## 物理代写|结构力学代写Structural Mechanics代考|Exact Behaviour

In Fig. 6.1(a.i), bars are either $L$ or $\sqrt{2} L$ long, have the same cross-sectional area $A$ and Young’s Modulus $E$. The symmetry of layout and the force, $F$, applied horizontally ensure that the outer bar tensions are the same and equal to $T_1$ : the middle bar tension is $T_2$. Equilibrium of the loaded pin-joint in Fig. 6.1(b) gives us a single statement for two unknown bar tensions: $F=\sqrt{2} T_1+T_2$.

The same joint displaces by amount $d$ horizontally. During the initial elastic response, the bar extensions are defined by $e_1=T_1(\sqrt{2} L) / A E$ and $e_2=T_2 L / A E$. The extension $e_2$ is equal to $d$ by definition: the other bars also rotate as well as extend with small displacement components respectively normal to and along the bar. The corresponding vector diagram in Fig. 6.1(c) shows that $e_1=d / \sqrt{2}$. The ratio $e_2 / e_1$ is thus $\sqrt{2}$; therefore, $T_2 / T_1=2$.

The same ratio can be found as quickly by using the Virtual Work method of Chapter 5 . Because of the indeterminate nature, we can set either virtual tension to be zero provided we satisfy equilibrium, in order to solve directly for the other. Setting $F^=1$ and $T_2^$ to be zero, $T_1^$ equals $1 / \sqrt{2}$, and with $e_1, e_2$ and $d$ as our real quantities, Virtual Work states that: $$\begin{gathered} \text { Eq. }(5.1) \rightarrow 1 \cdot d=2 T_1^ \cdot e_1+T_2^* \cdot e_2=2 \times \frac{1}{\sqrt{2}} \frac{\sqrt{2} T_1 L}{A E}+0 \ \rightarrow \quad T_1=\frac{A E}{2 L} \cdot d \end{gathered}$$
Setting $T_1^$ to be zero, $T_2^$ equals unity for $F^*=1$, and Virtual Work again returns $T_2=(A E / L) \cdot d$

The elastic response considers how $F$ and $d$ are related. Substituting both tensions into the original equilibrium equation and re-arranging non-dimensionally:
$$F=\sqrt{2} T_1+T_2 \quad \rightarrow \quad F=\frac{\sqrt{2}+1}{\sqrt{2}} \cdot \frac{A E}{L} \cdot d$$
and the displacement builds up linearly as the load increases, until yielding is reached.

## 物理代写|结构力学代写Structural Mechanics代考|Lower Bound Trials

Even for this simplest of indeterminate trusses, our working has not been trivial to account for exact displacements and true equilibrium tensions. The ultimate loading, however, turns out to be a simple equilibrium calculation when both bar tensions are set to be $T_Y$, and the benefit of considering equilibrium solutions by themselves is perhaps obvious.

These final bar tensions should, however, be on the cusp of reaching $T_Y$ so that the yielding condition is not strictly violated. Any other limiting equilibrium solution will obviously have either of $T_1$ or $T_2$ just equal to $T_{\mathrm{Y}}$ but not both, giving a lower ultimate load compared to the optimal value. For example, if we set $T_1=T_{\mathrm{Y}}$ and $T_2=0$, the ultimate load is $F=\sqrt{2} T_1+T_2=\sqrt{2} T_{\mathrm{Y}}$ or $f=\sqrt{2}$.

This result is plotted as the first dashed vertical line in Fig. 6.2(a). The corresponding true tensions from the intersection points are non-zero and can be read off or calculated from Eqs. $6.3$ to give $t_1=1 /(1+\sqrt{2})=0.414$ and $t_2=0.828$. Being less than unity, both tensions are elastic and our structure is safe but operating well below its ultimate capacity.

Imagine, however, that we are ignorant of the true ultimate load but wish to improve our conservative trial solution with a higher, elastic value for $T_2$ whilst retaining $T_1$ at $T_{\mathrm{Y}}$. In particular, set $T_2$ conveniently to $(2-\sqrt{2}) T_{\mathrm{Y}}\left(=0.585 T_{\mathrm{Y}}\right)$, which improves our load estimate, now $F=\sqrt{2} T_1+T_1=2 T_{\mathrm{Y}}(f=2)$.

When plotted as another dashed line in Fig. 6.2(a), $(f=2)$, the true tensions, in fact, behave oppositely: $t_1=1 / \sqrt{2}=0.707$, which is elastic, and $t_2=1$, at yielding.

## 物理代写|结构力学代写结构力学代考|精确行为

$$F=\sqrt{2} T_1+T_2 \quad \rightarrow \quad F=\frac{\sqrt{2}+1}{\sqrt{2}} \cdot \frac{A E}{L} \cdot d$$

## 物理代写|结构力学代写结构力学代考|下限试验

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