## 84 Optimizing substructures and individual elements

The connection points between the major structural elements (Fig. 8.7) are worthy of detailed attention so that the strength and stiffness of the connected members is preserved at the joint. At a joint between box members such as the important rear quarter to sill junction of a car, (a), consideration should be given to load transfer at the corner. In figures taken from the design of small car body structure, using aircraft thin-walled structural techniques, by T.K. Garrett6, a bending moment of 51 000 lbf in (5.76 kNm) induced end loads of 33 000 lbf (146.7 kN) in the top and bottom flanges of the 5 in (127 mm) sill. The 18.5 MN/m2 safe working stress steel used in this application dictated an area of 0.123 in2 (0.795 cm2) edge material - a 3.4 in (86.5 mm) wide flange of 20 gauge sheet. The webs of sill and quarter panels overlap as shown by the dotted lines to provide the necessary transfer of the 1900 lbf (8.45 kN) shear force applying in this example. It is also important to maintain the cross-section shape of thin-walled box beams, both along their length and, crucially, at the corner joint where a diagonal diaphragm (bulkhead) might be required.

The importance of structural continuity, in order to transfer loads from one part of a structure to another, is clear from the example shown. Here it is necessary to transfer both shear and bending moment between the sill and the rear end of the vehicle. The sill outer panel is joggled under the rear quarter panel and there is a vertical joint line CD. At the rear end the sill is closed by flanging outwards of the outer panel and its connection to the heel board at EF. End loads to be transferred from the top and bottom of the section to the rear quarter panel are found by dividing the section depth into the bending moment. These loads will govern the number of spot-welds required at the connection. The direction of the loads will determine whether a 'kink strut' is required at the joint, its dimensions being found by resolving the forces along the line of the strut.

In the case of a curved corner, or angular, joint between box beams, (b), the web is subject to direct stress M/hRt where R is the radius of curvature and t the web thickness. For the lightly angled joint shown in the same figure, a 'kink strut' may be required at the junction to maintain the cross-section against collapse. Load in the strut will be M/d(cos a + cos b). If the beam has particularly deep webs, it may be necessary to provide vertical stiffeners which help stabilize shear panels against buckling. One technique is to swage the panel rather than fasten on additional parts. General rules for swaging are that all swages should be straight and not intersect, and that they should run along the shortest distance between supported edges of the panel. In calculation they can be represented by end-load carrying bars.

Attachment of mechanical units to an integral body needs special attention to the spreading of load, in avoiding local distortion and failure also in limiting undesirable relative movement. The rear spring front hanger bracket is an important area in leaf sprung front-engine/rear-drive layouts on MPVs and minibuses. This mounting has to transmit braking and drive loads horizontally as well as vertical loads due to weight and braking/driving torque reaction, and side loads due to cornering. The example illustrated at (c) shows that the boxed (longitudinal) body member has an internal stiffener and side plates with through bolts.

Where mechanical units are cantilevered out from the main structure by triangulated frameworks, the theory of space trusses can be used to analyse loads and deflections. For a three-dimensional frame, resolving forces is a little more difficult than with two-dimensional frames and a method

Fig. 8.7 Joints and sub-structures: (a) load-carrying joints; (b) curved or angular joint; (c) structural attachment to box member; (d) single bay of space truss.

known as tension coefficients is applied. This is based on the fact that proportionality exists between both resolved components of length and of force. A tension coefficient

x y z for member force S, in a length L, has projecting force vectors and length components, Fx, Fy, Fz and X,Y,Z on perpendicular coordinate axes.

The view at (d) shows one bay of a space truss, that might be an extension used to support an engine/gearbox unit behind a monocoque bodyshell structure. For analysis, the frame is assumed to have rigid plates at b1 2 34 and c1 2 34 which offer no force reaction perpendicular to their own planes (zero axial warping constraint). When considering the torsional load case, for the bay, b1c1 and b4c4 member forces are zero by resolving at b1 and c4, observing zero axial constraint of the truss. Remaining members (the envelope) have equal force components F2 at the bulkheads and tension coefficients for all of them are F/Z. This common value can be determined by projecting envelope member forces onto the bulkhead planes, as shown. By taking moments about any axis O, for any one member (c3b4, say, with projected length d), contribution to torque reaction is = rtd (half the area of the triangle formed by joining O to c3 and b4). Thus total torque is 2tA, by summation.

### 8.4.1 STIFF-JOINTED FRAMES

Where structural members are curved or cranked over wheel arches or drive shaft tunnels the unit load method of analysis applied to stiff-jointed frames is particularly useful, Fig. 8.8. It is best understood by considering a small elastic element in a curved beam (a) which is otherwise assumed to be rigid. An imaginary load w, at P, is then considered to cause vertical displacement A; the beam deflects under the influence of the bending moment and it can be shown that A = ¡Mm/ EI. ds where m is bending moment due to imaginary unit load at P and Mdue to the real external load system. An example might be a car floor transverse crossbearer having a 'tunnel' portion for exhaust-system and prop-shaft clearance, (b). To find the downward deflection at B, the imaginary unit load is applied at that point and it develops reactions of two-thirds at A and one-third at D. The bending moments for the real and imaginary loads are shown at (c) which also illustrates the subsequent calculation to obtain the deflection due to the driver and seat represented by distributed load w over length l. Another example is the part of the frame at (d) subject to twisting and bending, deflection due to twist = ¡Tt dx/GJ. This deflection would then be added to that due to bending, obtained from the formula given in the earlier section. Usually the deflection due to axial straining of the elements is so small that it can be discounted.

A good example of a frame subject to twisting and bending is the portal shape frame at (e) having loading and support, in the horizontal plane, indicated. If it is required to find the vertical deflection at D for the load P applied at B, equating vertical forces and moments is first carried out to find the reactions at the supports. Next step is to break the frame up into its elements and determine the bending moments in each - ensuring compatibility of these, and that associated loads are 'transferred' across the artificially broken joints. Integrations have to be carried out for the three deflection modes as follows: . ds/AE + ¡Mm . ds/EI + ¡Ff. ds/AG to obtain the combined deflection. Both vertical and horizontal components can be obtained by applying vertical and horizontal imaginary unit loads, in turn, at the point where the deflection is to be determined. For the example shown in the figure, elastic and shear moduli, E and G, are 200 and 80 GN/m2 respectively. S is axial load and F the shear load, the latter being negative in compression. The values of these, with the bending moments, are as shown at (f right). These can readily be substituted in the expression (f left) to determine deflections as a factor of P.

Fig. 8.8 Stiff-jointed frames: (a) unit-load method; (b) seat support bearer; (c) unit-load calculation; (d) element subject to twisting and bending; (e) portal frame under combined loading; (f) loads in members; (g) semi-circular arch member under load: (h) front-end frame as portal; (i) battery-tray adjacent to wheel arch.

Fig. 8.8 Stiff-jointed frames: (a) unit-load method; (b) seat support bearer; (c) unit-load calculation; (d) element subject to twisting and bending; (e) portal frame under combined loading; (f) loads in members; (g) semi-circular arch member under load: (h) front-end frame as portal; (i) battery-tray adjacent to wheel arch.

The method can also be used for 'continuum' frames incorporating large curved elements provided curvature is high enough for the engineer's simple theory of bending to be applicable. In the semicircular arch at (g), at any point along it defined by variable angle 8, bending moment is given by Wr sin 8 due to the external load and r sin 8 due to the imaginary unit load at the position and in the direction of the deflection which is required. In this case it is the same as that of the external load and the deflection due to bending is equal to the integral of the product of these between q = 0 and p, divided by the flexural rigidity of the section EI, product of elastic modulus and section second moment of area.

A common example of a portal frame in a horizontal plane is the front crossmember and front ends of the chassis sidemembers of a vehicle imagined 'rooted' at the scuttle. The load case of a central load on the crossmember, perhaps simulating towing, can be visualized at (h). An example of a frame having both curved and straight elements is the half sill shown at (i) supporting a battery tray, with stout crossmembers at mid-span and above the wheel arch reacting the vertical load in this case. Again a table of bending moments can be written, as shown, and the vertical deflection at B calculated by using the unit load formula. Given the value of I for the sidemember of 0.25 x 104 m4 and E of 210 GN/m2, the deflection works out to be 53 mm. It is generally best to integrate along the perimetric coordinate for curved elements and, as above, when requiring a translational deflection, apply unit force, and for a rotational displacement, apply unit couple. In the case of redundancies, remove the redundant constraint by making a cut then equate the deflection of the determinate cut structure with that due to application of the redundant constraint.

### 8.4.2 BOX BEAMS

The panels and joints in box-membered structures can be treated differently, Fig. 8.9. In the idealized structure at (a), the effect on torsional stiffness of removing the panels from torsion boxes can be seen in the accompanying table, due to Dr J. M. Howe of Hertfordshire University. Torsional flexibility is 50% greater than that of the closed tube (or open tube with a rigid jointed frame of similar shear stiffness to the removed panel surrounding the cutout) if the contributions of flanges and ribs are neglected.

The effect ofjoint flexibility on vehicle body torsional resistance must also be taken into account. Experimental work carried out by P. W. Sharman at Loughborough University has shown how some joint configurations behave. The importance of adding diaphragms at intersections of box beams was demonstrated, (b). Without such stiffening, the diagram shows the vertical webs of the continuous member are not effective in transmitting forces normal to their plane so that horizontal flanges must provide all the resistance. The distortions shown inset were then found to take place if no diaphragms were provided.

### 8.4.3 STABILITY CONSIDERA TIONS

Applying beam theory to large box-section beams must, however, take account of the propensity of relatively thin walls to buckle. Smaller box sections such as windshield pillars may also be prone to overall column buckling. Classic examples of strut members in vehicle bodywork also include the B-posts of sedans in the rollover accident situation. Such a B-post section, idealized for analysis, is shown at (c). To determine its critical end load for buckling in the rollover situation, its neutral axis of bending has first to be found - using a method such as the tabular one at (d). Assuming the roof end of the pillar to impose 'pinned' end fixing conditions (so that L = 2l) and that the pillar is 1 metre in length, then critical load is 10.2 1 0.109.8.4.10-8/22. Taking E for steel as 210 x 103 N/mm2, the stress at this load is 44.103/2 80.106 = 157 MN/m2 since A = 250 mm2 -which is above the critical buckling stress. If, however, the cant rail is assumed to provide lateral

Element |
A |
y |
Ay |
ig |
h |
Ah2 |
ig + Ah2 | |

1. |
100 |
/ |
50 |
100/2 |
= 8.3 |
18.5 |
34 200 |
342 083 |

2. |
2 x 15 |
1 / |
45 |
30/12 |
= 2.5 |
17.5 |
9250 |
9252.5 |

3. |
2 x 51 |
26/ |
2750 |
2.502/12 |
= 1040 |
7.5 |
5620 |
6660 |

4. |
50 |
51 |
2550 |
50/12 |
= 4.14 |
32 |
34 000 |
34 004.17 |

100 mm

100 mm

Metal thickness

Metal thickness

Internal force due to T = 1 b |
Element flexibility matrix |
Number of identical elements |
Contribution to structural flexibility bTfb | |

1 |
o |
a P 1 ] |
8 |
8a3/12b2d2EAc |

2 |
a/2bd |
6EAc L 1 2 J | ||

3 |
a/2bd |
a [2 1 1 |
4 |
4a3/12b2d2EAc |

4 |
-a/2bd |
6EAc L 1 2 J | ||

5 |
3/4bd |
ab/Gt |
4 |
9ab/4b2d2Gt |

6 |
1/4bd |
ad/Gt |
4 |
ab/4b2d2Gt |

7 |
1 /bd |
ad/Gt |
2 |
2ad/b2d2Gt |

8 |
3/4bd |
db/Gt |
2 |
9db/8b2d2Gt |

.---1— |
A' ---- | |

A |
A | |

A |
A | |

B |
View along arrow (c) View along arrow (b) View along arrow (c) View along arrow (d) Fig. 8.9 Panels and joints in box members: (a) effect of panel removal on box tube; (b) use of diaphragms at beam intersections; (c) B-post section; (d) neutral axis determination. support at the top end of the pillar then L = 0.7l and critical buckling stress is 1280 MN/m2 and the strut would fail at the direct yield stress of 300 MN/m2. Other formulae, such as those due to Southwell and Perry-Robertson, will allow for estimation of buckling load in struts with initial curvature. |

## Post a comment