Figure 1. Front Suspension components.
The following Matlab code solves for the forces in each suspension link using the method of sections (Statics). We can use statics to solve this
problem because we assume that the suspension links are massless (to simplify the analysis). (in math notation: m 0 F 0 )
1 % define points (left front suspension, ADAMs coordinates,
2 % reference point is defined as:
3 % a. x = 0 at center of tire contact patch
4 % b. y = 0 at center of front axle
5 % c. z = 0 at groung level
6 % positive x is to the rear of the car
7 % positive y is toward the left for the driver
8 % positive z is up
9
10 clear all
11
12 %define suspension point (x, y, z coordinates in inches) for left wheel
13 %rear and front refer to inner points on control arms
14 tire =[ 0.00 -20.00 0.00 ]; %tire contact patch, for input forces
15 UCA_outer =[ 0.00 -20.00 15.00 ]; %upper control arm, upright upper ball joint
16 UCA_rear =[ 5.00 -8.00 15.00 ]; %rear inner ball joint for UPPER control arm
17 UCA_front =[ -5.00 -8.00 15.00 ]; %front inner ball joint for UPPER control arm
18 LCA_outer =[ 0.00 -20.00 5.00 ]; %lower control arm, upright lower ball joint
19 LCA_rear =[ 5.00 -8.00 5.00 ]; %rear inner ball joint for LOWER control arm
20 LCA_front =[ -5.00 -8.00 5.00 ]; %front inner ball joint for LOWER control arm
21 TieRod_out =[ -2.00 -20.00 10.00 ]; %tie rod connects to upright here
22 TieRod_in =[ -2.00 -8.00 10.00 ]; %tie rod connects to steering rack here
23 PushRod_out =[ 0.00 -20.00 5.00 ]; %pushrod connects to upright here
24 PushRod_in =[ 0.00 -12.00 20.00 ]; %pushrod connects to bell crank here
25
26
27 % define suspension links
28 geometry = [
29 UCA_outer UCA_front %UCA front tube
30 UCA_outer UCA_rear %UCA rear tube
31 LCA_outer LCA_front %LCA front tube
32 LCA_outer LCA_rear %LCA reat tube
33 TieRod_out TieRod_in %Tie Rod
34 PushRod_out PushRod_in %Push Rod
35 ];
36
37 %Notes: any rigid body has 6 degrees of freedom (DOFs). Thus 6 rows for properly constrained system (6-DOF, long, lat, vert, roll, pitch, yaw)
38 %the order of the points does matter a little bit because it determines the
39 %sign of the force (negative is compression right now).
40
41 % set loading condition
42 loadposition = tire;
43 force = [0.0 300.0 200.0]; % Fx, Fy and Fz in lbs
44 Long = force(1);
45 Lat = force(2);
46 Vert = force(3);
47
48 Force_on_tire_lbs = table(Long, Lat, Vert)
49
50 % method of sections code
51 vectors = geometry(:,4:6)-geometry(:,1:3); %force direction vectors
52 vectors = vectors./meshgrid(sqrt(sum(vectors.^2,2)),ones(3,1))'; %turns these vectors into unit vectors
53 positions = geometry(:,4:6); %force positions (for moment about origin)
54 M = [vectors'; cross(positions,vectors,2)']; %setup system of equations (M*linkforces + load = 0)
55
56 moment = cross(loadposition,force);
57 load = [force moment];
58
59 results = (M\-load')';
60 %determine angle between control arms
61 UCA_CosTheta = dot(vectors(1,:),vectors(2,:));
62 UCA_Angle = acosd(UCA_CosTheta);
63 LCA_CosTheta = dot(vectors(3,:),vectors(4,:));
64 LCA_Angle = acosd(LCA_CosTheta);
65 Control_Arm_Angle = table(UCA_Angle,LCA_Angle)
66
67 %label results
68 UCA_frontF = results(1);
69 UCA_rearF = results(2);
70 LCA_frontF = results(3);
71 LCA_rearF = results(4);
72 TieRodF = results(5);
73 PushRodF = results(6);
74 Force_lbs = table(UCA_frontF,UCA_rearF, LCA_frontF, LCA_rearF, TieRodF, PushRodF)
This code uses the Method of Sections to solve for the forces in the front suspension. Basically, you take a single section of your structure and
treat it as a single rigid body. In our particular case, our section will be the left upright, wheel, and tire. There are two sets of forces acting on this
rigid body, the tire load applied at the contact patch and the force from each of the 6 suspension links. Combined these 6 links provide the
necessary restraints to fully constrain the section relative to the body of the vehicle in all degrees of freedom (DOFs). Remember, a rigid body has
6 DOFs, the translational ones (longitudinal, lateral, and vertical) and rotational (yaw, pitch, and roll).
Each suspension link is treated as a Two Force Member. Thus the load that it carries is in the direction that the tube is pointing (technically, the
line of action of the force is in line with the two rod ends, in case you have any weirdly shaped links…). There are 6 forces corresponding to the front
suspension links, which we’ll define as a single column vector:
FUCA _ front _ tube
FUCA _ rear _ tube
F
LCA _ front _ tube
Flinks (1) suspension forces vector
FUCA _ rear _ tube
FTie _ rod
F
Push _ rod
The basic goal is to solve for the forces of the front suspension links (1) given a specified load condition (2) that is applied at the tire.
Fxapplied
Fapplied Fyapplied (2) forces applied on the tire vector. (longitudinal = x, lateral = y, and vertical = z).
applied
Fz
Note: We’re using ADAMS’s coordinate system. Positive x is toward the rear of the car. Positive y is to the right of the driver. Positive z is up.
We’ll be using linear algebra notation, because that greatly simplifies the problem. In order to add the forces together, you have to multiply the
forces by the direction unit-vectors which the tubes are pointing, because the links are not aligned to the coordinate system.
F u links Flinks Fapplied (3) sum of forces applied onto our section
Here is what equation (3) would look like using only scalars. Note, in this case, x, y, and z are the unit direction vectors:
FUCA _ front _ tube
FUCA _ rear _ tube
Fx xUCA _ front _ tube xUCA _ rear _ tube xLCA _ front _ tube xLCA _ rear _ tube xTie _ rod xPush _ rod Fxapplied
FLCA _ front _ tube applied
Fy yUCA _ front _ tube yUCA _ rear _ tube yLCA _ front _ tube yLCA _ rear _ tube yTie _ rod yPush _ rod Fy (3b)
F
Fz zUCA _ front _ tube zUCA _ rear _ tube z LCA _ front _ tube z LCA _ rear _ tube zTie _ rod zPush _ rod F Fzapplied
UCA _ rear _ tube
Tie _ rod
F
Push _ rod
We plug in (3) into Newton’s 2nd law (F=ma). By assuming that the suspension links are massless, we simplify the analysis). In math notation:
F ma , m 0 F 0 (4) Newton’s 2nd law applied to statics. Note: the arrow can be read out loud as “implies”
F u links Flinks Fapplied 0 (5) equations in the three translational DOFs
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
The three rotational DOFs are similar to the translational. The rotational equivalent to Newton’s 2 nd law is Euler’s equation. Similar to the
translational case, by assuming that the inertias of the section is zero, we can set the sum of moments to zero:
M I , I 0 M 0 (6) Euler’s equation.
We can actually take the moment about any arbitrary point and it should be zero. Taking the moment about the origin simplifies the math a little
bit, so that’s what we’ll do. This way, the moment arm becomes the coordinate of one of the rod ends, either one. Remember, each link is a two-
force member, so it would have two rod ends. It doesn’t matter which rod end you pick, they both give the same resultant moment, because they
both lie on the line-of-action of the force.
M p links ulinks Flinks pcontact _ patch Fapplied 0 (7) sum of moments.
The position vectors look like this. (x, y, and z are the coordinates). UBJ is the upper ball joint of the upright. LBJ is the lower.
xUBJ xUBJ xLBJ xLBJ xTie _ rod _ outer xPush _ rod _ outer
plinks yUBJ yUBJ yLBJ yLBJ yTie _ rod _ outer yPush _ rod _ outer (8)
z zUBJ zLBJ zLBJ zTie _ rod _ outer zPush _ rod _ outer
UBJ
xcontact _ patch
pcontact _ patch ycontact _ patch (9)
z
contact _ patch
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Combining equations (5) and (7) gives:
F ulinks Flinks Fapplied
0 (10)
M plinks ulinks Flinks pcontact _ patch Fapplied
Equation (10) can be simplified to…
ulinks Fapplied
p u Flinks 0 (10b)
links links pcontact _ patch Fapplied
ulinks Fapplied
p u Flinks (10c)
links links pcontact _ patch Fapplied
Fapplied
pcontact _ patch Fapplied
Flinks (10d) Notes: This is supposed to be a right divide by, Word doesn’t understand this.
ulinks
p u
links links
Equation (10d) is line 59 in the Matlab code.
1. Using the original MATLAB script, investigate suspension arm forces for the setup shown in Figure 2
(below) under the following conditions. Note you are only modifying force on tire (line 43).
Case Description Fx (lb) Fy (lb) Fz (lb)
Cornering (25ft 0 300 200
radius)
Cornering (50ft 0 375 250
radius)
Braking (-1.5g) 319 0 213
Trail Braking (-1g/1g) 250 250 250
Deliverable: For each condition, list the suspension component that has the highest tension and
compression forces.
2. Using the same chart conduct a sensitivity investigation to see the effects of A-arm included angle on
suspension arm forces. Vary the separation of the upper and lower control arm inner ball joints from 2
to 18 inches in increments of 2 inches. Do not alter outer ball joint, tie rod or push rod coordinates.
(Hint: modify the code to create a simple loop that writes the results to an Excel spreadsheet).
Deliverable: Create plots that show the force in each arm (y-axis) as a function of control arm included
angle (x-axis). You should have four plots (one for each case). Each plot will have 4 curves (one for
each control arm). Comment on the trends that you observe.
3. Repeat step two but this time vary outer ball joint vertical separation from 6 to 9 inches in increments
of 0.5 inches while maintaining a constant inner ball joint separation of 12 inches. You will need to
alter tie rod and push rod coordinates to match ball joint separation. Maintain parallel a-arm and tie
rod configuration. (Hint: modify the code to create a simple loop that writes the results to an Excel
spreadsheet). Deliverable: Create plots that show the force in each arm (y-axis) as a function of outer
ball joint separation (x-axis). You should have four plots (one for each case). Each plot will have 4
curves (one for each control arm). Comment on the trends that you observe.
4. Deliverable: Based on the trends observed in steps 2 and 3, determine the optimum (ie: lowest forces)
suspension arm configuration for a team using 10” wheels? What other factors might come into play?
Figure 2. Front view of simplified suspension.
