5.4 Equilibrium Examples

Example 1 – Beam with variable force direction

We want to determine how all reaction forces vary as the direction of the 4000 N force changes.

We isolate the beam and draw the free body diagram. The pin at \(A\) gives two reaction components (\(R_{Ax}\), \(R_{Ay}\)) and the roller at \(E\) gives one horizontal component (\(R_{Ex}\)). The 4000 N force at \(B\) acts at angle \(\theta\) and the 200 N force at \(C\) acts vertically. We take moments about \(A\).

Three unknowns and three independent equations (two force, one moment) so the system is solvable.

\[ \sum \bm F = 0: \quad \bm R_A + \bm F_B + \bm R_E + \bm F_C = \bm 0 \]

\[ \sum \bm M_A = 0: \quad \bm r_{AB} \times \bm F_B + \bm r_{AC} \times \bm F_C + \bm r_{AE} \times \bm R_E = \bm 0 \]

import sympy as sp
from mechanicskit import la

a = 4                                                    # [m] Distance ED
b = 3                                                    # [m] Distance AE
F_B = 4000                                               # [N] Force magnitude at B
F_C = 200                                                # [N] Force magnitude at C

theta = sp.Symbol('theta', real=True)                    # [rad] Angle at B
R_Ax, R_Ay, R_Ex = sp.symbols('R_Ax R_Ay R_Ex', real=True)

RR_A = sp.Matrix([R_Ax, R_Ay, 0])
FF_B = F_B * sp.Matrix([sp.cos(theta), sp.sin(theta), 0])
RR_E = sp.Matrix([R_Ex, 0, 0])
FF_C = F_C * sp.Matrix([0, 1, 0])

rr_AB = a * sp.Matrix([1, 0, 0])
rr_AC = sp.Matrix([2*a, -b, 0])
rr_AE = b * sp.Matrix([0, -1, 0])

N2 = sp.Eq(RR_A + FF_B + RR_E + FF_C, sp.Matrix([0, 0, 0]))
E2 = sp.Eq(rr_AB.cross(FF_B) + rr_AC.cross(FF_C) + rr_AE.cross(RR_E), sp.Matrix([0, 0, 0]))

sol = sp.solve([N2, E2], [R_Ax, R_Ay, R_Ex])
sol | la

\[ \begin{aligned} R_{Ax} &= \frac{16000 \sin{\left(\theta \right)}}{3} - 4000 \cos{\left(\theta \right)} + \frac{1600}{3} \\ R_{Ay} &= - 4000 \sin{\left(\theta \right)} - 200 \\ R_{Ex} &= - \frac{16000 \sin{\left(\theta \right)}}{3} - \frac{1600}{3} \end{aligned} \]

Code

import matplotlib.pyplot as plt
from mechanicskit import fplot
import numpy as np

plt.style.use('seaborn-v0_8-whitegrid')
fig, ax = plt.subplots(figsize=(8, 5))
fplot(sol[R_Ax], (theta, 0, 2*sp.pi), ax=ax, label='$R_{Ax}$', color='red')
fplot(sol[R_Ay], (theta, 0, 2*sp.pi), ax=ax, label='$R_{Ay}$', color='green')
fplot(sol[R_Ex], (theta, 0, 2*sp.pi), ax=ax, label='$R_{Ex}$', color='blue')
ax.set(xlabel=r'Angle $\theta$ [rad]', ylabel='Reaction force [N]')
ax.legend()
plt.show()

Example 2 – Beam with wire at variable position

A beam with mass \(m = 475\) kg is supported at \(A\) by a pin and by a wire connected at \(B\). The wire attachment point \(B\) can slide along the beam. A 10 kN force acts downward at \(C\). We want to determine the reaction forces as a function of the wire position \(r_{OBx}\).

The free body diagram and vector definitions are shown below. We take moments about \(O\) (the left end of the beam). The wire direction \(\bm e_{BD}\) is computed from the geometry, and the wire force is \(\bm F_B = F_B \bm e_{BD}\).

\[ \sum \bm F = 0: \quad \bm R_A + \bm W + \bm F_C + \bm F_B = \bm 0 \]

\[ \sum \bm M_O = 0: \quad \bm r_{OA} \times \bm R_A + \bm r_{OG} \times \bm W + \bm r_{OC} \times \bm F_C + \bm r_{OB} \times \bm F_B = \bm 0 \]

g = 9.81                                                 # [m/s^2]
m = 475                                                  # [kg]

r_OBx = sp.Symbol('r_OBx', positive=True)
R_Ax, R_Ay, F_B = sp.symbols('R_Ax R_Ay F_B', real=True)

rr_OA = sp.Matrix([0.12, 0, 0])
rr_OG = sp.Matrix([2.5, 0, 0])
rr_OC = sp.Matrix([3.5, -0.25, 0])
rr_OB = sp.Matrix([r_OBx, 0.25, 0])

r_ODy = sp.tan(sp.rad(25)) * 5
rr_OD = sp.Matrix([0, r_ODy, 0])

# Wire direction from B toward D
ee_BD = (rr_OD - rr_OB).normalized()

RR_A = sp.Matrix([R_Ax, R_Ay, 0])
WW   = m * g * sp.Matrix([0, -1, 0])
FF_C = sp.Matrix([0, -10000, 0])
FF_B = F_B * ee_BD

sumF = sp.Eq(RR_A + WW + FF_C + FF_B, sp.Matrix([0, 0, 0]))
sumM = sp.Eq(
    rr_OA.cross(RR_A) + rr_OG.cross(WW) + rr_OC.cross(FF_C) + rr_OB.cross(FF_B),
    sp.Matrix([0, 0, 0])
)

sol = sp.solve([sumF, sumM], [R_Ax, R_Ay, F_B])
sol | la

\[ \begin{aligned} F_{B} &= \frac{8978041.0 \sqrt{16.0 r_{OBx}^{2} + 69.3248264953997}}{1865.23063261999 r_{OBx} - 199.827675914399} \\ R_{Ax} &= \frac{8978041.0 r_{OBx}}{466.307658154999 r_{OBx} - 49.9569189785998} \\ R_{Ay} &= \frac{27343814.766551 r_{OBx}}{1865.23063261999 r_{OBx} - 199.827675914399} - \frac{77681968.2425773}{1865.23063261999 r_{OBx} - 199.827675914399} \end{aligned} \]

Code

plt.style.use('seaborn-v0_8-whitegrid')
fig, ax = plt.subplots(figsize=(8, 5))
fplot(sol[F_B],  (r_OBx, 0.5, 5), ax=ax, label='$F_B$',   color='black')
fplot(sol[R_Ax], (r_OBx, 0.5, 5), ax=ax, label='$R_{Ax}$', color='blue')
fplot(sol[R_Ay], (r_OBx, 0.5, 5), ax=ax, label='$R_{Ay}$', color='red')
ax.set(xlabel='Placement of B [m]', ylabel='Force [N]')
ax.legend()
plt.show()

When the wire attachment approaches \(A\) the forces blow up because \(\bm F_B\)’s line of action passes through the support, making the moment arm vanish. The wire force \(F_B\) becomes negative when the line of action intersects between \(O\) and \(A\), which is not physical for a wire (wires can only pull).

Example 3 – Multi-body structure

The structure is supported at \(A\), \(B\) and \(C\) from the floor. A person with mass \(m = 80\) kg stands at point \(D\).

We want to determine the reaction forces at \(A\), \(B\), \(C\) and the joint forces at \(D\), \(E\), \(G\).

By symmetry, points \(D\) and \(G\) are centered between \(A\) and \(B\), so the reactions at \(A\) and \(B\) are equal. We can reduce this to a 2D problem where the computed reaction represents the sum of \(A\) and \(B\).

We need three free body diagrams: body ODG, body CEG, and body DE. Newton’s third law ensures that the joint forces at connections are equal and opposite between bodies.

Before writing equations we must decide what type of support acts at each floor contact. The problem figure does not specify this, so we have to make a modeling choice (see Static determinacy for background). If all three supports are pins (rough surfaces), we get \(2 + 2 + 2 = 6\) reaction components from the floor but only \(3 \times 3 = 9\) equations for \(6 + 6 = 12\) unknowns (6 floor reactions + 6 joint forces at D, E, G). That is 12 unknowns and 9 equations, the system is statically indeterminate.

To make the system determinate we model \(C\) as a roller (smooth surface, only a normal force in the \(y\)-direction). Now \(O\) contributes 2 unknowns and \(C\) contributes 1, giving \(2 + 1 + 6 = 9\) unknowns and 9 equations. This is the minimum number of supports that still prevents the structure from moving, and it makes the problem solvable with equilibrium alone.

Body 1 (ODG):

\[ \sum \bm F = 0: \quad \bm R_O + \bm R_D + \bm F + \bm R_G = \bm 0 \]

\[ \sum \bm M_O = 0: \quad \bm r_{OD} \times \bm R_D + \bm r_{OD} \times \bm F + \bm r_{OG} \times \bm R_G = \bm 0 \]

Body 2 (CEG):

\[ \sum \bm F = 0: \quad \bm R_C + \bm R_E + (-\bm R_G) = \bm 0 \]

\[ \sum \bm M_C = 0: \quad \bm r_{CE} \times \bm R_E + \bm r_{CG} \times (-\bm R_G) = \bm 0 \]

Body 3 (DE):

\[ \sum \bm F = 0: \quad (-\bm R_D) + (-\bm R_E) = \bm 0 \]

\[ \sum \bm M_D = 0: \quad \bm r_{DE} \times (-\bm R_E) = \bm 0 \]

theta = sp.rad(65)
m = 80                                                   # [kg]
g = 9.82                                                 # [m/s^2]

rr_OD = sp.Matrix([480/sp.tan(theta), 480, 0])
rr_OG = sp.Matrix([905/sp.tan(theta), 905, 0])
rr_CE = sp.Matrix([-480/sp.tan(theta), 480, 0])
rr_CG = sp.Matrix([-905/sp.tan(theta), 905, 0])
rr_OE = rr_OG - (rr_CG - rr_CE)
rr_DE = rr_OE - rr_OD

R_Ox, R_Oy, R_Dx, R_Dy, R_Gx, R_Gy, R_Ex, R_Ey, R_Cy = sp.symbols(
    'R_Ox R_Oy R_Dx R_Dy R_Gx R_Gy R_Ex R_Ey R_Cy', real=True)

RR_O = sp.Matrix([R_Ox, R_Oy, 0])
RR_D = sp.Matrix([R_Dx, R_Dy, 0])
RR_G = sp.Matrix([R_Gx, R_Gy, 0])
RR_E = sp.Matrix([R_Ex, R_Ey, 0])
RR_C = sp.Matrix([0, R_Cy, 0])
FF   = sp.Matrix([0, -m*g, 0])

# Body 1: ODG
sumF1 = sp.Eq(RR_O + RR_D + FF + RR_G, sp.Matrix([0, 0, 0]))
sumM1 = sp.Eq(rr_OD.cross(RR_D) + rr_OD.cross(FF) + rr_OG.cross(RR_G), sp.Matrix([0, 0, 0]))

# Body 2: CEG
sumF2 = sp.Eq(RR_C + RR_E + (-RR_G), sp.Matrix([0, 0, 0]))
sumM2 = sp.Eq(rr_CE.cross(RR_E) + rr_CG.cross(-RR_G), sp.Matrix([0, 0, 0]))

# Body 3: DE
sumF3 = sp.Eq((-RR_D) + (-RR_E), sp.Matrix([0, 0, 0]))
sumM3 = sp.Eq(rr_DE.cross(-RR_E), sp.Matrix([0, 0, 0]))

sol = sp.solve(
    [sumF1, sumM1, sumF2, sumM2, sumF3, sumM3],
    [R_Ox, R_Oy, R_Dx, R_Dy, R_Gx, R_Gy, R_Ex, R_Ey, R_Cy]
)
sol | la

\[ \begin{aligned} R_{Cy} &= 208.33591160221 \\ R_{Dx} &= 206.869437880414 \\ R_{Dy} &= 0.0 \\ R_{Ex} &= -206.869437880414 \\ R_{Ey} &= 0.0 \\ R_{Gx} &= -206.869437880414 \\ R_{Gy} &= 208.33591160221 \\ R_{Ox} &= 0.0 \\ R_{Oy} &= 577.26408839779 \end{aligned} \]

The reaction forces from the floor (\(R_{Oy}\) and \(R_{Cy}\)) are positive, confirming they point upward as expected. Since there is no external horizontal force we get \(R_{Ox} = 0\). The negative \(R_{Ex}\) and positive \(R_{Dx}\) tell us that body DE is in tension.