We write the static equilibrium equations; the first two equations are independent of whether the force is in front of 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-V_{C} \cdot 6=0 \\ &V_{C}=x / 6 \\ &\sum y=0 \\ &V_{A}-1+V_{C}=0 \\ &V_{A}=\frac{6-x}{6} \end{aligned}

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

\begin{aligned} &x \in(0 ; 4) \\ &\sum M_{P}^{P}=0 \\ &-H_{C} \cdot 6-V_{C} \cdot 2=0 \\ &H_{C}=-\frac{x}{18} \\ &\sum x=0 \\ &H_{A}+H_{C}=0 \\ &H_{A}=\frac{x}{18} \end{aligned} \begin{aligned} &x \in(4 ; 6) \\ &\sum M_{P}^{P}=0 \\ &-H_{C} \cdot 6-V_{C} \cdot 2+1(x-4)=0 \\ &H_{C}=\frac{3(x-4)}{3 \cdot 6}-\frac{x}{18}=\frac{2 x-12}{18} \\ &\sum x=0 \\ &H_{A}+H_{C}=0 \\ &H_{A}=\frac{12-2 x}{18} \end{aligned}

To calculate the cross-sectional force functions, we consider two cases - the force in front or behind section "l".

Section l

Interval I \( x \in (0,6) \)

\begin{aligned} Q_{l}=-H_{A} \end{aligned}

as the influence line Ql, we redraw the line with reversed sign

\begin{aligned} M_{l}=-2 H_{A} \end{aligned}

as the influence line Ml, we redraw the line with reversed sign and doubled value

\begin{aligned} N_{l}=-V_{A} \end{aligned}

as the influence line Nl, we redraw the line with reversed sign

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

Section k

Interval I \( x \in (0,2) \)

\begin{aligned} Q_{k}=-V_{C} \end{aligned}

as the influence line Qk, we redraw the line with reversed sign (only in the interval 0;2)

\begin{aligned} &M_{k}=V_{C} \cdot 4+H_{C} \cdot 6 \\ &M_{k}=\frac{4 x}{6} \cdot \frac{3}{3}-\frac{6 x}{18}=\frac{x}{3} \\ &M_{k}(0)=0\\ &M_{k}(2)=0.667\\ \end{aligned} \begin{aligned} N_{k}=H_{C} \end{aligned}

as the influence line Nk, we redraw the line with reversed sign (only in the interval 0;2)

Interval II \( x \in (2,6) \)

\begin{aligned} Q_{k}=V_{A} \end{aligned}

in the interval 2;6 as the influence line Qk, we redraw the line with reversed sign

\begin{aligned} M_{k}=V_{A} \cdot 2-H_{A} \cdot 6 \end{aligned} \begin{aligned} &x \in(2 ; 4) \\ &M_{k}=\frac{2(6-x)}{6}-\frac{6 x}{18}=\frac{6-2 x}{3} \\ &M_{k}(2)=0.667\\ &M_{k}(4)=-0.667 \end{aligned} \begin{aligned} &x \in(4 ; 6) \\ &M_{k}=\frac{2(6-x)}{6}-\frac{12-2 x}{3}=\frac{x-6}{3} \\ &M_{k}(2)=-0.667\\ &M_{k}(4)=0 \end{aligned} \begin{aligned} N_{k}=H_{A} \end{aligned}

Charts

Influence lines

Structural Analysis - Influence lines - Example 3

Example 3