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Discussion on programmed calculation model of bus frame longitudinal beam strength

Abstract: This paper establishes a calculation method of bending moment and stress of bus frame longitudinal beam suitable for computer programmed processing, compares and analyzes the advantages and disadvantages of the new calculation model and the traditional calculation model, and compiles the calculation program in C language. Through a real vehicle calculation example, the basic method of load simplification is expounded, and the new calculation model is verified

key words: strength calculation model of bus frame longitudinal beam

for non load-bearing and semi load-bearing bus bodies, the frame longitudinal beam is the basic bearing component in the whole body structure, and the design of frame longitudinal beam is a crucial link in the design of bus. In the strength analysis of the longitudinal beam of the frame, in addition to the finite element method, the classical mechanical solution is generally used. Especially in the preliminary scheme design stage of the frame, because the more mature structural scheme has not yet been formed, the finite element method cannot be used for a comprehensive strength analysis. The designer must rely on the classical mechanical calculation to determine the structural size of the longitudinal beam

the load borne by the bus frame is relatively complex. In addition to the support force of the front and rear suspension, it also bears the load of the engine, fuel tank, spare tire and other components, the load of passengers and seats, the load of the front and rear walls, left and right side walls and ceilings, the dead weight load of the floor and frame, etc. The task of strength analysis is to calculate the bending moment of any section of the longitudinal beam under the action of the above loads, and calculate the stress of this section from the bending moment and bending section coefficient through the preliminarily selected longitudinal beam section form

1 simplification of mechanical model and its comparison with traditional model

1.1 simplification of calculation model

assumptions are as follows:

(1) the left and right longitudinal beams of the frame bear the same kind of load at the same position, and all forces pass through the bending center of the section, that is, the influence of local torque is ignored

(2) the left and right longitudinal beams are symmetrical in material, structure and support

(3) each assembly suspended on the frame is simplified into several concentrated forces acting on the frame according to its actual position

(4) the weight of the front and rear walls, the left and right side walls and the ceiling is simplified into a concentrated force acting on the bracket (or shoulder pole beam)

(5) when simplifying the loads of floor, seats, passengers, etc., if the floor directly contacts the frame, these loads can be regarded as several segmented uniformly distributed loads acting on the frame. This is because the number of seats is unevenly arranged in the direction of vehicle length, so it cannot be simplified as uniformly distributed loads along the total length of the frame. If the floor is laid on the pole beam, and the pole beam is in contact with the frame, the loads of the floor, seats, passengers and so on can be simplified into a concentrated force acting on the pole beam

(6) when the front and rear section dimensions of the frame remain unchanged, the self weight of the frame is simplified into a uniformly distributed load form, otherwise, it is simplified into a segmented local load form

under the above conditions, the strength analysis mechanical model of the frame longitudinal beam can be simplified to the results shown in Figure 1

Figure 1 frame longitudinal beam strength calculation model

in Figure 1, the positive direction of X axis is taken to point to the rear of the bus, and the coordinate origin is selected at the front axle position. The meanings of the parameter symbols in the figure are as follows:

l -- wheelbase, m or mm

t1, T2 -- the distance between the front and rear suspension lugs, m or mm respectively

lf, LR -- front and rear suspension length of the frame, m or mm respectively

fi (i=1, 2,..., n) -- the ith concentrated load, N

n-- total number of concentrated loads acting on the frame

xi (i=1, 2,..., n) -- the action position coordinate of the ith concentrated load, m or mm

qi (i=1, 2,..., m) -- uniform load coefficient of the ith uniform load, n/m or n/mm2

m-- total number of uniformly distributed loads

xfi, XRI -- coordinates of the starting and ending positions of the ith uniformly distributed load in the field of high-end new material industry, m or mm respectively

z1, Z2 -- the supporting force on the front and rear suspension, N

fn+1, fn+2-- support loads at the front and rear lifting lugs of the front suspension, N

fn+3, fn+4-- support loads at the front and rear lifting lugs of the front suspension, n

1.2 comparison with the traditional calculation model

the traditional calculation model (as shown in Figure 2) simplifies the load on the frame into a continuous uniformly distributed load along the length of the vehicle or a continuous segmented uniformly distributed load, and the whole frame longitudinal beam is simplified into a simply supported beam system with supports at the front and rear axles. Although this simplification is convenient for calculation, it cannot fully reflect the actual bearing condition of the frame due to the monotonous load types, And the support mode is not completely consistent with the actual situation, so the deviation of the calculation results is large

Figure 2 traditional calculation model of frame longitudinal beam strength

compared with the traditional calculation model, the newly simplified model in Figure 1 has the following advantages:

(1) the model includes concentrated force and local uniform load, which is more in line with the actual stress condition of frame longitudinal beam. As shown in Figure 3, this local uniformly distributed load can be discontinuous piecewise uniformly distributed load (Figure 3-A), continuous piecewise uniformly distributed load (Figure 3-B), or more complex combined uniformly distributed loads such as mutual overlap (Figure 3-C). As a uniformly distributed load, the self weight of the frame must form a cross overlap with other uniformly distributed loads. This model is more convenient to deal with this kind of cross overlap local uniformly distributed load

a)

b)

c)

Figure 3 types of locally uniformly distributed load

(2) in a specific calculation, the concentrated force and uniformly distributed load often exist at the same time. The model can also have no concentrated force but only local uniformly distributed load, or it can have no local uniformly distributed load but only concentrated force. Therefore, this model is more convenient for designers to deal with problems according to different actual load conditions

(3) the support conforms to the actual situation and improves the calculation accuracy. According to the actual support condition of the suspension, four support reactions such as fn+1, fn+2, fn+3 and fn+4 are applied at the lifting lug position of the model, making the local stress condition of the frame closer to the actual situation. Especially near the rear suspension support where the longitudinal beam bending moment is the largest, the analysis accuracy of the new model is higher than that of the traditional model

(4) it is convenient for computer programming. The model allows n concentrated forces to act on the upper part of the frame, and each concentrated force has a corresponding coordinate value. When dealing with local uniformly distributed loads, m local uniformly distributed loads are allowed, and each local uniformly distributed load corresponds to the start and end coordinates. The values of M and N are set by the user according to the actual load

2 calculation of longitudinal beam bending moment

in Figure 1, in the section c-c'at the coordinate value x, the bending moment of longitudinal beam is a function of X. Let the torque change be t (x) and derive its calculation formula

as the derivation order, the formula of suspension reaction force is obtained first, and then the calculation formula of bending moment is obtained further

2.1 calculation of support reaction

2.1.1 support reaction Z1 at the front and rear axles, Z2

the vertical force z2f generated by the concentrated force on the rear axle is: (1)

the vertical force z2q generated by the uniformly distributed load on the rear axle is: (2)

where xqi is the central position coordinate of the ith uniformly distributed load. (3)

support reaction at the rear axle:

Z2 =z2f +z2q (4)

support reaction at the front axle:

Z1 = w-z2 (5)

where W is the total load on the upper part of the frame (including the self weight of the frame). (6)

2.1.2 support reaction at lifting lug position

for semi elliptical symmetrical leaf spring suspension, it can be considered that the support reaction force of a pair of suspension at two lifting lugs is equal. Therefore, in Figure 1:

fn+1 =fn+2 =0.5z1 (7)

fn+3 =fn+4 =0.5z2 (8)

2.2 bending moment t (x) calculate the bending moment of

c-c section, which is composed of two parts. One part is the bending moment caused by concentrated force, set as TF (x); The other part is the bending moment caused by locally uniformly distributed load, set as TQ (x), then:

t (x) = TF (x) + TQ (x) (9)

2.2.1 bending moment caused by concentrated force on C-C section TF (x)

set the bending moment caused by the ith concentrated load on C-C section as TFI (x), Then: (10)

(i=1, 2,..., N, n+1, n+2, n+3, n+4)

the bending moment of all concentrated forces (including support reactions fn+1, fn+2, fn+3, fn+4) on the C-C section is: (11)

2.2.2 the bending moment TQ (x)

caused by the locally uniformly distributed load on the C-C section from various angles is set as TQI (x), Then: (12)

the bending moment caused by the whole uniformly distributed load on the C-C section is: (13)

after calculating TF (x) and TQ (x) with formula (11) and formula (13), the value of T (x) can be obtained from formula 9. It is pointed out here that t (x) is the bending moment borne by two longitudinal beams

3 stress calculation and strength check

3.1 stress calculation

if the effect of shear stress is ignored, the static bending stress in any C-C section δ (10) Is (14)

where w (x) is the bending section coefficient of the left or right longitudinal beam at the C-C section. The bending section coefficient of the longitudinal beam section about the horizontal axis

because the longitudinal beam stress changes along the direction of the vehicle length, different sections can be selected at a certain interval for actual calculation

3.2 strength check

after calculation, the maximum stress in the whole longitudinal beam and its location are selected for strength check. The maximum stress under static load is δ Max, take the support coefficient as -4.7 [1], and consider that under the symmetrical vertical support load condition, the frame is in the fatigue stress state, and the general fatigue safety factor is 1.15 ~ 1.4 [2]. Then the maximum dynamic stress δ Dmax is

δ dmax ≈(3.5～6.5) δ Max (15)

when selecting the coefficient in equation 15, it should be selected according to the purpose of the bus, the use environment, and the type of vehicle. The lower limit is taken for tourist buses and urban buses, the upper limit is taken for buses with extremely poor service conditions, and other buses are selected by comparison. According to the author's experience, this coefficient can be corrected to 2.5 ~ 5.5. This is because the road conditions have been significantly improved in recent years, and the passenger car operating environment has changed fundamentally compared with previous decades. The husband type selection coefficient is still used, which is conservative. The actual calculation of many models also shows this point. Another reason is that it is not common for modern buses to have completely unloaded bodies, which more or less share part of the load. Based on the above two considerations, the coefficient can be reduced through application

the stress obtained according to equation 15 should be less than the fatigue limit of the material, that is:

δ dmax ＜ δ- 1 (16)

16mn steel symmetrical cyclic fatigue limit stress δ- 1 =220～260MPa。

4. Computer programmed processing and calculation examples

the meaning of programmed processing is to apply the derived formula to compile a general program for calculating the longitudinal beam of the frame

4.1 input data parameters

l, LF, LR, T1, T2, N, m, fi (i=1, 2,..., n), Xi (i=1, 2,..., n), Qi (i=1, 2,..., n), XFI (i=1, 2,..., n), XRI (i=1, 2,..., n)

see 1.1 for the above parameter descriptions. In addition to these parameters, it is very convenient to operate. Other input parameters are as follows:

dx- calculate the interval increment, that is, calculate the bending moment and stress of the section every DX distance, m or mm; K-the number of section types of unequal section longitudinal beam

wi (i=1, 2,..., K) - the bending section coefficient of the i-th section, m3 or mm3, which is the value of the reinforced longitudinal beam section; Xwfi (i=1, 2,..., K) - the starting position coordinate of the ith section, m or mm

xwri (i=1, 2,..., K) - the coordinate of the end position of the ith section, m or mm

4.2 programming

according to the above algorithm, a general program for calculating the strength of the longitudinal beam of the bus frame can be compiled. The program does not need to specify the number of concentrated forces, the number of uniformly distributed loads, load positions, etc

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