Crane Girder Design 2dq4g
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CRANE GIRDER DESIGN
Pv1
MEMBER :
Pv2
3.90 0.00
0.00
Max Wheel Load
=
No of Wheels
=
Distance between wheels (du1) Fh1
CG1
4 wheels
=
3.90 m
b
=
0.00 m
c
=
0.00 m
Length of Crane Bridge
=
9.40 m
Length of Crane Girder (L)
=
9.00 m
Fh2
a
155 kN
L
y x z
1. DETERMINE LOAD a.) VERTICAL LOAD Pv1
=
155 kN
without impact
Pv1a
=
155 x 1.25 =
193.75 kN
Pv2
=
155 kN
without impact
Pv2a
=
155 x 1.25 =
193.75 kN
with impact with impact
Pv3
=
0 kN
without impact
Pv3a
=
0 x 1.25 =
0.0 kN
with impact
Pv4
=
0 kN
without impact
Pv4a
=
0 x 1.25 =
0 kN
with impact
b.) HORIZONTAL LOAD Fh1
=
0.20
x
[ Pv1 ] =
31.00 kN
Fh2
=
0.20
x
[ Pv2 ] =
31.00 kN
Fh3
=
0.00
x
[ Pv3 - (Bridge weight) / 8 ] =
0.00 kN
Fh4
=
0.00
x
[ Pv4 - (Bridge weight) / 8 ] =
0.00 kN
c.) DEAD LOAD - SELFWEIGHT W
=
2.20 kN/m
d.) LOAD FACTOR f
Dead Load
=
Vertical and Horizontal Crane Load considered separately f = Vertical and Horizontal Crane Load considered acting together f = 2. MAXIMUM MOMENT AND SHEAR A. MAXIMUM MOMENT DUE TO WHEEL LOAD (W/O IMPACT)
Pv1
Pv2
center
y
1.575
3.90
x
Mv
=
427.98 kN-m
M1
=
0 kN-m
M1
=
297.01 kN-m
M2
=
427.98 kN-m
M4
=
0.00 kN-m
M5
=
0 kN-m
M6
=
0 kN-m
3.525
L = 9.0 m
Kn-m 427.98 kN-m MOMENT DIAGRAM
(MAX. )
B. MAXIMUM SHEAR DUE TO WHEEL LOAD (W/O IMPACT) Pv1
y x
Qv
=
242.66 kN
Q1
=
242.66 kN
Q2
=
87.66 kN
Q3
=
-67.34 kN
Q4
=
-67.34 kN
Q5
=
-67.34 kN
(MAX. )
Pv2
3.90
5.10 L = 9.0 m
0.00 242.66 kN
88 kN -67.3 kN -67.34 kN
-67.34 kN
SHEAR DIAGRAM C. MAXIMUM MOMENT DUE TO LATERAL WHEEL LOAD Fh1
Mh
=
29.06 kN-m
Np = Mh/1.0
=
29.06 kN
(MAX. )
M4
=
0.00 kN-m
M5
=
0.00 kN-m
M6
=
0 kN-m
Fh
=
31.00 kN-m
(MAX. )
Mv2
=
535 kN-m
(MAX. )
M1
=
0 kN-m
M1
=
371.27 kN-m
M2
=
535.00 kN-m
M4
=
535.00 kN-m
M5
=
535.00 kN-m
z x 4.50
4.50 L = 3.75 m Fh1
59.41 Kn-m Le = 3.75
29.06 kN-m MOMENT DIAGRAM
D. MAXIMUM SHEAR DUE TO LATERAL WHEEL LOAD Fh1
z x
L = 3.75 m Fh1
48.53 kN
18 kN 0.0 kN Le = 3.75 SHEAR DIAGRAM
E. MAXIMUM MOMENT DUE TO WHEEL LOAD (W/ IMPACT)
Pv1a
Pv2a
center
y
1.575
3.90
3.525
L = 9.0 m
x
371.27 Kn-m 535.00 kN-m MOMENT DIAGRAM
M6
0 kN m
F. MAXIMUM SHEAR DUE TO WHEEL LOAD (W/ IMPACT) Pv1a
0.01
y
Qv2
=
303.33 kN
Q1
=
303.33 kN
Q2
=
109.58 kN
Q3
=
-84.17 kN
Q4
=
-84.17 kN
Q5
=
-84.17 kN
Pv2a
8.99 L = 9.0 m
x
0.00 303.33 kN
110 kN -84.17 kN
-84.17 kN
-84.17 kN
SHEAR DIAGRAM G. MAXIMUM MOMENT AND SHEAR DUE TO SELFWEIGHT Q
WL
=
=
M
=
=
2
2.20
x
9.00
2
9.90 kN WL^2
=
=
8
2.20
x
9.00
^2
8
22.28 kN-m
H. FACTORED MOMENT AND SHEAR a.) Vertical crane load with impact and no horizontal crane load Maximum moment Mx
=
1.4 x M + 1.6 x Mv2
=
( 1.4 x 22.28 ) + ( 1.6 x 535.00) 887.19 kN-m
=
Maximum shear Fa
=
1.4 x Q + 1.6 x Qv2
=
( 1.4 x 9.90 ) + ( 1.6 x 303.33) 499.19 kN
=
b.) Vertical crane load with no impact and horizontal crane load Vertical maximum moment Mx
Maximum axial load
=
1.4 x M + 1.6 x Mv1
=
( 1.4 x 22.28 ) + ( 1.6 x 427.98) 715.96 kN-m
=
N
=
1.6 x Np
=
1.6 x 29.06
=
46.50 kN
Horizontal maximum moment My
=
1.6 x Mh
=
1.6 x 29.06 46.50
=
kN-m
c.) Vertical crane load with impact and horizontal crane load acting together Vertical maximum moment Mx
Maximum axial load
=
1.4 x M + 1.4 x Mv2
=
( 1.4 x 22.28 ) + ( 1.4 x 535.00) 780.19 kN-m
=
Horizontal maximum moment My
=
1.4 x Mhp
=
1.4 x 29.06 40.69
=
kN-m
N
=
1.4 x Np
=
1.4 x 29.06
=
40.69 kN
(MAX. )
3. DESIGN OF CRANE GIRDER a.) SECTION MEMBER PROPERTIES : ( See Section Properties Calculation ) FIG. 1
UB914x305x224
B T b
D
d
t
D
=
91.04 cm
d
=
86.26 cm
B
=
30.41 cm
A
=
282.51 cm^2
tw
=
1.59 cm
Af
=
72.68 cm^2
Tf
=
2.39 cm
x
=
38.09 cm
Ix
=
370702.52 cm^4
u
=
Iy
=
11230.90 cm^4
ry
=
Sx
=
9400.79 cm³
py
=
Zy
=
738.63 cm³
pyw
=
2 265.0 N/mm
Zfy
=
368.37 cm³
Le
=
b.) BUCKLING RESISTANCE MOMENT FOR THE XX-AXIS
( Table 11 ) b T b T d t
=
=
=
=
14.4 2.39 14.4 2.39 86.3 1.59
275 / py
=
6.03
< 9.0
= 9.0
=
6.03
<28
= 28.0
=
54.25
<80
= 80.0
275 265.0 275 265.0 275 265.0
=
1.02
=
9.17
=
28.52
=
81.5
SECTION IS PLASTIC SLENDERNESS CHECK
=
=
( 4.3.6.7 ) Le
=
ry
375.00
59.43
=
6.31
( 200.00 = back truss lattice pitch )
x
=
0.5
For symmetrical H shape ( 4.3.6.7 )
59.43 38.09
v
=
0.968
LT
=
u.v. w
=
52
=
1.6
1.6
From Table 19 BS 5950 Part I 2000
w =
1.0
( Section is plastic ) from 4.3.6.9
pb
=
200.2
N/mm2
From Table 17 BS 5950 Part I 2000
pc
=
230
N/mm2
From Table 24 c BS 5950 Part I 2000
Mb
=
Sx.pb
=
=
1882.04 kn-m
9401
x
200.20 /10^3
c.) MOMENT CAPACITY FOR THE SECTION FOR YY-AXIS Mzy
= =
Zfy.py
=
368.37
x
265
Mb
=
/10^3
97.62 kn-m
4. CHECK IN BENDING a.) Vertical moment with impact and no horizontal moment Mx
=
887.19
kn-m
<
b.) Vertical moment with no impact and horizontal moment N AfPc 46500.00 1671637.7
+ +
Mx Mb 715.96 1882.04 0.89
+ + <
My Mzy 46.50 97.62 1.0
0.9 6.31 cm 2 265.0 N/mm
<
1.0
<
1.0
c.) Vertical moment with impact and horizontal moment < N Mx My + + APc Mb Mzy < 40687.50 780.19 40.69 + + 1671637.7 1882.04 97.62 < 1.0 0.86
O.K. 1.0 1.0 O.K.
1882.04
kn-m
O.K.
375 cm
5. CHECK FOR DEFLECTION a.) VERTICAL L
9000
=
1000
1000
=
9.00
mm
=
7.50
mm
a.) HORIZONTAL L
=
500
3750 500
Consider 2 wheels inside the 9.0m beam span , v
4x[155x2550x(9000^2-2550^2-4500^2)+155x6450x(9000^2-6450^2-4500^2)
= +
6x205000x3707025200x9000/1000000
5 x 28251 x 78.5 / 1000000 x 9000^4 384 x 205000 x 3707025200
= h
=
<
4.21 mm
9.00 mm
O.K.
7.50 mm
O.K.
31000 x 3750^3 48 x 205000 x 112309000
=
<
1.48 mm
6. SHEAR CAPACITY Py
Py
=
0.60 x d x t py
=
0.60 x 863 x 16 x 265 /10^3
=
2180.74 kN
=
2180.74 kN
>
Fa
=
499.19 kN
O.K.
7. WEB BUCKLING AND BEARING d/t
=
54.25
<
Shear buckling strength of web qw
=
158
62 =
63.24
( from 4.4.5.1 )
From Table 21 of BS 5950 Part I 2000
N/mm2
Buckling Resistance Vb
=
Vb
=
d t qw 863 x 16 x 158 /1000 499.19 kN
= 2167.02 kn > Bearing Capacity of web ( 4.5.2 of BS 5950 Part I 2000 ) Pbw
=
( b1 + n k ) t pyw
- except at the end of the member : b1 + nk = Pbw = =
O.K.
t
=
16 mm
T
=
28 mm
k
=
28 mm ( k = T )
n
=
t +2T + 5T
5
=
212 mm
212 x 16 x 265 /10^3 898.88 kN
- at end of the member :
> n
kN
620.00 =
O.K.
2 + 0.6 be / k
be
=
( B - t - 2T ) / 2
=
116 mm
n
=
2 + 0.6 x 116 / 28
=
4.49
t + 2T + 4.5T
=
b1 + nk = Pbw = =
5.0
n =
197.6 x 16 x 265 /10^3 837.9512 kN
Fx =
Fx - Pbw =
499.2 - 838.0
Ps
=
As net x Py
=
20 x (400 -20 -20 ) x 265/1000
=
>
197.63 mm
1908 kN
499.19 kN =
>
O.K.
-338.76 kN
-338.76
kN
O.K.
4.5
Local compression under wheels The local compressive stress in the web due to crane wheel load is obtained by distributing it to over the length x R given by 4.11.4 of BS 5950-1 : 2000. xR = 2 (HR + T)
<
sw
=
3900 mm
HR =
25 mm P = 193.8 x 1.6 = 310.00 kN
xR =
106
pcw
=
P / ( tw * xR )
=
183
mm
N/mm2
<
pyw
=
xR N/mm2
265
(
O.K.
)
DIAGONAL MEMBER SECTION PROPERTY (T152x76x19) L
A
0.80
Nh
1.75
Nh
= =
b t b+d t
=
23.3 cm^2
rx,ry =
=
=
31.0 kN
x
1.924
x
2.00 cm
ru
=
rv
=
3.83 cm 1.95 cm
L
=
192.40 cm
1.6
0.8 119.29 kN 100.00 10.00 200.00 10.00
=
10
< 15
= 15
=
20
< 24
= 24
275 275 275 275
=
15
=
24
SECTION IS SEMI COMPACT SLENDERNESS CHECK
= =
pc
=
0.70 L rx
+ 30 =
0.70 x 192.4 2.002
+ 30
97 L ry
=
96
=
150
>
192.40 2.002
N/mm2
119.29
x
10
23.3
x
150 0.34
Compressive Strength
<
1.0
<
1.0
O.K.
From Table 24 curve c BS 5950 Part I 2000
Pv1
MEMBER :
Pv2
3.90 0.00
0.00
Max Wheel Load
=
No of Wheels
=
Distance between wheels (du1) Fh1
CG1
4 wheels
=
3.90 m
b
=
0.00 m
c
=
0.00 m
Length of Crane Bridge
=
9.40 m
Length of Crane Girder (L)
=
9.00 m
Fh2
a
155 kN
L
y x z
1. DETERMINE LOAD a.) VERTICAL LOAD Pv1
=
155 kN
without impact
Pv1a
=
155 x 1.25 =
193.75 kN
Pv2
=
155 kN
without impact
Pv2a
=
155 x 1.25 =
193.75 kN
with impact with impact
Pv3
=
0 kN
without impact
Pv3a
=
0 x 1.25 =
0.0 kN
with impact
Pv4
=
0 kN
without impact
Pv4a
=
0 x 1.25 =
0 kN
with impact
b.) HORIZONTAL LOAD Fh1
=
0.20
x
[ Pv1 ] =
31.00 kN
Fh2
=
0.20
x
[ Pv2 ] =
31.00 kN
Fh3
=
0.00
x
[ Pv3 - (Bridge weight) / 8 ] =
0.00 kN
Fh4
=
0.00
x
[ Pv4 - (Bridge weight) / 8 ] =
0.00 kN
c.) DEAD LOAD - SELFWEIGHT W
=
2.20 kN/m
d.) LOAD FACTOR f
Dead Load
=
Vertical and Horizontal Crane Load considered separately f = Vertical and Horizontal Crane Load considered acting together f = 2. MAXIMUM MOMENT AND SHEAR A. MAXIMUM MOMENT DUE TO WHEEL LOAD (W/O IMPACT)
Pv1
Pv2
center
y
1.575
3.90
x
Mv
=
427.98 kN-m
M1
=
0 kN-m
M1
=
297.01 kN-m
M2
=
427.98 kN-m
M4
=
0.00 kN-m
M5
=
0 kN-m
M6
=
0 kN-m
3.525
L = 9.0 m
Kn-m 427.98 kN-m MOMENT DIAGRAM
(MAX. )
B. MAXIMUM SHEAR DUE TO WHEEL LOAD (W/O IMPACT) Pv1
y x
Qv
=
242.66 kN
Q1
=
242.66 kN
Q2
=
87.66 kN
Q3
=
-67.34 kN
Q4
=
-67.34 kN
Q5
=
-67.34 kN
(MAX. )
Pv2
3.90
5.10 L = 9.0 m
0.00 242.66 kN
88 kN -67.3 kN -67.34 kN
-67.34 kN
SHEAR DIAGRAM C. MAXIMUM MOMENT DUE TO LATERAL WHEEL LOAD Fh1
Mh
=
29.06 kN-m
Np = Mh/1.0
=
29.06 kN
(MAX. )
M4
=
0.00 kN-m
M5
=
0.00 kN-m
M6
=
0 kN-m
Fh
=
31.00 kN-m
(MAX. )
Mv2
=
535 kN-m
(MAX. )
M1
=
0 kN-m
M1
=
371.27 kN-m
M2
=
535.00 kN-m
M4
=
535.00 kN-m
M5
=
535.00 kN-m
z x 4.50
4.50 L = 3.75 m Fh1
59.41 Kn-m Le = 3.75
29.06 kN-m MOMENT DIAGRAM
D. MAXIMUM SHEAR DUE TO LATERAL WHEEL LOAD Fh1
z x
L = 3.75 m Fh1
48.53 kN
18 kN 0.0 kN Le = 3.75 SHEAR DIAGRAM
E. MAXIMUM MOMENT DUE TO WHEEL LOAD (W/ IMPACT)
Pv1a
Pv2a
center
y
1.575
3.90
3.525
L = 9.0 m
x
371.27 Kn-m 535.00 kN-m MOMENT DIAGRAM
M6
0 kN m
F. MAXIMUM SHEAR DUE TO WHEEL LOAD (W/ IMPACT) Pv1a
0.01
y
Qv2
=
303.33 kN
Q1
=
303.33 kN
Q2
=
109.58 kN
Q3
=
-84.17 kN
Q4
=
-84.17 kN
Q5
=
-84.17 kN
Pv2a
8.99 L = 9.0 m
x
0.00 303.33 kN
110 kN -84.17 kN
-84.17 kN
-84.17 kN
SHEAR DIAGRAM G. MAXIMUM MOMENT AND SHEAR DUE TO SELFWEIGHT Q
WL
=
=
M
=
=
2
2.20
x
9.00
2
9.90 kN WL^2
=
=
8
2.20
x
9.00
^2
8
22.28 kN-m
H. FACTORED MOMENT AND SHEAR a.) Vertical crane load with impact and no horizontal crane load Maximum moment Mx
=
1.4 x M + 1.6 x Mv2
=
( 1.4 x 22.28 ) + ( 1.6 x 535.00) 887.19 kN-m
=
Maximum shear Fa
=
1.4 x Q + 1.6 x Qv2
=
( 1.4 x 9.90 ) + ( 1.6 x 303.33) 499.19 kN
=
b.) Vertical crane load with no impact and horizontal crane load Vertical maximum moment Mx
Maximum axial load
=
1.4 x M + 1.6 x Mv1
=
( 1.4 x 22.28 ) + ( 1.6 x 427.98) 715.96 kN-m
=
N
=
1.6 x Np
=
1.6 x 29.06
=
46.50 kN
Horizontal maximum moment My
=
1.6 x Mh
=
1.6 x 29.06 46.50
=
kN-m
c.) Vertical crane load with impact and horizontal crane load acting together Vertical maximum moment Mx
Maximum axial load
=
1.4 x M + 1.4 x Mv2
=
( 1.4 x 22.28 ) + ( 1.4 x 535.00) 780.19 kN-m
=
Horizontal maximum moment My
=
1.4 x Mhp
=
1.4 x 29.06 40.69
=
kN-m
N
=
1.4 x Np
=
1.4 x 29.06
=
40.69 kN
(MAX. )
3. DESIGN OF CRANE GIRDER a.) SECTION MEMBER PROPERTIES : ( See Section Properties Calculation ) FIG. 1
UB914x305x224
B T b
D
d
t
D
=
91.04 cm
d
=
86.26 cm
B
=
30.41 cm
A
=
282.51 cm^2
tw
=
1.59 cm
Af
=
72.68 cm^2
Tf
=
2.39 cm
x
=
38.09 cm
Ix
=
370702.52 cm^4
u
=
Iy
=
11230.90 cm^4
ry
=
Sx
=
9400.79 cm³
py
=
Zy
=
738.63 cm³
pyw
=
2 265.0 N/mm
Zfy
=
368.37 cm³
Le
=
b.) BUCKLING RESISTANCE MOMENT FOR THE XX-AXIS
( Table 11 ) b T b T d t
=
=
=
=
14.4 2.39 14.4 2.39 86.3 1.59
275 / py
=
6.03
< 9.0
= 9.0
=
6.03
<28
= 28.0
=
54.25
<80
= 80.0
275 265.0 275 265.0 275 265.0
=
1.02
=
9.17
=
28.52
=
81.5
SECTION IS PLASTIC SLENDERNESS CHECK
=
=
( 4.3.6.7 ) Le
=
ry
375.00
59.43
=
6.31
( 200.00 = back truss lattice pitch )
x
=
0.5
For symmetrical H shape ( 4.3.6.7 )
59.43 38.09
v
=
0.968
LT
=
u.v. w
=
52
=
1.6
1.6
From Table 19 BS 5950 Part I 2000
w =
1.0
( Section is plastic ) from 4.3.6.9
pb
=
200.2
N/mm2
From Table 17 BS 5950 Part I 2000
pc
=
230
N/mm2
From Table 24 c BS 5950 Part I 2000
Mb
=
Sx.pb
=
=
1882.04 kn-m
9401
x
200.20 /10^3
c.) MOMENT CAPACITY FOR THE SECTION FOR YY-AXIS Mzy
= =
Zfy.py
=
368.37
x
265
Mb
=
/10^3
97.62 kn-m
4. CHECK IN BENDING a.) Vertical moment with impact and no horizontal moment Mx
=
887.19
kn-m
<
b.) Vertical moment with no impact and horizontal moment N AfPc 46500.00 1671637.7
+ +
Mx Mb 715.96 1882.04 0.89
+ + <
My Mzy 46.50 97.62 1.0
0.9 6.31 cm 2 265.0 N/mm
<
1.0
<
1.0
c.) Vertical moment with impact and horizontal moment < N Mx My + + APc Mb Mzy < 40687.50 780.19 40.69 + + 1671637.7 1882.04 97.62 < 1.0 0.86
O.K. 1.0 1.0 O.K.
1882.04
kn-m
O.K.
375 cm
5. CHECK FOR DEFLECTION a.) VERTICAL L
9000
=
1000
1000
=
9.00
mm
=
7.50
mm
a.) HORIZONTAL L
=
500
3750 500
Consider 2 wheels inside the 9.0m beam span , v
4x[155x2550x(9000^2-2550^2-4500^2)+155x6450x(9000^2-6450^2-4500^2)
= +
6x205000x3707025200x9000/1000000
5 x 28251 x 78.5 / 1000000 x 9000^4 384 x 205000 x 3707025200
= h
=
<
4.21 mm
9.00 mm
O.K.
7.50 mm
O.K.
31000 x 3750^3 48 x 205000 x 112309000
=
<
1.48 mm
6. SHEAR CAPACITY Py
Py
=
0.60 x d x t py
=
0.60 x 863 x 16 x 265 /10^3
=
2180.74 kN
=
2180.74 kN
>
Fa
=
499.19 kN
O.K.
7. WEB BUCKLING AND BEARING d/t
=
54.25
<
Shear buckling strength of web qw
=
158
62 =
63.24
( from 4.4.5.1 )
From Table 21 of BS 5950 Part I 2000
N/mm2
Buckling Resistance Vb
=
Vb
=
d t qw 863 x 16 x 158 /1000 499.19 kN
= 2167.02 kn > Bearing Capacity of web ( 4.5.2 of BS 5950 Part I 2000 ) Pbw
=
( b1 + n k ) t pyw
- except at the end of the member : b1 + nk = Pbw = =
O.K.
t
=
16 mm
T
=
28 mm
k
=
28 mm ( k = T )
n
=
t +2T + 5T
5
=
212 mm
212 x 16 x 265 /10^3 898.88 kN
- at end of the member :
> n
kN
620.00 =
O.K.
2 + 0.6 be / k
be
=
( B - t - 2T ) / 2
=
116 mm
n
=
2 + 0.6 x 116 / 28
=
4.49
t + 2T + 4.5T
=
b1 + nk = Pbw = =
5.0
n =
197.6 x 16 x 265 /10^3 837.9512 kN
Fx =
Fx - Pbw =
499.2 - 838.0
Ps
=
As net x Py
=
20 x (400 -20 -20 ) x 265/1000
=
>
197.63 mm
1908 kN
499.19 kN =
>
O.K.
-338.76 kN
-338.76
kN
O.K.
4.5
Local compression under wheels The local compressive stress in the web due to crane wheel load is obtained by distributing it to over the length x R given by 4.11.4 of BS 5950-1 : 2000. xR = 2 (HR + T)
<
sw
=
3900 mm
HR =
25 mm P = 193.8 x 1.6 = 310.00 kN
xR =
106
pcw
=
P / ( tw * xR )
=
183
mm
N/mm2
<
pyw
=
xR N/mm2
265
(
O.K.
)
DIAGONAL MEMBER SECTION PROPERTY (T152x76x19) L
A
0.80
Nh
1.75
Nh
= =
b t b+d t
=
23.3 cm^2
rx,ry =
=
=
31.0 kN
x
1.924
x
2.00 cm
ru
=
rv
=
3.83 cm 1.95 cm
L
=
192.40 cm
1.6
0.8 119.29 kN 100.00 10.00 200.00 10.00
=
10
< 15
= 15
=
20
< 24
= 24
275 275 275 275
=
15
=
24
SECTION IS SEMI COMPACT SLENDERNESS CHECK
= =
pc
=
0.70 L rx
+ 30 =
0.70 x 192.4 2.002
+ 30
97 L ry
=
96
=
150
>
192.40 2.002
N/mm2
119.29
x
10
23.3
x
150 0.34
Compressive Strength
<
1.0
<
1.0
O.K.
From Table 24 curve c BS 5950 Part I 2000
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