11 Crane Frame Structures

8/18/2019 11 Crane Frame Structures

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Mechanical Engineering Department Material Handling Equipment

Teaching Material Jose M. Dominguez E.99

Material Handling Equipment. Hoisting Equipments.

11 – Crane frame structures.

A crane frame carries all operating mechanisms, electrical equipment, motors and controlling gear. The

frame sustains externals dead load, live load, wind pressure, inertia forces and transmits these forces either

onto a fixed foundation or through traveling wheels and runway rails to the foundation or the supportingmembers of a building. The frame should ensure adequate strength and stability of the entire construction.

Stresses in its separate elements should not exceed the safe values while the strains should be recoverable.

Should be so negligible that variable loads acting on the crane do not cause vibrations either in the frame as

a whole or in its components. Adequate rigidity of the crane structure is the prime requirement for

dependable and faultless operation of all the working mechanisms. Design and solution of a crane

framework should be done with great accuracy.

Structure of overhead traveling cranes.

Depending of the lifting capacity and length of span the bridges of overhead traveling cranes are made up

of plate girders (double tees), trusses or box, double wed, girder.

Plate girders of overhead traveling cranes. Structures of overhead traveling crane with plate girders is

composed of two main longitudinal girders fastened at the ends to two cross girders which accommodatesthe traveling wheels.

Fig. No. 73 Curves for approximate determination of the weight of the main longitudinal girders in

overhead traveling cranes. A Plate girders; B trusses.


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The other elements are auxiliary or side girder, cross bracing with platform flooring and an operator’s cab

(not for all cranes).

Min factors in the solution of plate girder are the safe unit bending stress and the permissible girder

deflection. Vertical loads on the girders are: dead weight (constant load) and the force exerted by the

wheels of a trolley carrying the maximum load. The dead weight is composed of the weight of the girder

one half the weights of the traveling mechanisms (without wheels) and half the weight of the cross bracing

and flooring.

Fig. No 74 Curves of moments and lateral forces for plate girders.

Approximates values of the design dead weight of the main girder of electrically driven overhead traveling

crane with a lifting capacity from 5 to 50 tons and with span from 10 to 30 meters shown in (Fig. No 74).


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Bending moments and lateral forces due to constant load.

If L in (Fig. No. 74 a) is the span of the main longitudinal girder in meters; G its weight (constant load intons) uniformly distributed over the entire length of the span; q dead weight in tons per running meter,

then the bending moment at a distance x from the left support due to the dead weight will be:

M q =q*(L/2)*x – q*(x/2) = q*(x/2)*9L- x) ton-m

Maximum bending moment when x = L/2 is

M qmax = q(L2 /8) = G(L/8) ton-m

Bending moment a parabola plotted on length L with a maximum ordinate of M qmax

(Fig. No. 74 b)

Lateral force at a distance s from the left support

T’ x = q*[(L/2) – x] tons

Maximum lateral force on right or left support at x = 0 will be

T’ max = q*(L/2) tons (fig. No. 74 c)

Bending moments and lateral forces due to moving load (trolley and load). When the weight of a loaded

trolley is distributed uniformly on the wheels the load acting on one wheel will be P= (Q + G0 )/4;where Q weight of the load carried and G0 weigh of the trolley.

Is assumed that the girder carries two identical loads P spaced a distance b (Fig. No. 74 a). The bending

moment in the section under the left wheel stationed at a distance x from the left support: M p = (2P/L)*[{L – (b/2)} – x]*x in ton-m

Maximum bending moment in the section at a distance (b/4) from the middle of the girder span:

M pmax

= (P/2L0*[L – (b/2)] ton-m

Bending moment is in the form of a parabola with the largest ordinate on the length [L – (b/2)] Curve of moments for the most unfavorable position of the loads will be: 0 – 1 – 2 – 3; obtained from

plotting of two curves of moments from the movement of the loads from the left to the right and from the

right to the left. Maximum bending moment is reduced as the distance between the wheels increases. If the

distance between wheels is small in respect to the length of the span (b = 0) then:

M pmax

= P*(L/2)If the trolley wheels are unequally loaded, unequal loads P 1 and P 2 travel along the girder then the resultantof both loads is V = P 1 + P 2 acting on the wheels and b1 = P 2b/V and b2 = P 1b/V

In this case bending moments in the section under the left wheel at a distance x from the left support will

be: M’ p = (V/L)*[(L – b1 ) – x]*x ton-m. Maximum bending moment under the load P 1 at a distance

b1 /2 from the middle of the girder

M’ pmax

= (V/4L)*(L – b1 )2 ton-m.

This is the equation of a parabola plotted on the length (L – b1 ) with the maximum ordinate M pmax

The moments in the section under the load P 2 at a distance x from the right support are determined by:

M p’’ = (V/L)*[(L – b2 ) – x] ton-m and M p’’ max

= (V/4L)*(L – b2 )2 ton-m

The later is the equation of a parabola on the length (L – b2 ) with the ordinate M p’’ max

For identical loads P and P the lateral force at a distance x from the left support (Fig. No. 74 f)

T’’ x = (2P/L)*[{L – (b/2)} – x] tonsFor x = 0 maximum lateral force on left support

T’’ max = (2P/L)*[L – (b/2)] = 2P – P*(b/L) ton

Lateral force at a distance (L – b) from the left support

T’’ b = P*(b/L) tons

With load P 1 and P 2 lateral force at a distance x:

T x P1 – P2

+ (V/L)*[ (L – b) – x] tonsWhen x = 0

T x(max) P1 – P2

= A = (V/L)* (L – b) = (P 1 – P 2 ) – P 2*(b/L) tons

Lateral force at a distance (L – b) from the left support T b = P 1*(b/L)


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Adding together the curves of moments due to the action of dead weight (Fig. No. 74b) and the moving

loads (Fig. No. 74 c) we obtain the resultant curve of moments (fig. No. 74 d). The resultant diagram of the

lateral forces in (Fig. No. 74 g). Fig. No. 74 represents moments for the main longitudinal girders of

overhead traveling cranes with lifting capacities fro 5 to 50 tons and for spans from 10 to 30 meters.

Fig. No. 75 Curves for approximate determination of the maximum moments due to constant (M qmax

) and

moving (M pmax

) loads acting on the main longitudinal girders of overhead traveling cranes.

A Plate girders; B Trusses

Unit stress due to bending moments.

a)

From main loads = [ T add M q + M p ]/W net < [ ] b)

From the main and additional loads = [ M q + M p + M add ]/ W net < [ ]

Unit stress from lateral forces

a) From the main loads = [ T’ + T’’] S gr / I gr < [ ]

b) From the main and additional loads = [ T’ + T’’ + T add ] S gr / I gr < [ ]

The following notation is accepted in the preceding formulas:

actual normal unit stress (bending) kg/cm2

actual tangential unit stress (shear) kg/cm2

[ ] [ ] safe unit bending and shearing stresses kg/cm2

dynamic coefficient for forces due to constant load dynamic coefficient for forces due to moving loads

W net net section moment (without rivets holes)

M add sum of the bending moments in the section due to additional loads

S gr gross static moment in the corresponding part of the section (with rivets holes)

thickness of the section wall I gr gross moment of inertia in respect to the main axis

T add sum of the lateral forces in the section due to additional loads


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Table No. 41 Dynamic coefficient for forces due to constant load

Type of crane Duty Crane traveling speed m/min Coefficient Hand driven -- Up to 60 1,0

Power driven Light Up to 60 1,0Medium Up to 60 1,1

Heavy

Table No. 42 Dynamic coefficient for forces due to the moving load

Type of crane Duty Coefficient Notes

Hand power -- 1,0 --

Power driven Light 1,1 --

Medium 1,2 This group also includes foundry cranes

Heavy 1,4 This group includes special cranes employed in

metallurgical plants

Forces arise in the metal structures of the crane under the action of constant loads should be reduced to

actual forces by multiplying them by the coefficient which takes into account the dynamic effects of themoving structure and which depends on the traveling effects of the moving structure depending of the

traveling speed. Coefficient is used to correct the forces due to a moving load.Main loads in the structure are: dead weight, movable load and inertia load caused by the forces of

acceleration of the structure together with the trolley and load. Additional loads are: wind pressure, load

due to temperature effects as well as the horizontal braking forces. Along the bridge when the crane trolley

is braked (1/7 of the total forces exerted by all the driving wheel of the trolley) and across the bridge when

the crane itself is braked.

Deformation (Deflection) of girders.

Excessive deflection of the main girder causes the entire bridge to oscillate. To keep the deflection within

safe limits, main longitudinal girder should be sufficiently deep and posses an adequate moment of inertia.

Maximum deflection from its own weight ’ = (G/EI)*(5L/384) cm

G weight of the girder; E modulus of elasticity (2200000 kg/cm2)

Fig. No. 76 Diagrams for calculating the deflection to the main girder of overhead traveling crane.

Deflection due to moving loads can be determined on the assumption that the two loads P and P (loads onthe wheels of a loaded trolley) are located symmetrically in the middle of the crane (Fig. No. 76) then is

obtained with sufficient accuracy ’’ = (P/48EI)*(L – b)*[L2 + (L + b

2 )] cm

The full deflection = ’ + ’’


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Checking the strength of the lower flanges on monorail tracks.

In single girder overhead cranes a hand or power driven hoist is often designed to move on a roller beam.

Fig. No. 77 Lower flange of an I-beam

The lower flange of a beam shown is subject to a bending down force. In calculation it is regarded as a

plate rigidity fixed on one (long) side and with a load moving along the other free end. In this case the

bending moment (per unit length) M = 0,51 P ; where P is the force exerted by a runner which is a four

wheel design will be P = (Q + G0 )/4.Where: G0 weight of the movable hoist; Q is the lifting capacity.

If the thickness of the flange is t ; then the moment (per unit length) is W = lt 2 /6 = t

2 /6

Then the stress in the flange will be σ =3,05 P/l 2

11 Crane Frame Structures

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