lecture-8 (laterally unsupported beams-crane girders)

Design and Behavior of Steel Structures

Prof. Ahmed Abdelsalam El-Serwi 1

8- DESIGN OF LATERALLY UNSUPORTED BEAMS

APPLICATION ON CRANE TRACK GIRDERS

8.1 GENERAL

The function of the crane girders is to support the rails on which the traveling cranes move. These are subjected to vertical loads from crane, horizontal lateral loads due to surge of the crane, that is, the effect of acceleration and braking of the loaded crab and swinging of the suspended load in the transverse direction, and longitudinal force due to acceleration and braking of the crane as a whole. In addition to the weight of the crane, impact and horizontal surge must be considered. Vertical load, of course, includes the additional load due to impact.

The crane girder spans from column to column, usually having no lateral support at intermediate points excepting when a walkway is formed at the top level of the girder which restrains the girder from lateral bending. Thus, under normal circumstances, the crane girder must be designed as laterally unsupported beam carrying vertical and horizontal load at the level of the top flange. Apparently a girder with heavier and wider compression flange is more economic. Figure (8-1) shows some typical sections adopted for crane girders an elevation of a portal frame supporting an overhead crane, while Figure (8-2) shown some typical cross-sections used in the design of crane girders.

Generally, the vertical wheel loads of the overhead vrane are increased by considering a dynamic coefficient equal to 25% for electrically operated cranes and 10% for manually operated cranes. The value of the transverse lateral shock is considered 10% of vertical wheel loads without impact. Braking force (in longitudinal direction) is considered 1/7 of wheel loads.

For simplicity, the major axis bending moment is considered to be resisted by the whole cross-section, while the lateral bending moment (from lateral shock) is considered to be resisted by the compression flange only.



Figure (8-1) Overhead Crane



Figure (8-2) Sections Used in Crane Girders

8.2 LOADS AFFECTING THE CRANE TRACK GIRDER

The loads affecting the crane track girders are the following loads:

1. The vertical loads from the crane bridge (Wheel Loads) which include the crane load, the bridge own weight, the trolley, etc..

2. The dynamic effect of the moving loads which is included in the dynamic magnification factor (In the Egyptian Code of of Calculation of Loads and Forces Structures and Buildings, Impact Factor is considered 25% in case of electrical cranes and 10% in case ofmanual cranes)

3. Horizontal transverse loads due to the lateral shock of the crane acting at the top level of the rail and is equal to 10 % of the wheel loads without impact.

4. Horizontal longitudinal force along the rail due to brakes and is equal to 1/7 of the wheel loads without impact (to be neglected in the crane girder design and to be considered in the bracing and column design).

Crane loads are usually defined by the crane designer (mechanical part) where the wheel loads are given as two moving vertical loads along the span of the crane girder with a fixed spacing between the two loads. This spacing depends on the crane capacity and span of the frame, which the crane serves. A set of cases of loading is used to get the case of maximum moment and maximum shear affecting the crane track girder.



Figure (8-3) Loads Affecting Crane Girders

Figure (8-4) Loads and Statical System of Crane Girders

When considering the lateral shock in the design, the allowable stress should be increased by 20% (Case II). Dynamic effect (Impact) should be considered as primary (case I) load.



8.3 CHECK OF CRANE TRACK GIRDER

The following checks must be done for the crane track girder:

8.3.1 Check of Flexure stress:

The section must be checked to resist the bending moment affecting the carane girder. The lower flange is checked to resist the bending moment from dead load, vertical wheel load in addition to impact "MD+L+I" (MX). The upper flange is checked under the effect of "MD+L+I" (MX) in addition to the bending moment resulting from the lateral shock (My). The stresses affecting the lower flange are considered case I and the stresses affecting the upper flange are considered case II.

8.3.2 Check of Shear stress:

The section must be checked to resist the maximum shear force affecting the crane girder (QD+L+I).

8.3.3 Check of Crippling stress:

The web of the section must be checked to resist shear stresses resulting from the direct wheel load as mentioned in lecture 2.

8.3.4 Check of Deflection:

The deflection of the crane girder due to live load only (without impact) shall not exceed the values given in the Egyptian code of practice (span/800).

8.3.5 Check of Fatigue Stresses:

The structural members subjected to repeated fluculations of stresses shall be checked according to the Egyptian code of practioce, chapter 3.



The following definitions are important to start the fatigue study:

Fatigue: Damage in a structural member the gradual crack propagation caused by repeated stress fluculations.

Design life: The peiod in which a structure is required to perform safely with an acceptable propability that it will not fail or require repair.

Stress range: The algebric difference between the extreme values of stresses due to fatigue loads through standard elastic analysis.

Fatigue strength: The stress range determined from test data for a given number of stress cycles.

Faqtigue limit: The maximum stress range of constant for constant amplitude cycles that will not form fatigue crack.

Detail category: The detail take into consideration the type of member and joint to specify its detail.

How to Calculate fatigue strength "Fsr":

1-From table 3-1c in the Egyptian code of practice, specify the number of constant stress cycles "N"

2-From table 3.3 in the Egyptian code of practice, specify the detail category of the used cross-section (foe example rolled sections are detail category A and welded sections are detail category B)

3- From table 3-2 in the Egyptian code of practice, and using the previously specified number of constant stress cycles "N" and detail category, get the allowable stress range "Fsr".

Average Daily Application for 50 years design life

Field of application Number of constant stress cycles "N"

5 Occasional use 100,000



25 Regular use with intermittent operation

500,000

100 Regular use with continuous operation

2,000,000

>100 Sever continuous operation

According to actual use

8.4 DESIGN PROCEDURE OF CRANE GIRDERS USING ROLLED

“I” SECTIONS

1. Calculation of straining actions: The crane girder is considered as a simple beam of span equal to the spacing “S” between the main trusses or frames. The main loads considered are as follows:

Dead Loads: The dead load affecting the crane girder is only the own weight, which is considered 0.1 t/m’ to 0.15 t/m’.

Get MD.L.=WD.L. x S2/8 = ---m.t.

QD.L. = WD.L. x S/2 =--- ton



Live Loads: The Live loads affecting the crane girder are two wheel loads.

Get ML.L.max.= ---m.t. & QL.L.max. =--- ton

Impact The dynamic coefficient “I” is considered 25% in case of electrical cranes and 10% in case of manual cranes.

ML.L+I.= ML.L.max x (1+I)= ---m.t.

QL.L.+I = QL.L.max. x(1+I)= --- ton

Lateral Shock The lateral shock is considered 0.1 of maximim wheel loads (without impact), so it causes bending moment My equal to 0.1 ML.L.max

The straining actions affecting the crane girder are as follows:



Mx = MD.L.+ML.L.+I

Qx = QD.L.+QL.L.+I

My = 0.1 ML.L.max

2. Estimation of the cross-section

Assume the section as non-compact (class II) section, “Fb=0.58 Fy =1.40 t/cm

2 for steel 37”

The crane girder is subjected to bi-axial bending moment, the following interaction equation is used:

01.F

S/M

F

S/M

bcy

yy

bcx

xx

Neglecting the effect of “My” on the stress to eliminate the number of unknowns and assume Fb = 1.20 t/cm

2.

x

xb

S

MF 3

2cm ----

201

cm/t.

"ton.cm"MS x

requiredx

From tables, choose suitable wide flanged I-section.

3. Check of the Class of Section:

If :

yw

w

F

127

t

d

&

yf F

9.16

t

C

The cross-section is compact (Class I)



If :

yw

w

y F

190

t

d

F

127

&

yfy F

23

t

C

F

9.16

The cross-section is non-compact (Class II)

If :

yw

w

F

190

t

d

&

yf F

23

t

C

The cross-section is slender (Class III), a new cross-section shall be chosen.

4. Check of Lateral-Torsional Buckling of Compression Flange

Lu-act=S (cm),

2fff cmxtbA

0.1Cb

cmF

b20L

y

fmaxu

or cmCdF

A1380L b

y

fmaxu

maxuL is the minimum of the two calculated values.

If maxuactu LL )section I class 640 ofcasein(F.F ybcx



)section II class 580 ofcasein(F.F ycbx

If maxuactu LL Calculate Fltb

AT = bf x tf + (dw /6) x tw = ---- cm2

43f

fy cm12

b.tI

cmA

Ir

T

y

T

(I) ybfu

1ltb F58.0CA/d.L

800F

(II)When y

b

T

u

F

C84

r

L y2ltb F58.0F

When y

b

T

u

y

b

F

C188

r

L

F

C84 yy

b5

y2

Tu2ltb F58.0F)

C10x176.1

F)r/L(64.0(F

When y

b

T

u

F

C188

r

L y

Tu2ltb F58.0Cb

)r/L(

12000F

ltb2ltb1 F F ltbF

5. Check of Stresses Flexure stresses: It is considered that the bending oment “Mx” is resisted by the whole section and “My” is resisted by the upper flange only. Flexure stresses must be checked at point "1" at the bottom flange (resisting Mx only) and point "2" at top flange (resisting Mx and My). Straining actions affecting point



"1" are considered of case of loading "I" while straining actions affecting point "2" are considered of case "II" of loading (allowable stresses are increased by 20%).

a.

sections)compact -nonF 0.58

sections) 640

y

2

1

(

compact(F.cm/tS

.)t.cm(Mf y

x

x

b. Check the stresses in point 2:

20.10.172.0

)2//(

64.0

/x

F

SM

F

SM

y

yy

y

xx (Compact sections and Lu-act <Lu-max)

20.10.158.0

)2//(

58.0

/x

F

SM

F

SM

y

yy

y

xx (Non-compact sections and Lu-act <Lu-

max)



20.10.158.0

)2//(/x

F

SM

F

SM

y

yy

ltb

xx (Lu-act >Lu-max)

Crippling stresse: Calculate the actual crippling stress using n = 10 cm and make sure that this stress does not exceed the allowable crippling stress given by code "Fcrp = 0.75 Fy"

Fatigue stresse:

sr

x

DILD

x

sr FcmtS

MM

S

MMf

2minmax /

Shear stresses:

y2

w

max F35.0cm/tdxt

Qq

Deflection: The deflection due to live load (without impact) shall be calculated and compared with the values given in code:

800

SL.L



Example 1

Design a crane girder supporting a crane of capacity 10 tons using HEB section. The maximum reactions of the crane are two loads 8.5 tons each and spaced 3.0 m. Spacing between main columns is 6.0 m.

(Consider the dynamic coefficient “I” =25% and the lateral shock = 10%)

Solution:

1- Calculation of the maximum straining actions:

Dead Loads: Assume dead load affecting the crane girder =0.15 t/m’.

Get MD.L.=0.15 x 62/8 = 0.675 m.t.

QD.L. = 0.15 x 6/2 =0.45 ton

Live Loads: The Live loads affecting the crane girder are two wheel loads.



Get ML.L.max.= 14.34 m.t.

QL.L.max. = 12.75 ton

ML.L+I.= ML.L.max x (1+I)= 14.34 x (1+0.25) = 17.925 m.t.

QL.L.+I = QL.L.max. x(1+I)= 12.75 x (1+0.25) = 15.938 ton


Mx = MD.L.+ML.L.+I = 0.675 + 17.925 = 18.6 m.t.

Qx = QD.L.+QL.L.+I = 0.45 + 15.93 = 16.38 t

My = 0.1 ML.L.max = 0.1 x 14.34 = 1.434 m.t.

2. Estimation of the cross-section:


2 for steel 37”


2.

x

xb

S

MF 3

2cm 1550

/20.1

1006.18

cmt

xSx

From tables, choose HEB 300 (Sx=1680 cm3)


If : 9.81F

1279.18

11

208

t

d

yw

w



& 9.10F

9.166.7

9.1

2/)1.130(

t

C

yf


4. Check of the Lateral-Torsional buckling:

Lu-act=600 cm,

2fff cm579.1x30xtbA

0.1Cb

cm3.3874.2

30x20

F

b20L

y

fmaxu

or cm5.10920.1x4.2x30

57x1380C

dF

A1380L b

y

fmaxu

maxuL is 387.3 cm

maxuactu LL

AT = bf x tf + (dw /6) x tw = 30x1.9+4.367x1.1=61.8 cm2

433

ffy cm4275

12

30.9.1

12

b.tI

cm 3178861

4275.

.A

Ir

T

y

T

yb

fu

ltb F.cm/t.../x

CA/d.L

F 5807520186130600

800800 2

1



No need to calculate Fltb2 because Fltb1 governs the design (taken 0.58 Fy)

5. Check of Stresses Flexure stresses:

Check the stresses at point “1”

10711680

10060181 .

x.

S

Mf

x

x t/cm2 <0.58 Fy =1.40 t/cm

2

Check the stresses at point “2”:

4.1

)5715.0/(10043.1

4.1

1680/1006.18

58.0

)2//(/ xxx

F

SM

F

SM

y

yy

ltb

xx

20.10.115.1 x O.K.

Fatigue stress: From table 3.1c and considering the regular use with continuous operation, N=2,000,000 From table 3.3 the detail category of the rolled section is A From table 3.2, the allowable fatigue strength "Fsr" is 1.68 t/cm

2

O.K. / 68.1 t/cm066.11680

100)675.06.18( 22 cmtx

S

MMf

x

DILDsr

Crippling stress: The crippling stress affecrting the web of the HEB 300 is calculated as follows:

O.K. / 8.14.275.0 t/cm48.01.1)3210(

5.8

)2(

22 cmtxxxxtKn

Rf

w

acrp



Shear stresses:

o.k. cm/t84.0F35.0cm/t 496.01.1x30

38.16

dxt

Qq 2

y2

w

max

Deflection:

Unsafecm.

cm.x

x.L.L

750800

600

9950251702100

105952 6

Use HEB 340 and check deflection again,

safecm.

cm.x

x.L.L

750800

600

6830366602100

105952 6

8.5 DESIGN PROCEDURE OF CRANE GIRDERS USING



UNSYMMETRICAL (BUILT-UP SECTION)

1. Calculation of straining actions: The straining actions are calculated using the same steps previously discussed in section 5.8.3.


Mx = MD.L.+ML.L.+I

Qx = QD.L.+QL.L.+I

My = 0.1 ML.L.max

2. Estimation of the cross-section:


2 for steel 37”


2.

Assume 1510

SShw according to crane capacity (see

section 12-4),

cm)F35.0.(h

Qt

yw

maxw

Noting that )F/(

ht

y

ww

830 (section 12-4)

The tensile force in the lower flange “T” and the compressive force in the upper flange “C” are calculated as follows:

w

x

h

tcmMCT

98.0

).(----- ton



The area of one flange is then calculated:

2/2.1 cmt

TorCA fl ------- cm

2

The total area of the two flanges is 2 x Afl

The area of the upper flange is assumed 2/3 of the total area of the two flanges and the area of the lower flange is assumed 1/3 of the total flange area.

flflu xAA 2.3

2

flfll xAA 2.3

1

Assume bf /tf =15-20 Au-fl = (20 tf x tf )

Get bf and tf

The cross-section is as shown in figure.


If :

yw

w

F

127

t

d

&

yf F

9.16

t

C




If :

yw

w

y F

190

t

d

F

127

&

yfy F

23

t

C

F

9.16

The cross-section is non-compact (Class II)

If :

yw

w

F

190

t

d

&

yf F

23

t

C

The cross-section is slender (Class III), a new cross-section shall be chosen.

4. Check the Lateral-Torsional Buckling of Compression

Flange

For the estimated cross-section shown in figure, calculate the following properties:

Area= ---- cm2,

Y ------cm

Ix = ------cm4

Iy-upper flange =-------cm4

Lu-act=S (cm),



Au-fl= bu x tu = ----- cm2

0.1Cb

cmF

bL

y

umaxu

20

or cmCdF

AL b

y

flu

maxu

1380

maxuL is the minimum of the two calculated values.

If maxuactu LL I) classsection 640 ofcasein(F.F ybcx

II) classsection 580 ofcasein(F.F ybcx

If max uactu LL Calculate Fltb

AT = bu x tu +dwxtw/6= ---- cm2

4

3

12. cmb

tIf

fy

cmA

Ir

T

y

T

(I) yb

Tu

ltb F.CA/d.L

F 580800

1

(II)When y

b

T

u

F

C84

r

L yltb FF 58.02



When y

b

T

u

y

b

F

C

r

L

F

C18884 yy

b

yTu

ltb FFCx

FrLF 58.0)

10176.1

)/(64.0(

5

2

2

When y

b

T

u

F

C

r

L188 y

Tu

ltb FCbrL

F 58.0)/(

120002

y

22

2

2

1 F 580.cm/tFFF ltbltbltb

5. Check of Stresses Flexure stresses:

a. Check of stresses in point "1"

YI

Mf

x

x1 ------ t/cm

2

Sections)Compact -(Non 580

Sections)(Compact 640

y

y

F.

F.

b. Check the stresses in point "2":

)LL.x.F.

b.

I

M

F.

YI

M

maxuactu

y

u

flangeuppery

y

y

x

x

and Sections(Compact 20101

720

2

640

20101580

2

580.x.

F.

b.

I

M

F.

YI

M

y

u

flangeuppery

y

y

x

x

) and SectionsCompact -(Non max uactu LL

)LL.x.F.

b.

I

M

f

YI

M

maxuactu

y

u

flangeuppery

y

ltb

x

x

( 20101

580

2

Crippling stresse:



Calculate the actual crippling stress using n = 10 cm and make sure that this stress does not exceed the allowable crippling stress given by code "Fcrp = 0.75 Fy"

Fatigue stresse:

sr

x

DILD

x

sr FcmtS

MM

S

MMf

2minmax /

Shear stresses:

ww

max

xtd

Qq ---- t/cm

2 bq

For yw

w

Ft

d 105 yb F.q 350 t/cm

2

For yw

w

y Ft

d

F

105159 yy

yww

b F.]F.][F)t/h(

.[q 350350212

51

t/cm2

For yw

w

Ft

d 159 yy

yww

b F.]F.][F)t/h(

q 350350119

t/cm2

Deflection:

800

SL.L



Example 2

Design a crane girder supporting a crane of capacity 10 tons using a built-up section. The maximum reactions of the crane are two loads 8.5 tons each and spaced 3.0 m. Spacing between main columns is 6.0 m.

(Consider the dynamic coefficient “I” =25% and the lateral shock = 10%)

Solution:

1. Calculation of the maximum straining actions is the same as in

example 1


Mx = MD.L.+ML.L.+I = 0.675 + 17.925 = 18.6 m.t.

Qx = QD.L.+QL.L.+I = 0.45 + 15.93 = 16.38 t

My = 0.1 ML.L.max = 0.1 x 14.34 = 1.434 m.t.

2. Estimation of the cross-section


2 for steel 37”


2.

Assume dw= 40 cm (S/15)

cm..x.x

.

F.xd

Qt

yw

w 4904235040

3816

350 taken 8 mm

834542

83083050

80

40.

.F.t

d

yw

w O.K.

454740980

1006018

980.

x.

x.

d.

MCT

w

x ton



54.392.1

45.47

20.1

TA fl cm

2

Total area of the two flanges=2x39.54=79.08 cm2

72.5208.793

2 xA flu cm

2 = 220 ft

20/72.52ft 1.623 cm taken 16 mm

0.336.1

72.52fb cm

The upper flange 330x16

The lower flange 165x16

3. Check of the Class of Section

981127

508

400.

Ft

d

yw

w

9.109.16

06.106.1

2/)8.033(

yf Ft

C

The cross-section is compact (class I)

4. Check of Lateral-Tortional Buckling of Compression Flange

2.1118.0406.15.166.133 xxxA cm2

2.111

6.218.040)8.0406.1(6.1338.06.15.16 xxxxxxY 26.54 cm



42

223

cm 8.35765)8.054.262.43(6.133

)8.054.26(6.15.16)6.2154.26(408.012

408.0

xx

xxxxxI x

6.479112

336.1

3

xI upperfly cm4

actuL 600 cm

6.133xA fl 52.8 cm2

bC 1.0

42

332020

.

x

F

bL

y

fu

maxu 426 cm

or 7590.14.240

8.5213801380max

x

xC

dF

AL b

y

f

u cm

426max uL cm

max uactu LL

52.98.52

6.4791

fl

eupperflangy

TA

Ir cm

AT = 33x1.6+40x0.8/6 = 58.13 cm2

937101135840600

800800..

./.C

A/d.LF b

Tu

Iltb t/cm2 >0.58 Fy

No need to calculate Fltb-II because Fltb-I governes the design (taken 0.58 Fy)

5. Check of Stresses



Flexure Stress

At point 1:

3815426835765

10060181 ...

.

x.Y.

I

Mf

x

x t/cm2 <0.58 Fy = 1.40 t/cm

2

At point 2:

20101970

41

2

33

64791

1004341

41

6616835765

1006018

580

2

580

.x..

.

..

x.

.

..

x.

F.

b.

I

M

F.

YI

M

y

flfu

flangeuppery

y

y

x

x

Fatigue stress: From table 3.1c and considering the regular use with continuous operation, N=2,000,000 From table 3.3 the detail category of the rolled section is B From table 3.2, the allowable fatigue strength "Fsr" is 1.26 t/cm

2

/ 26.1 t/cm33.154.268.35765

100)675.06.18( 22 cmtx

yI

MMf

x

DILDsr

The section is unsafe Use larger section Web 400x8 Upper flange 360x16 Lower flange 180x16 Ix= 38616 cm

4 Iy = 6220 cm

4



/ 26.1 t/cm24.17.2638616

100)675.06.18( 22 cmtx

yI

MMf

x

DILDsr

Crippling stress: The crippling stress affecrting the web of the HEB 300 is calculated as follows:

O.K. / 8.14.275.0 t/cm74.08.0)2.2210(

5.8

)2(

22 cmtxxxxtKn

Rf

w

acrp

Shear Stress

yb

yw

w F.q..F.t

d350 767

42

10510550

80

40

51108040

3816.

.x

.

t.d

Qq

ww

max t/cm2 <0.35 Fy 0.84 t/cm

2

deflection:

.K. 75.0800

60065.0

386162100

1059.52 6

. Ocmcmx

xLL



8.6 DESIGN OF MONORAIL BEAMS SUPPORTING CRANES

Monorail

Monorail beams are the beams supporting a moving hoist load over its lower flange. The beam is designed to resist a mojor axis bending moment "Mx" from the dead load, live load and impact affecting the whole section, a minor axis bending moment "My" from the lateral shock affecting the lower flange only in addition to the shear. The monorail beam itself is supported below the roof by bolts subjected to tensile forces.

lecture-8 (laterally unsupported beams-crane girders)

Documents

crane loads

crane bridge wheel loads

crane girder design

vertical loads

design of crane girders

overhead crane design

crane track girders

vertical wheel loads