Example 1: The beam, by rank

We write \(\bm E\) for the six sets of supports in Figure 5.4.1, with \(A\) at the origin and \(B\) at \((L, 0)\) as in Example 4 of Static equilibrium, and let SymPy find the rank. Each support contributes the columns of the reactions it gives, and the null spaces of \(\bm E^\mathsf{T}\) and \(\bm E\) are the free motions and the self-stresses.

L = sp.symbols('L', positive=True)

def col(P, n):
    """Force and moment about the origin of a unit reaction along n at P."""
    return sp.Matrix([n[0], n[1], P[0]*n[1] - P[1]*n[0]])

moment = sp.Matrix([0, 0, 1])          # the moment of a fixed support
ex, ey = (1, 0), (0, 1)
A, B = (0, 0), (L, 0)

beam = {
    'pin + roller':           [col(A, ex), col(A, ey), col(B, ey)],
    'fixed':                  [col(A, ex), col(A, ey), moment],
    'roller + roller':        [col(A, ey), col(B, ey)],
    'pin + roller on a wall': [col(A, ex), col(A, ey), col(B, (-1, 0))],
    'pin + pin':              [col(A, ex), col(A, ey), col(B, ex), col(B, ey)],
    'fixed + roller':         [col(A, ex), col(A, ey), moment, col(B, ey)],
}

def counts(cases, n_rows=3):
    """Unknowns, rank, free motions and self-stresses of each set of supports."""
    rows = []
    for name, cols in cases.items():
        E = sp.Matrix.hstack(*cols)
        m, r = E.shape[1], E.rank()
        rows.append({'Supports': name, '$m$': m, 'rank': r,
                     'free motions': n_rows - r, 'self-stresses': m - r})
    return rows

beam_counts = counts(beam)

The unknowns, the rank and the two counts for each row of Figure 5.4.1 are

The table repeats the verdicts of Figure 5.4.1, and the row of the pin with the roller on the wall shows why counting alone missed it: three reactions, but rank two. Its two null spaces say what happens.

E_wall = sp.Matrix.hstack(*beam['pin + roller on a wall'])
v_wall = E_wall.T.nullspace()[0]
lam_wall = E_wall.nullspace()[0]
lam_pins = sp.Matrix.hstack(*beam['pin + pin']).nullspace()[0]

For the pin at \(A\) and the roller on the wall at \(B\), with the reactions in the order \(R_{Ax}\), \(R_{Ay}\), \(R_{Bx}\),

The free motion \(\bm v\) leaves the origin in place and turns the beam, so it turns about \(A\). The self-stress \(\bm\lambda_0\) has equal sizes for \(R_{Ax}\) and \(R_{Bx}\), and since the wall pushes along \(-x\), the pin and the wall squeeze the beam between them. They can do so with any force, and equilibrium does not see it. For the two pins, with the reactions in the order \(R_{Ax}\), \(R_{Ay}\), \(R_{Bx}\), \(R_{By}\), the self-stress is

that is \(R_{Ax} = -R_{Bx}\), the result of Case 5 in Example 4, now read as a set of reactions that can grow without any load.

Example 2: A block on links

A rectangular block, 3.2 m wide and 1.8 m high, is held by links pinned at both ends, in the four arrangements of Figure 5.7.1, which are those of Meriam and Kraige [1]. A link pinned at both ends carries a force along its own line, so each link gives one reaction along the link. We place \(A\), the lower left corner, at the origin.

w, h = sp.Rational(16, 5), sp.Rational(9, 5)            # the block, 3.2 m by 1.8 m
toward_A = -sp.Matrix([w, h])/sp.Matrix([w, h]).norm()   # link 3 of case (b) points at A

block = {
    '(a)': [col(A, ex), col(A, ey), col((w, 0), ey)],
    '(b)': [col(A, ex), col(A, ey), col((w, h), toward_A)],
    '(c)': [col((0, sp.Rational(3, 2)), ex), col((0, sp.Rational(3, 10)), ex),
            col((w, sp.Rational(9, 10)), (-1, 0))],
    '(d)': [col(A, ex), col(A, ey), col((w, 0), ey),
            col((sp.Rational(9, 5), h), (-sp.Rational(3, 5), -sp.Rational(4, 5)))],
}
block_counts = counts(block)
E_b, E_c, E_d = (sp.Matrix.hstack(*block[k]) for k in ('(b)', '(c)', '(d)'))

The counts for the four cases are

Cases (b) and (c) have the right number of links and rank two. With the link forces \(S_1\), \(S_2\), \(S_3\) in order, the free motion and the self-stress of case (b) are

The block turns about \(A\), where the three link lines meet, and the self-stress is link 3 pushing towards \(A\) while links 1 and 2 push back, in the ratio \(16 : 9 : \sqrt{337}\) set by the slope of link 3. For case (c),

The block slides up and down, across all three links, and the right link can push against the two left ones with twice the force of each. Case (d) is held, rank three, and its fourth link adds one self-stress,

a set of link forces that the block cannot feel. Their size depends on how the links stretch, and only an analysis of the deformation can give it.

Example 3: A three-hinged frame

Two bars, \(AC\) and \(CB\), are pinned together at \(C\) and to the ground at \(A\) and \(B\) (Figure 5.7.2 (a)). The load \(F\) acts downwards at the middle of \(AC\). In metres, \(A\) is at the origin, \(C\) at \((2, 1.5)\) and \(B\) at \((4, 0)\). The unknowns are the pin forces \(A_x\) and \(A_y\) on \(AC\), the joint force \(C_x\), \(C_y\) on \(AC\), which acts on \(CB\) with the opposite sign, and the pin forces \(B_x\) and \(B_y\) on \(CB\).

F = sp.symbols('F', positive=True)

def col2(P, n, body, sign=1):
    """A reaction in the six rows of the two bars, on bar 0 (AC) or bar 1 (CB)."""
    c = sp.zeros(6, 1)
    c[3*body:3*body + 3, 0] = sign*col(P, n)
    return c

A, C, B = (0, 0), (2, sp.Rational(3, 2)), (4, 0)
frame = [col2(A, ex, 0), col2(A, ey, 0),
         col2(C, ex, 0) + col2(C, ex, 1, -1),       # the joint: on AC, and opposite on CB
         col2(C, ey, 0) + col2(C, ey, 1, -1),
         col2(B, ex, 1), col2(B, ey, 1)]
E_frame = sp.Matrix.hstack(*frame)
f_frame = sp.Matrix([0, -F, -F, 0, 0, 0])            # F down at (1, 0.75): moment -F about the origin
lam_frame = E_frame.LUsolve(-f_frame)

E_roller = sp.Matrix.hstack(*frame[:4], frame[5])    # a roller at B: B_x is gone
v_roller = E_roller.T.nullspace()[0]

The equilibrium matrix has six rows, three for each bar, and six columns. Its rank and the reactions, in the order \(A_x\), \(A_y\), \(C_x\), \(C_y\), \(B_x\), \(B_y\), are

Six reactions and rank six: the frame is determinate. The pins push the two feet inwards, \(A_x = F/3\) and \(B_x = -F/3\), and the two vertical reactions carry \(3F/4\) and \(F/4\) of the load, as the lever arms of \(F\) about \(B\) and \(A\) require.

With a roller at \(B\) the column of \(B_x\) goes. Five reactions and rank five: no self-stress, and \(6 - 5 = 1\) free motion, which in the order \(u_x\), \(u_y\), \(\theta\) of \(AC\) and then of \(CB\) is

Bar \(AC\) turns clockwise about \(A\), and bar \(CB\) moves and turns so that its end at \(C\) follows \(AC\) and its end at \(B\), which moves by \((3, 0)\) for the turn \(1\), slides along the ground. The feet spread and the frame sags, as the dashed line in Figure 5.7.2 (b) shows: the horizontal pin force at \(B\) was what held the feet together.