Mechanicshard🕐 50 min

Off-Road Vehicle Suspension System Design for Obstacle Traversal

Design a suspension system for an off-road vehicle to safely traverse a 100mm obstacle while meeting passenger comfort limits. Calculate spring stiffness, damping coefficients, maximum safe velocity, and required suspension travel using quarter-car modeling principles.

Key Concepts

suspension-systemsspring-damperquarter-car-modelmotion-ratiodamping-rationatural-frequencyharmonic-oscillatorobstacle-traversalenergy-transformation

Off-Road Vehicle Suspension System Design for Obstacle Traversal

A Classical Mechanics Engineering Problem (Quarter-Car Model)

Problem Statement

An automotive engineering firm is developing a heavy-duty off-road vehicle designed to traverse challenging terrain while maintaining passenger safety and comfort. The vehicle has the following specifications (representative of real-world UTVs and off-road vehicles):

Vehicle Specifications

  • Total vehicle mass: m=900 kgm = 900\ \text{kg}
  • Weight distribution: 60%60\% front / 40%40\% rear
  • Front corner mass: mf=270 kgm_f = 270\ \text{kg} per wheel
  • Rear corner mass: mr=180 kgm_r = 180\ \text{kg} per wheel
  • Wheelbase: L=1.6 mL = 1.6\ \text{m}
  • Front track width: 1.26 m1.26\ \text{m}
  • Rear track width: 1.39 m1.39\ \text{m}
  • Suspension type: Double wishbone independent suspension (front and rear)

Consistency check (mass distribution):
Front axle mass =0.6900=540 kg270 kg/wheel=0.6\cdot 900=540\ \text{kg}\Rightarrow 270\ \text{kg/wheel}.
Rear axle mass =0.4900=360 kg180 kg/wheel=0.4\cdot 900=360\ \text{kg}\Rightarrow 180\ \text{kg/wheel}.
This matches the given corner masses.


Design Requirements

The vehicle must safely traverse a 100 mm (0.1 m0.1\ \text{m}, 3.94 in) tall obstacle at various speeds while meeting the following criteria (typical of industry targets):

  • Human comfort limit (ISO 2631-1): Maximum passenger vertical acceleration amax1.5g=14.7 m/s2a_{\max} \le 1.5g = 14.7\ \text{m/s}^2
  • Suspension travel requirement: Minimum 50 mm total suspension travel in each direction (compression and rebound)
  • Ride frequency target (front): 1.52.0 Hz1.5\text{–}2.0\ \text{Hz} (off-road applications)
  • Damping ratio target: ζ=0.30.5\zeta = 0.3\text{–}0.5 (comfort + control balance)

Questions

  1. What is the maximum safe forward velocity at which the vehicle can traverse the 100 mm100\ \text{mm} obstacle without exceeding the 1.5g1.5g acceleration limit on passengers?
  2. What spring stiffness (spring rate) is required for the front suspension to achieve a target natural frequency of 1.8 Hz1.8\ \text{Hz}, assuming a motion ratio of MR=0.54\text{MR}=0.54 (typical for double wishbone)?
  3. What damping coefficient is needed for the shock absorbers to achieve a damping ratio of ζ=0.4\zeta = 0.4?
  4. What is the maximum force experienced by the suspension components when the vehicle strikes the obstacle at the calculated maximum safe velocity?
  5. How much suspension compression travel is required to absorb the impact, and how does it compare to the 50 mm minimum travel requirement?

Analytical Reasoning

Before diving into mathematical formulation, we must understand the physical phenomena at play.

Energy Transformation

When the vehicle encounters a step obstacle, the wheel must rise by the obstacle height. This sudden vertical displacement injects energy into the suspension system, which must be handled via:

  • Spring compression: elastic potential energy storage
  • Damper dissipation: conversion of kinetic energy to heat via viscous damping

Two-Phase Impact Analysis

Phase 1 — Initial impact (road input):
The tire contacts the obstacle and the wheel center trajectory rapidly changes. A forward velocity vv generates an effective vertical velocity component vzv_z associated with climbing the step geometry.

Phase 2 — Energy transfer to the body (sprung mass response):
The suspension transmits forces to the chassis. The peak acceleration experienced by passengers depends primarily on:

  • Impact input severity (captured here by vzv_z)
  • Effective stiffness at the wheel (wheel rate)
  • Damping characteristics
  • Corner (quarter-car) mass distribution

Critical Physics Considerations

  • Geometric constraint: For a wheel of radius RR to climb a step of height hh, the geometry sets a rapid change in direction at the contact point; higher vv increases the severity of the vertical input.
  • Force amplification: Transient events can create forces substantially above static wheel loads; damper forces can dominate during rapid compression events.
  • Natural frequency selection: The suspension natural frequency should avoid resonance with road input frequencies and human sensitivity bands.
  • Motion ratio effect: With MR=0.54\text{MR}=0.54, the spring/damper sees reduced motion relative to the wheel; thus the hardware rates must be higher than the wheel-equivalent rates by approximately 1/MR21/\text{MR}^2.

Mathematical Formulation and Resolution

Modeling Assumptions (Quarter-Car)

This analysis uses a single-degree-of-freedom (SDOF) quarter-car approximation at a front corner:

  • Corner mass mfm_f represents the effective sprung mass supported by one front wheel.
  • The suspension is modeled as a linear spring and linear viscous damper at the wheel (wheel-equivalent), mapped to the shock via motion ratio.
  • Tire stiffness, unsprung mass, and detailed tire–step contact dynamics are neglected (first-order design sizing).

Step 1 — Determine Required Front Spring Stiffness

The natural frequency of the suspension at the wheel is:

fn=12πkwheelmff_n = \frac{1}{2\pi}\sqrt{\frac{k_{\text{wheel}}}{m_f}}

where kwheelk_{\text{wheel}} is the wheel rate (effective stiffness at the wheel).

The relationship between spring rate and wheel rate through motion ratio MR\text{MR} is:

kwheel=kspringMR2k_{\text{wheel}} = k_{\text{spring}}\,\text{MR}^2

Therefore:

kspring=(2πfn)2mfMR2k_{\text{spring}} = \frac{(2\pi f_n)^2\,m_f}{\text{MR}^2}

Given:

  • fn=1.8 Hzf_n = 1.8\ \text{Hz}
  • mf=270 kgm_f = 270\ \text{kg}
  • MR=0.54\text{MR}=0.54

Compute:

ωn=2πfn=2π(1.8)=11.3097 rad/s\omega_n = 2\pi f_n = 2\pi(1.8)=11.3097\ \text{rad/s} MR2=0.542=0.2916\text{MR}^2 = 0.54^2 = 0.2916 kspring=ωn2mfMR2=(11.3097)2(270)0.29161.184×105 N/mk_{\text{spring}}=\frac{\omega_n^2\,m_f}{\text{MR}^2} =\frac{(11.3097)^2(270)}{0.2916} \approx 1.184\times 10^5\ \text{N/m} kspring1.184×105 N/m=118.4 kN/m\boxed{k_{\text{spring}} \approx 1.184\times 10^5\ \text{N/m} = 118.4\ \text{kN/m}}

Unit conversion:

1 N/m=0.005710 lb/inkspring118435×0.005710676 lb/in1\ \text{N/m} = 0.005710\ \text{lb/in} \quad\Rightarrow\quad k_{\text{spring}} \approx 118435\times 0.005710 \approx 676\ \text{lb/in} kspring676 lb/in\boxed{k_{\text{spring}} \approx 676\ \text{lb/in}}

Wheel rate:

kwheel=kspringMR2118435(0.2916)34536 N/mk_{\text{wheel}} = k_{\text{spring}}\text{MR}^2 \approx 118435(0.2916)\approx 34536\ \text{N/m} kwheel3.45×104 N/m\boxed{k_{\text{wheel}} \approx 3.45\times 10^4\ \text{N/m}}

Step 2 — Calculate Required Damping Coefficient for ζ=0.4\zeta = 0.4

Critical damping for an SDOF system is:

ccrit=2kwheelmfc_{\text{crit}} = 2\sqrt{k_{\text{wheel}}m_f}

Using kwheel34536 N/mk_{\text{wheel}}\approx 34536\ \text{N/m} and mf=270 kgm_f=270\ \text{kg}:

ccrit=2(34536)(270)6107 Ns/mc_{\text{crit}} = 2\sqrt{(34536)(270)} \approx 6107\ \mathrm{N \cdot s/m}

For target damping ratio ζ=0.4\zeta=0.4:

cwheel=ζccrit=0.4(6107)2443 Ns/mc_{\text{wheel}} = \zeta\,c_{\text{crit}} =0.4(6107)\approx 2443\ \mathrm{N \cdot s/m} cwheel2.44×103 Ns/m\boxed{c_{\text{wheel}} \approx 2.44\times 10^3\ \mathrm{N \cdot s/m}}

Motion ratio mapping to the shock hardware (same MR2\text{MR}^2 scaling):

cwheel=cshockMR2cshock=cwheelMR2=24430.29168378 Ns/mc_{\text{wheel}} = c_{\text{shock}}\,\text{MR}^2 \quad\Rightarrow\quad c_{\text{shock}} = \frac{c_{\text{wheel}}}{\text{MR}^2} =\frac{2443}{0.2916}\approx 8378\ \mathrm{N \cdot s/m} cshock8.38×103 Ns/m\boxed{c_{\text{shock}} \approx 8.38\times 10^3\ \mathrm{N \cdot s/m}}

Step 3 — Obstacle Impact Kinematics (Effective Vertical Velocity)

For a wheel of radius RR climbing a step of height hh, a simple geometric approximation for the instantaneous vertical velocity component is:

vzv(2hR)v_z \approx v\left(\frac{2h}{R}\right)

Assume a typical off-road tire radius:

R=0.35 m,h=0.10 mR = 0.35\ \text{m},\qquad h=0.10\ \text{m} 2hR=0.200.35=0.5714\frac{2h}{R}=\frac{0.20}{0.35}=0.5714 vz0.571v\boxed{v_z \approx 0.571\,v}

Step 4 — Maximum Acceleration Constraint and Maximum Safe Velocity

Peak compression estimate (underdamped SDOF with initial velocity input)

With damping ratio ζ\zeta, the damped natural frequency factor is:

ωn1ζ2\omega_n\sqrt{1-\zeta^2}

Using ωn=11.3097 rad/s\omega_n=11.3097\ \text{rad/s} and ζ=0.4\zeta=0.4:

1ζ2=10.16=0.84=0.9165\sqrt{1-\zeta^2}=\sqrt{1-0.16}=\sqrt{0.84}=0.9165 ωn1ζ2=11.3097(0.9165)=10.3655 s1\omega_n\sqrt{1-\zeta^2}=11.3097(0.9165)=10.3655\ \text{s}^{-1}

Approximate maximum compression from an initial velocity vzv_z:

xmaxvzωn1ζ2=vz10.3655x_{\max}\approx \frac{v_z}{\omega_n\sqrt{1-\zeta^2}} =\frac{v_z}{10.3655}

Peak force at the wheel

At peak response (approximation), wheel force is the sum of spring and damper contributions:

Fmaxkwheelxmax+cwheelvzF_{\max} \approx k_{\text{wheel}}x_{\max} + c_{\text{wheel}}v_z

Substitute xmax=vz/10.3655x_{\max}=v_z/10.3655:

Fmax(kwheel10.3655+cwheel)vzF_{\max} \approx \left(\frac{k_{\text{wheel}}}{10.3655}+c_{\text{wheel}}\right)v_z

Compute the coefficient:

kwheel10.3655=3453610.3655=3331.8 Ns/m\frac{k_{\text{wheel}}}{10.3655}=\frac{34536}{10.3655}=3331.8\ \mathrm{N \cdot s/m} (kwheel10.3655+cwheel)=3331.8+2442.9=5774.7 Ns/m\left(\frac{k_{\text{wheel}}}{10.3655}+c_{\text{wheel}}\right) =3331.8+2442.9=5774.7\ \mathrm{N \cdot s/m}

So:

Fmax5774.7vzF_{\max} \approx 5774.7\,v_z

Acceleration limit

Passenger vertical acceleration is constrained by:

amax=Fmaxmf14.7 m/s2a_{\max}=\frac{F_{\max}}{m_f}\le 14.7\ \text{m/s}^2 5774.7vz27014.7vz14.7(270)5774.7=0.6873 m/s\frac{5774.7\,v_z}{270}\le 14.7 \quad\Rightarrow\quad v_z \le \frac{14.7(270)}{5774.7} =0.6873\ \text{m/s} vz,max0.687 m/s\boxed{v_{z,\max} \approx 0.687\ \text{m/s}}

Convert to forward velocity using vz0.571vv_z\approx 0.571v:

vmax=vz,max0.5714=0.68730.5714=1.203 m/sv_{\max}=\frac{v_{z,\max}}{0.5714} =\frac{0.6873}{0.5714} =1.203\ \text{m/s} vmax1.20 m/s4.33 km/h2.69 mph\boxed{v_{\max} \approx 1.20\ \text{m/s} \approx 4.33\ \text{km/h} \approx 2.69\ \text{mph}}

Step 5 — Suspension Travel Required (Compression)

xmax=vz,max10.3655=0.687310.3655=0.0663 mx_{\max}=\frac{v_{z,\max}}{10.3655} =\frac{0.6873}{10.3655} =0.0663\ \text{m} xmax0.0663 m=66.3 mm\boxed{x_{\max}\approx 0.0663\ \text{m} = 66.3\ \text{mm}}

Comparison to the minimum travel requirement (50 mm compression):

  • Required compression at the safe speed: 66 mm\approx 66\ \text{mm}
  • Minimum requirement stated: 50 mm50\ \text{mm} compression
To satisfy the 1.5g limit at vmax, the design must provide 66 mm compression travel.\boxed{\text{To satisfy the 1.5g limit at }v_{\max},\ \text{the design must provide }\ge 66\ \text{mm compression travel.}}

A suspension designed with only 50 mm50\ \text{mm} compression travel would bottom out before meeting the 1.5g1.5g objective at that speed (so either travel must increase or speed must decrease).


Step 6 — Maximum Forces on Components at the Safe Speed

At vz,max=0.6873 m/sv_{z,\max}=0.6873\ \text{m/s} and xmax=0.0663 mx_{\max}=0.0663\ \text{m}:

Spring force

Fspring=kwheelxmax=34536(0.0663)=2290 NF_{\text{spring}} = k_{\text{wheel}}x_{\max} =34536(0.0663)=2290\ \text{N}

Damper force (wheel-equivalent)

Fdamper=cwheelvz=2442.9(0.6873)=1679 NF_{\text{damper}} = c_{\text{wheel}}v_z =2442.9(0.6873)=1679\ \text{N}

Total peak force

Ftotal2290+1679=3969 NF_{\text{total}} \approx 2290 + 1679 = 3969\ \text{N} Ftotal3.97 kN892 lbf\boxed{F_{\text{total}} \approx 3.97\ \text{kN} \approx 892\ \text{lbf}}

Factor above static load

Static load at the front corner:

Fstatic=mfg=270(9.81)=2648.7 NF_{\text{static}} = m_fg = 270(9.81)=2648.7\ \text{N}

Load factor:

FtotalFstatic=39692648.7=1.50\frac{F_{\text{total}}}{F_{\text{static}}} =\frac{3969}{2648.7}=1.50 Peak load1.5×static wheel load\boxed{\text{Peak load} \approx 1.5\times \text{static wheel load}}

Conclusions

Based on a classical mechanics quarter-car model (spring–damper SDOF mapped through motion ratio), the results are:

  1. Maximum Safe Velocity (1.5g limit):

    vmax1.20 m/s4.33 km/h2.69 mph\boxed{v_{\max}\approx 1.20\ \text{m/s} \approx 4.33\ \text{km/h} \approx 2.69\ \text{mph}}

    This is consistent with typical crawling speeds for technical off-road obstacles.

  2. Required Spring Stiffness (front, fn=1.8 Hzf_n=1.8\ \text{Hz}, MR=0.54):

    kspring118.4 kN/m676 lb/in\boxed{k_{\text{spring}} \approx 118.4\ \text{kN/m} \approx 676\ \text{lb/in}} kwheel34.5 kN/m\boxed{k_{\text{wheel}} \approx 34.5\ \text{kN/m}}
  3. Required Damping Coefficient (ζ=0.4\zeta=0.4):
    Wheel-equivalent:

    cwheel2.44×103 Ns/m\boxed{c_{\text{wheel}} \approx 2.44\times 10^3\ \mathrm{N \cdot s/m}}

    Shock hardware (mapped through MR):

    cshock8.38×103 Ns/m\boxed{c_{\text{shock}} \approx 8.38\times 10^3\ \mathrm{N \cdot s/m}}
  4. Maximum Component Forces at Safe Speed:

    Ftotal3.97 kN892 lbf\boxed{F_{\text{total}} \approx 3.97\ \text{kN} \approx 892\ \text{lbf}}

    This corresponds to roughly 1.5×1.5\times the static corner load.

  5. Suspension Compression Travel Required:

    xmax66 mm\boxed{x_{\max}\approx 66\ \text{mm}}

    Therefore, to meet the 1.5g1.5g requirement at vmaxv_{\max}, the design must provide at least 66 mm\sim 66\ \text{mm} compression travel. The stated 50 mm50\ \text{mm} minimum is not sufficient at that speed (unless the speed is reduced further or additional compliance is introduced).


Engineering Implications (Design Takeaways)

  • The low safe speed for a 100 mm step illustrates why off-road vehicles must crawl over sharp obstacles when passenger acceleration limits are strict.
  • Motion ratio strongly drives hardware sizing: spring/damper hardware rates scale approximately with 1/MR21/\text{MR}^2.
  • To increase obstacle-crossing speed while keeping acceleration limits, typical options are:
    • Increased available travel (and/or progressive-rate springs)
    • Tire compliance modeling and tuning (effective first-stage "spring")
    • Semi-active/active damping strategies
    • Accepting higher peak accelerations beyond comfort thresholds

This problem demonstrates how classical mechanics—Newton's laws, harmonic oscillator dynamics, and motion ratio mapping—directly informs practical suspension design tradeoffs between performance, comfort, and constraints.