��ก�##������" �$�� ��%���$� &$'�(������#$�� W350x159 �)(��*+!��ก �����ก 150 � ��,��$� &��� 15 �-$� � ".�$�/ก A36 ��, F
b=
0.75(2,500) = 1,875 กก./��.2 ��ก�* ����ก�*��� = 210 กก./��.2 ��, Fp
= 0.35(210) = 73.5 กก./��.2
� *('+� W350x159 (d = 35.6 ��., tw = 14 ��., bf = 35.2 ��., tf = 22 ��.)
�,5,$5�+��6�5& e = M /P = 15(100)/150 = 10 ��.
�����'7&�5��8�5"9:ก$�� '5�5��;��"��5��".��� 1/3 ก���������$�/ก
2pNB
M6NB
Pf ±=
2
53
8050
10156805010150
×
××±
××
=
13.285.37 ±=
=2
2
kg/cm 9.37
kg/cm 63.65 < Fb = 73.5 kg/cm2 OK
> 0 OK
9.37
65.63
80 cm
33.4 cm23.3 cm23.3 cm
49.27
�����������ก 7.5 x 50 x 80 "�.
����������������ก'()&��ก��
8 8(16,237)7.21 cm
2,500y
Mt
F= = =
�����&-�����.����/)�:
2 249.27(23.3) 65.63(23.23)16,237 kg-cm
6 3+ =
Max Moment without UpliftMax Moment without Uplift
0
fp,max
P
M
N
P
e
N/3
0NB
M6NB
Pf
2min,p =−=
6NP
M ==eP
6N
e =
Criteria for full plate bearing:6N
e6N
≤≤−
Load P within middle third of plate length N
−′=+
3
NN
2
BNFMNP ppp
T
2
BNFPT pp=+
Large Moment with UpliftLarge Moment with Uplift
B
dN
bf
TFp
P
M
bedge
N’Np
NT
2N
e6N
≤≤Eccentricity limit:
[ ΣFy = 0 ]
[ ΣMT = 0 ]
where T = Tensile force in anchor rod
NT = Distance between anchor rodand column center
Np = Bearing Length ( < N )
N’ = Distance between anchor rodand plate edge
−′=+
3
NN
2
BNFMNP ppp
T[ ΣMT = 0 ]From
Solve quadratic function to determine the bearing length Np :
where f’ = Fp B N’ / 2
3/BF
)MNP()6/BF(4ffN
p
Tp2
p
+−′±′=
Tensile force in anchor rod: P2
BNFT pp −=
&������'() 14-7 %!� ���5���%*( 14-6 �+!���5".�$�� W350x159 �$�=� ��#+!��ก P = 50
� ��,��$� &��� M = 10 �-$� � ����ก�* �*���$%�����$�/ก f’c = 240 ksc
� *('+� W350x159 (d = 35.6 ��., tw = 14 ��., bf = 35.2 ��., tf = 22 ��.)
B
dN
bf
bedge
���$���ก������$�/ก;�ก���$��#�ก�,5,��#���5)�
N > d + 2 x 8 = 35.6 + 16 = 51.6 cm
B > bf + 2 x 8 = 35.2 + 16 = 51.2 cm
������$�/ก N = 52 cm, B = 52 cm
�,5,$5�+��6�5& e = M/P = 10x100/50 = 20 cm
N/6 = 8.67 cm < e = 20 < N/2 = 26 cm
∴∴∴∴ �����&-��ก��'+�����ก ����1��.���1�
TFp
P
M
N’Np
NT
2 ���3��������4ก'��:
f’ = Fp B N’ / 2
= 84 x 52 x 48 / 2
= 104832 kg
Np = 24.0 cm
Fp = 0.35 x 240 = 84 kg/cm2
3/BF
)MNP()6/BF(4ffN
p
Tp2
p
+−′±′=
2 ���3����1��.���1�:
T = 84 x 24.0 x 52 / 2 – 50x103
= 2416 kg
Trod = T/2 = 1208 kg
2 ���3���������������ก:
�,5,5�( m = (N – 0.95d)/2
= (52 – 0.95x35.6)/2
m = 9.1 cm
Tfp = 84 ksc
P
M
24.0
m = 9.1 cm
fpm
fpm = 84 x (24.0 – 9.1) / 24.0 = 52.2 ksc
P2
BNFT pp −=
cm 1.32500
30398t =
×=
31.984
61.92.52
M22
pl×
+×
=
84
m = 9.1 cm
52.2
�����&-�������������ก:
�����������ก PL 3.1 x 52 x 52 "�.
= 3039 kg-cm
Alternative Shorter MethodAlternative Shorter Method
B
dN
bf
TFp
P
M
bedge
N’Np
NT
2
BNFRPT pp==+[ ΣFy = 0 ]
�=7*�5������5"ก���!��>����5���#ก%���5���? ="�6�5&@��������" ��� ��ก�#9:ก��#������
Np/3
R
where T = Tensile force in anchor rod
NT = Distance between anchor rodand column center
Np = Bearing Length ( < N )
N’ = Distance between anchor rodand plate edge
R = Resultant force from bearing pressure
[ ΣMT = 0 ]
−′=+
3
NNRMPN p
T
&������'() 14-7 %!� ���5���%*( 14-6 �+!���5".�$�� W350x159 �$�=� ��#+!��ก P = 50
� ��,��$� &��� M = 10 �-$� � ����ก�* �*���$%�����$�/ก f’c = 240 ksc
� *('+� W350x159 (d = 35.6 ��., tw = 14 ��., bf = 35.2 ��., tf = 22 ��.)
B
dN
bf
bedge
������$�/ก N = 52 cm, B = 52 cm
N/6 = 8.67 cm < e = 20 < N/2 = 26 cm
����5���#ก%� Np = 3 (52 – 35.6 + 2.2) / 2
= 27.9 cm
T = 84 x 27.9 x 52 / 2 – 50x103 = 10934 kg
Trod = T/2 = 5467 kg
2 ���3����1��.���1�: P2
BNFT pp −=
2 ���3���������������ก:
�,5,5�( m = (N – 0.95d)/2
= (52 – 0.95x35.6)/2
m = 9.1 cm
Tfp = 84 ksc
P
M
27.9
m = 9.1 cm
fpmfpm = 84 (27.9 – 9.1) / 27.9
= 56.6 ksc
84
m = 9.1 cm
56.6
�����&-�������������ก:
�����������ก PL 3.2 x 52 x 52 "�.
= 3100 kg-cm
31.984
61.96.56
M22
pl×
+×
=
cm 2.32500
31008t =
×=
BiBi--Axial Bending Base PlateAxial Bending Base Plate
B
dN
bf
Small moment: e < N/6 ?
My
Mx
fp,minfp,max
0NB
M6
BN
M6
BNP
f2
x2y
min,p ≥−−=
p2x
2y
max,p FNB
M6
BN
M6
BNP
f ≤++=
B
dN
bf
fp,max
fpm
Ix = NB3/12, Iy = BN3/12
x
fx
y
ypm
)2/b80.0(M)2/d95.0(M
BNP
fΙΙ
++=x
fpm fp,max
x
3
xf
6
xfM
2max,p
2pm
pl +=
yFM8
t =
BiBi--Axial Bending Base PlateAxial Bending Base Plate
Large moment :
B
dN
bf
My
MxB
dN
bf
M