Tie – Rod Under the Uniform load and Axial Pull, Pinned at both the Ends
Study: A horizontal tie-rod is freely pinned at each end. It carries a uniform load w lb per unit length and has a horizontal pull. Find the central deflection and the maximum bending moment, taking the origin at one of its ends.
Introduction
A steering tie-rod is typically designed as a slender, high-strength steel member that primarily carries axial tension and compression. However, in real automotive or in the motorsport environments the tie-rod may also experience small transverse loads due to the aerodynamics especially in F1 car, suspension setup, the road load forces and it may also introduced through the manufacturing tolerance and the alignment loads
from the given figure of the question for the pinned-pinned beam is under:
a Uniform transverse load : w (N/m) and the axial tensile force P (N)
From the beam theory of axial loading, the deflection y(x) satisfies:
where:
E = Young’s Modulus
I = second moment of area
L = Tie-rod length between the joints
P = axial tension
w = uniform lateral or transverse load
The axial tension raises the stiffness, reducing deflection compared to the normal beam ( just like stretching the small thread)
The boundary condition of a pinned tie-rod:
y(0) = 0 and y(L) = 0
solving the differential equation yields a closed-form expression for y(x), including hyperbolic functions cosh (ax), sinh(ax), wher e:
CODE FORMULATION
PLOTTING
Central deflection y(mid) at mid point along the length of the tie-rod = 0.21mm
Maximum bending moment M(max) = 1.18 N/m
RESULTS
Noting point is: that the Bending stresses are small but not negligible, this is due to the axial tension significantly reduces the bending deflection.
This example also demonstrates how a steering tie-rod (normally an axially loaded member) can be analysed for lateral bending by using classic beam theory of Bernoulli-Eular law of the uniform cross-section with the axial reinforcement.
The axial tension P increases the effective stiffness, reducing transverse deflection and bending moments.
EXTENDED EXAMPLE WITH F1 PERSPECTIVE
PULL-ROD SUSPENSION – THE BEAM THEORY APPLICATION
The Pull-Rod suspension is widely used in Formula 1 and high-performance vehicles. It uses a slender rod connecting the wheel upright to a rocker or bell-crank mounted low in in the chassis. When the wheel moves upward under load, it pulls the rod in tension, transmitting forces to the spring-damper assembly.
Although pull-rods are primarily designed as tension members, they can experience the lateral (Bending) loads due to the vehicle dynamics on the tracks. The proper deep analysis ensures, durability, stiffness, and safety under combined loading.
Mechanical Model:
Pinned-pinned slender beam carrying a large axial tensile force (P) generated by the wheel vertical load.
The lateral loads:
wheel side forces (cornering), toe/camber compliance, small geometric offsets, rocker misalignment, road-induced vibrations and manufacturing tolerances. In the example mentioned, the this load is represented as (w)
Length : L = 0.4 m
Diameter: d = 10-12 mm
axial tension under braking/ cornering: P = 4,000 -7,000 N
Transverse load: w = 30 -40 N/m
# Beam (tie-rod) under uniform load w and axial pull P
# Based on the maximum bending moment: EI*y'' - P*y = (w/2)*(x^2 -l*x)
# Automotive Application : tie rod idealized as a slender steel member
# Under Axial tension (P) and a small uniform transverse load (w)
# the code computes the deflection, slope, bending moment, and central deflection.
import numpy as np
import matplotlib.pyplot as plt
#------------------- Inputs (SI Units)-----------------
E = 210e9 # youngs modulus for mild steel [pa]
d = 0.012 # Tie-rod diameter [m]
L = 0.60 # Span between ends [m]
P = 5000 # Axial Tension [N]
w = 50 # Uniform transverse load per length [N/m]
#---------------------Section properties---------------
I = (np.pi*d**4)/64 # Second moment of area of circular section [m^4]
#---------------constants from closed form solution -----------
a = np.sqrt(P/(E*I))
# y(x) = c1*cosh(ax) + c2*sinh(ax) - (w/(2P))*(x^2 - L*x + 2/a^2)
# Applying pinned-end deflection boundary conditions: y(0) = 0 and y(L) = 0
c1 = w/(P*a**2) # from y(0) = 0
c2 = (w/(P*a**2))*(1 - np.cosh(a*L))/np.sinh(a*L) # from Y(L) = 0
def y_deflection(x):
return (c1*np.cosh(a*x) + c2*np.sinh(a*x) - (w/(2*P))*(x**2 - L*x +2/a**2))
def y_slope(x):
return (c1*a*np.sinh(a*x) + c2*a*np.cosh(a*x) - (w/(2*P)) * (2*x - L))
def y_curvature(x):
return (c1*a**2*np.cosh(a*x) + c2*a**2*np.sinh(a*x) - (w/(P)))
# Bending moment from governing equation:
# EI*y'' = p*y - (w/2) *(L*x - x**2)
def M_bending(x):
return P*y_deflection(x) - 0.5*w*(L*x - x**2)
# --------------descretise the span--------------------
x = np.linspace(0.0, L, 600)
y = y_deflection(x)
M = M_bending(x)
#---------------Central deflection and max bending moment------------
x_mid = L/2.0
Y_mid = float(y_deflection(x_mid))
# Find maximum bending moment location
dMdx = np.gradient (M,x)
critical_idx = np.where(np.sign(dMdx[:-1]) != np.sign(dMdx[1:]))[0]
candidates = list(critical_idx) +[0, len(x)-1]
i_star = max(candidates, key=lambda i: abs(M[i]))
x_star = x[i_star]
M_star = float(M[i_star])
# -------------- Printing key results ------------
print("===Inputs===")
print(f"E = {E/1e9:.1f} GPa, d = {d*1e3:.1f} mm, L = {L*1e3:.0f} mm")
print(f"P (axial pull) = {P:.0f} N, w (uniform load) = {w:.1f} N/m")
print(f"I = {I:.3e} m^4, a = {a:.3e} 1/m")
print("\n === results (automotive tie-rod====")
print(f" Central deflection y (L/2) = {Y_mid*1e3:.4f} mm")
print(f"Max bending Moment |M| = {abs(M_star):.3f} N-m at x = {x_star*1e3:.1f} mm")
# ----------- Plots ---------------------
plt.figure()
plt.plot(x,y, color = 'orange')
plt.xlabel("x [m]")
plt.ylabel("Deflection y(x) [m]")
plt.title("Tie-rod deflection under uniform load with axial pull")
plt.grid(True)
plt.show()
plt.figure()
plt.plot(x,M, color="orange")
plt.xlabel("x [m]")
plt.ylabel("Bending moment M(x) [N-m]")
plt.title("Bending moment diagram")
plt.show()