# 1.5.K

## Video Given symbols: $$a, F, q$$

Please calculate the support reactions at A and B and the forces in hinge G. Proceed as follows.

## Steps

### 1. Free Body Diagram

• at G: Through the hinge between the left beam and the right beam.

• at A: At the boundary between the left beam and its support.

• at B: At the boundary between the right beam and its support.

Draw two free body diagrams: One for the left beam - and one for the right beam.

Solution ### 2. Resultant and Equilibrium Conditions

For the given symbols: Please calculate the resultant of the distributed load. And - for each beam - provide 3 equations (equilibrium conditions). Enumerate the equations, and count the unknowns. Show that you get 6 equations for 6 unknowns.

Solution ### 3. Solution using Symbols

Please calculate all unknowns in terms of the given symbols.

Solution

Solution for the six unknowns $$A_h, A_v, M_A, B_v, G_h, G_v$$:

\begin{align*} A_{h} &= 0 \\ A_{v} &= - F \\ B_{v} &= 2 F + 2 a q \\ G_{h} &= 0 \\ G_{v} &= F \\ M_A &= F a \end{align*}

### 4. Solution using Quantities

Please calculate all unknowns in terms of the quantities:

$\begin{split}a&= 1 \, \mathrm{m}\\ F&= 1 \, \mathrm{N}\\ q&= 1 \, \tfrac{\mathrm{N}}{\mathrm{m}}\end{split}$

Solution

$\begin{split}A_{h} &= 0 \\ A_{v} &= -1 \,\mathrm{N} \\ B_{v} &= 4 \,\mathrm{N} \\ G_{h} &= 0 \\ G_{v} &= 1 \,\mathrm{N} \\ M_A &= 1 \,\mathrm{Nm}\end{split}$

### 5. Support Reactions Only

Show how it is possible to find the support reactions - without calculating the joint forces.

Solution Hint

• For one body, only one equilibrium condition is used.

• A system of four equations for for unknowns is solved.

Solution with Python: Copy - Paste - Play

• Copy: Source Code (see below) view and copy.

• Paste: Paste as Python-Notebook on:

• Colab (Google-Account needed)

• Play: Run.

Source Code

from sympy import *
from sympy.physics.units import kg, m, s

Newton = kg*m/s**2

a, F, q = var("a, F, q")

sub_list=[
( a, 1 *m         ),
( F, 1 *Newton    ),
( q, 1 *Newton/m  ),
]

Ah, Av, MA, Bv, Gh, Gv \
= var("Ah, Av, MA, Bv, Gh, Gv")

R = 2*a*q

# equilibrium conditions:
eq1 = Eq(  Gh )
eq2 = Eq(  - Bv + F + Gv + R  )
eq3 = Eq(  2*a*F + a*(-Bv + R)  )
eq4 = Eq(  Ah + Gh)
eq5 = Eq(  Av + Gv)
eq6 = Eq(  Gv*a - MA )

unks = [Ah, Av, MA, Bv, Gh, Gv]
eqs = [eq1, eq2, eq3, eq4, eq5, eq6]

sol = solve(eqs, unks)
pprint(sol)

Ah, Av, MA, Bv, Gh, Gv \
= sol[Ah], sol[Av], sol[MA], sol[Bv], sol[Gh], sol[Gv]

pprint("\nAh, Av, Bv, Gh, Gv, MA in SI units:")
for s in [Ah, Av, Bv, Gh, Gv]:
tmp = s
tmp = tmp.subs(sub_list)
tmp /= Newton
pprint(tmp)

tmp = MA
tmp = tmp.subs(sub_list)
tmp /= Newton*m
pprint(tmp)

pprint("\nAh, Av, Bv, MA in SI units:")
Ah, Av, MA, Bv= var("Ah, Av, MA, Bv")
eq1 = Eq(  2*F*a + a*(-Bv + R) )
eq2 = Eq(  Ah )
eq3 = Eq(  Av + Bv - R - F )
eq4 = Eq(  2*a*Bv -2*a*R -3*a*F - MA )

unks = [Ah, Av, MA, Bv]
eqs = [eq1, eq2, eq3, eq4]

sol = solve(eqs, unks)
pprint(sol)

# {Ah: 0, Av: -F, Bv: 2⋅F + 2⋅a⋅q, Gh: 0, Gv: F, MA: F⋅a}
#
# Ah, Av, Bv, Gh, Gv, MA in SI units:
# 0
# -1
# 4
# 0
# 1
# 1
#
# Ah, Av, Bv, MA in SI units:
# {Ah: 0, Av: -F, Bv: 2⋅F + 2⋅a⋅q, MA: F⋅a}