A steel portal frame has dimensions, plastic moment capacities and applied loads as shown in the figure. If the collapse load of the beam is $α \frac{M_{p}}{L}$ , the value of $α$ is ____.

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Solution

The possible locations of plastic hinges are at A, B, C, D and E.

Number of possible plastic hinges,
$N = 5$
Degree of indeterminacy,
$r = (6 - 3) = 3$
Number of possible independent mechanisms,
$n = (N - r) = (5 - 3) = 2$

The two independent mechanisms are
1. Beam mechanism
2. sway mechanism

These two independent mechanism can also form a combined mechanism.

1. Beam Mechanism

$M_{p} θ + 2 M_{p} θ + 2 M_{P} θ + M_{p} θ = 2 P \times L θ$
$\Rightarrow 6 M_{p} θ = 2 P L θ$
$\Rightarrow P = (\frac{3 M_{p}}{L})$

2. Sway Mechanism:

Since the axial deformations are neglected, both the columns will sway by same amount.

$2 L ϕ = L θ = Δ$

$\Rightarrow θ = 2 ϕ$

Hence, equating intrenal work done to external work, we have

$M_{p} θ + M_{p} θ + M_{p} ϕ + M_{p} ϕ = P \times L θ$

$\Rightarrow M_{p} θ (1 + 1 + \frac{1}{2} + \frac{1}{2}) = P L \times θ$

$\Rightarrow 3 M_{p} θ = P L θ$

$\Rightarrow P = (\frac{3 M_{p}}{L})$

3. Combined Mechanism:

$\Rightarrow 2 L ϕ = L θ = Δ$

$\Rightarrow θ = 2 ϕ$

By the principal of virtual work, Internal work done = External work done

$\Rightarrow M_{p} θ + 2 M_{p} θ + 2 M_{p} θ + M_{p} θ + 2 M_{p} ϕ = P \times L θ + 2 P \times L θ$

$\Rightarrow 6 M_{p} θ + 2 M_{p} ϕ = 3 P L θ$

$\Rightarrow 6 M_{p} θ + M_{p} θ = 3 P L θ$

$\Rightarrow P = \frac{7 M_{p}}{3 L}$

$∴ The collapse load (p) = m i n [\frac{3 M_{p}}{L}, \frac{3 M_{p}}{L}, \frac{7 M_{p}}{3 L}]$

$= \frac{7 M_{p}}{3 L}$

$= 2.33 (\frac{M_{p}}{L})$

$∴ The required value of α = 2.33$

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