We are recording the static equilibrium equations; the first two equations are independent of whether the force is in front or behind the joint - the next two, however, depend on this, so we need to consider two cases.

\begin{aligned} &\sum M_{A}=0 \\ &1 \cdot(x-2)-V_{B} \cdot 6=0 \\ &V_{B}=\frac{x-2}{6} \\ &\sum y=0 \\ &V_{A}-1+V_{B}=0 \\ &V_{A}=\frac{8-x}{6} \end{aligned}

The above equations are independent of where the unit force is, the following depend on:

\begin{aligned} &x \in(0 ; 6) \\ &\sum M_{P}^{P}=0 \\ &-H_{B} \cdot 5-V_{B} \cdot 2=0 \\ &H_{B}=-\frac{x-2}{15} \\ &\sum x=0 \\ &H_{A}+H_{B}=0 \\ &H_{A}=\frac{x-2}{15} \end{aligned} \begin{aligned} &x \in(6 ; 11) \\ &\sum M_{P}^{L}=0 \\ &-H_{A} \cdot 5+V_{A} \cdot 4=0 \\ &H_{A}=\frac{16-2 x}{15} \\ &\sum x=0 \\ &H_{A}+H_{B}=0 \\ &H_{B}=-\frac{16-2 x}{15} \end{aligned}

To calculate the shear force functions, we consider two cases - the force is in front or behind the section "k".

section k

Interval I \(x\in (0;4) Q_k=-V_B \)

in the interval (0;4) the influence line Qk is equal to the negative of Vb

\begin{aligned} &M_{k}=V_{B} \cdot 4+H_{B} \cdot 5=\frac{x-2}{3} \\ &M_{k}(0)=-2 / 3 \\ &N_{k}=H_{B} \quad M_{k}(4)=2 / 3 \end{aligned}

in the interval (0;4) the influence line Nk is equal to Hb

Interval II \( x\in (4;11) Q_k=V_A\)

in the interval (4;11) the influence line Qk is equal to Va

\begin{aligned} M_{k}=V_{A} \cdot 2-H_{A} \cdot 5 \end{aligned} \begin{aligned} &x \in(4 ; 6) \\ &M_{k}=\frac{8-x}{3}-\frac{x-2}{3}=\frac{10-2 x}{3} \\ &M_{k}(4)=2 / 3\\ &M_{k}(6)=-2 / 3 \end{aligned} \begin{aligned} &x \in(6 ; 11) \\ &M_{k}=\frac{8-x}{3}-\frac{16-2 x}{3}=\frac{x-8}{3} \\ &M_{k}(2)=-2 / 3\\ &M_{k}(11)=1 \end{aligned} \begin{aligned} N_{k}=-H_{A} \end{aligned}

in the interval (4;11) the influence line Nk is equal to the negative of Ha

Charts

Influence lines

Structural Analysis - Influence lines - Example 5

Example 5