4. Structural Dimensioning
4.9 Bolted Joints
There are two types of bolted joints to be studied: the ones connecting the tripod to the tubes that support the rear wheels (there are two of these that are equal) and the one linking the central element to the front wheel connection tube. They can be seen in figure 41.
Figure 41 – Location of the bolted joints
The former is constituted by 4 steel bolts and the latter by 2: their characteristics are presented in table 17, along with the used nuts and washers [60]:
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Standard Class Total Length (mm)
Number of elements used
Bolts ISO 4017 M10 35 10
Nuts ISO 4032 M10 10 10
Washers ISO 7089 M10 2 20
Table 17 – Data of the used connection elements
Both joint types 1 and 2 use the elements of the previous table. Considering the thickness of the flanges to connect (each being 16mm in total in every situation) and the fact that 2 washers of 2 mm each are used in every joint, the connection lengths are given by:
𝑙1 = 𝑙2 = 16+2+2 = 20 mm
Considering the data obtained in the static analyses of the chassis, is possible to verify that all bolted joints in the structure are subjected to normal loads and sheer loads (as mentioned in the respective section these loads were obtained using the post analysis capacities of the Siemens NX software).
Following once again the methods of reference [55], we have that the tensile stress
𝜎
𝑏 in a bolt is given by:𝜎𝑏= 𝐹𝐴𝑏
𝑡 = 𝐹𝑖𝐴+𝐶𝐹
𝑡 , (13) Here 𝐴𝑡 is the tensile stress area of the bolt, 𝐹𝑏 is the resultant load on the bolt, C is the stiffness constant of the joint, F is the external separation load and 𝐹𝑖 the applied pre-load. C is given by:
C = 𝑘𝑏
𝑘𝑏+𝑘𝑚, (14) In equation (14), 𝑘𝑏 is the effective stiffness of the bolt in the clamped zone and 𝑘𝑚 the stiffness of the members of the connection in that same zone. These constants are calculated using the following equations [55]:
𝑘𝑏 = 𝐴𝑑 𝐴𝑡𝐸𝑏
𝐴𝑑𝑙𝑡+𝐴𝑡𝑙𝑑 (15) 𝑘𝑚 = A𝐸𝑚d𝑒𝐵𝑑𝑙 (16) In the equations above 𝐸𝑏, 𝐸𝑚 are the respective Young’s modulus, 𝐴𝑑 is the bolt major diameter area, 𝑙𝑑 is the length of the unthreaded portion in grip, 𝑙𝑡 the length of the threated portion in grip, l the total length, A and B are constants.
Table 18 – Constants A and B for different materials [55]
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To perform the analyses is necessary to determine the forces occurring in the bolts and for that are going to be used the values presented in section 4.6. In each case are being considered only the maximum moments and forces in every direction. Therefore, the considered combined loads are conservative. It is also considered that the loads go through the centroid of the connection.
Bolted Joint 1 (back) Bolted Joint 2 (front)
𝐅𝐱 (N) 527 76
𝐐𝐱𝐲 (N) 2794 2595
𝐐𝐱𝐳 (N) 1224 558
𝐌𝐱 (Nm) 745 116
𝐌𝐲 (Nm) 507 428
𝐌𝐳 (Nm) 505 122
Table 19 – Maximum loads acting on joints 1 and 2
The axial force on the bolts depends on their location due to the existence of the moments 𝑀𝑦 and 𝑀𝑧. These moments have as a consequence for half of the flange to tend to rotate away from the other half about an axis (hinge line). This is illustrated in figures 42 and 43 for joint 1 but the it is also applicable to joint 2.
Figure 42 – Bending moments cause different axial loads according to the position of the bolt
Figure 43 – Rotation of half of the flange around the hinge line
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In the situation of figure 42, a moment along y would exert the same axial stress in bolts 1/2 and another in 3/4, while a moment along z would do it in 1/3 and 2/4. To make sure the joint is not vulnerable, it would only be needed to verify the bolt under the highest load, which is designated the critical bolt. The direction of the moment determines which of the bolts will be more affected: as an example, in figure 43 the bolts in the upper part of the joint have to support greater stresses because there is a moment applied in the plane of the sheet of paper that points inwards. Therefore, to determine where the hinge line is located in a certain situation it is required to observe the direction of My/Mz.
To determine which one of the bolts in the joint is the critical one, it may be necessary to calculate the loads caused in each bolt by each moment, and sum them. However, in the situation of joint 1 is sufficient to notice that attending to the directions of the moments applied, bolts 1 and 2 are the ones the furthest away from the hinge line when considering My (thus the ones under higher loads) and 1 and 3 when considering Mz. Therefore, bolt 1 must the critical one.
For both types of joints the axial and shear loads are going to be analyzed separately.
Joint 1
The load on each bolt caused by the moments is given by the following equations (𝑟𝑖 is the distance between the bolt i and the hinge line in the respective situation):
𝐹𝑀𝑦 = 𝑀∑𝑟𝑦𝑟𝑖
𝑖2 ; 𝐹𝑀𝑧 = 𝑀∑𝑟𝑧𝑟𝑖
𝑖2 (17) This equation is also valid to calculate the shear force caused by the torque, in which case 𝑟𝑖 is the distance between the centroid and the bolt.
The shear force originated by Qxy and Qxz on each bolt is:
𝐹𝑧’ = 𝑄𝑛𝑥𝑦 = 2794
4 = 699 N (18) 𝐹𝑦’ = 𝑄𝑛𝑥𝑧 = 1224
4 = 306 N (19) On the other hand, Mx causes a shear force on each bolt that is equal in magnitude but changes in direction, as illustrated in figure 44. This means that the sum of the vectors of the shear forces varies from bolt to bolt. The force caused by Mx is:
𝐹𝑀𝑥 = 𝑀𝑥𝑟𝑖
∑𝑟𝑖2 = 𝑀𝑥
4𝑟 = 745
4.0,0407 = 4576 N (20) In order to obtain the critical one regarding shear, was used the parallelogram rule. It was found the critical bolt is number 1, and the total force is 𝐹𝑠1= 5337 N.
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Figure 44 – Shear forces in joint 1
The shear stress on the critical bolt is given by:
𝜏𝑠1= 𝐹𝐴𝑠1
𝑠 = 533758 = 92 MPa (21) The shear area 𝐴𝑠 of the bolt is equal to 𝐴𝑡 (58𝑚𝑚2). The shear strength is obtained considering [55]:
𝑆𝑥𝑦= 0.577 𝑆𝑦 = 0.577 x 640 = 369 MPa (22)
𝑛
𝑠1=
𝑆𝜏𝑥𝑦𝑠1
=
36992 = 4 (23)Bearing in mind the location of the bolts in relation to the hinge lines in each case and the way in which the respective moment tends to rotate the flange, and substituting in equations (17), we have for bolt 1 of joint 1:
𝐹𝑀𝑦1 = 𝑀∑𝑟𝑦𝑟𝑖
𝑖2 = 507 x 0,085
0,0852+0,0852+0,0132+0,0132 = 0,014843,095 = 2914 N (24) 𝐹𝑀𝑧1 = 𝑀∑𝑟𝑧𝑟𝑖
𝑖2 = 505 x 0,049
0,0492+0,0492+0,0112+0,0112 = 24,7450,005 = 4906 N (25) Assuming that Fx is distributed equally by each bolt, its contribution to the load in bolt 1 is:
𝐹𝑥1= 527/4 = 132 N (26) The total axial force in the critical bolt of the joint is therefore:
𝐹𝑡1 = 𝐹𝑀𝑦1+ 𝐹𝑀𝑥1+ 𝐹𝑥1 = 7952 N (27) M10 steel bolts are used, so A=0.787, B=0.628, 𝐴𝑡1= 58𝑚𝑚2
, 𝑙
𝑑= 4mm, 𝑙𝑡 = 16mm, 𝐴𝑑1= 78,5 𝑚𝑚2 . We have:𝑘𝑏1 = 78,5.10
−6x 58.10−6 x 207.109
78.5.10−6 x 0,016+58.10−6 x 0,004 = 6.33 x 108 N/m (28)
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𝑘𝑚1 = 0,787x 207. 109x 0.01 x 𝑒0.628𝑥1020 = 2.23 x 109 N/m (29)
C =
𝑘 𝑘𝑏𝑏+𝑘𝑚
=
0.221 (30) The recommended pre-load for a reusable connection is given by the following equation (𝑆
𝑝is the proof strength):𝐹
𝑖= 0.75 x 𝐴
𝑡x
𝑆𝑝 (31)Consulting the appropriate data tables [55], we see that for the used bolts
𝑆
𝑝= 580 MPa𝐹𝑖 = 0.75 x 58. 10−6 x 580.106 = 25230N (32) The factor of safety to axial stress for the critical bolt is therefore:
𝑛
𝑏=
𝐴𝑡𝑆𝐶𝑝−𝐹𝑖𝐹𝑡1
=
8410/1757 = 4.79 (33)Joint 2
Although this joint is geometrically different and only has 2 bolts, the same approach used for joint 1 applies. In this case, the distance between the centroid and each bolt is 41mm.
Figure 45 – Shear forces in joint 2
Repeating the same procedure used for joint 1, we have:
𝐹𝑧’ = 𝑄𝑥𝑦𝑛 = 25952 = 1298 N (34) 𝐹𝑦’ = 𝑄𝑛𝑥𝑧 = 5582 = 279 N (35)
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𝐹𝑀𝑥 = 𝑀𝑥𝑟𝑖∑𝑟𝑖2 = 𝑀𝑥
2𝑟 = 116
2.0,041 = 1415 N (36) Once again, using the parallelogram rule to determine the total load each bolt has to withstand, was determined that the critical bolt is subjected to 𝐹𝑠2= 2134 N
𝜏𝑠2= 𝐹𝐴𝑠2
𝑠 = 213458 = 36.8 MPa (37)
𝑛
𝑠2=
𝑆𝜏𝑥𝑦𝑠2
=
36,8369 = 10 (38)As for the axial loads, the force in the critical bolt may be obtained by:
𝐹𝑀𝑦2 = 𝑀𝑦𝑟𝑖
∑𝑟𝑖2 = 428 x 0,03 0,032+0,032 = 12,8
0,0018 = 7111 N (39) 𝐹𝑀𝑧2 = 𝑀𝑧𝑟𝑖
∑𝑟𝑖2 = 122 x 0,094
0,0942+0,0122 = 11,47
0,009 = 1274 N (40) The influence of Fx may be ignored since it is of much smaller magnitude.The total axial force in the critical bolt of the joint is therefore:
𝐹𝑡2 = 𝐹𝑀𝑦2+ 𝐹𝑀𝑥2 = 8385 N (41)
Having in mind that the same type of bolt is used in both joints, we have:
𝑛
𝑏=
𝐴𝑡𝑆𝐶𝑝−𝐹𝑖𝐹𝑡2 = 8410/1853 = 4.5 (42) Therefore, it is possible to state that in all joints the factors of safety for axial and shear stresses are met.