# Tension members (6.2.3)

• Subjected to axial tensile forces,
• quite common: truss members, bracing members.

There are no restrictions for cross-section type, when a member is loaded by a tension. The most important is area of the section. The section has to be chosen such that

• structure is easy to assemble (for example circular hollow section is difficult to connect; profiles T, I, L, U are stiffer, easier to connect and look better than a rod);
• nice (if exposed to spectators).

The value watched is stress from tension (member axially loaded): $f=P/A$. According to the Eurocode:

$$\begin{equation} N_{Ed} \le N_{t,Rd}, \end{equation}$$

where $N_{Ed}$ is design normal force from a load, $N_{t,Rd}$ is design tension resistance and is taken as smaller of

\begin{align} N_{pl,Rd} &= \frac{A \color{red}{f_y}}{\color{red}{\gamma_{M0}}} \\ N_{u, Rd} &= \frac{0.9 A_{net} \color{red}{f_u}}{\color{red}{\gamma_{M2}}} \end{align}

$N_{pl,Rd}$ is plastic resistance computed with yield strength $f_y$ (there is still a reserve to utlimate strength). $N_{u, Rd}$ is the ultimate resistance computed from the ultimate strength of the material; there is no reserve and net area (holes for bolts are subtracted) is computed with.

## Partial factors

Recommended values for the basic material (usually a plate, a rolled profile):

 Common, general cases $\gamma_{M0}$ 1.0—1.15 Cases involving loss of stability $\gamma_{M1}$ 1.0—1.15 Cases involving tension: ultimate state, sections weakened by joints, bolts $\gamma_{M2}$ 1.25—1.3

## Holes

In ultimate state, the holes for bolts needs to be subtracted from the section area. The exact diameter, which needs to be subtracted depends on the fabrication procedure. Usually 1 to 3 mm (depends also on bolt size).

The symbols used are

 $A_g$ gross area of the member $A_{eff}$ effective area (usually $A_{eff} = A_{net}$) $A_{net}$ net area (without the holes for bolts)

#### Example (net area of a plate)

Find the net area of the plate exposed to a tension.  Bolts M16, holes ø18
\begin{align} A_g &= 120 \times 20 = 2400 \ \text{mm}^2 \nonumber \\ A_{net} &= 2400 - 18 \times 20 \times 2 = 1680 \ \text{mm}^2 \nonumber \end{align}

Note: it is important to place bolts in such way, that an eccentricity is avoided. Otherwise moments will produced $\implies$ additional stresses to deal with.

### Effective area for stagerred fasteners

The holes for bolts might be not placed in two perpendicular/regular lines but can be placed according to the picture:  Possible failures when the holes are staggered

In such cases, all possibilities of failure have to be taken into account and evaluated. The net area has to be computed for each variant and the worst scenario is taken.

$$A_{net} = \color{red}{A- \sum d_0 t} + \sum \frac{s^2 t}{4p}$$

The term ${A- \sum d_0 t}$ is net area as in the case from the above example (cross section area $A -$ holes and the last term is effect of staggering.

#### Example (staggered fasteners)

Find the net area of the plate exposed to a tension.  $$A = bt = 180 \times 10 = 1800\ \text{mm}^2$$  All paths of fracture has to be considered. Net area for each case is determined and finally the smallest one is taken as $A_{net}$

The net area is taken as the smallest one from the four cases marked in the above picture.

\left. \begin{align} A_{net1} = A - d_0 t = 1800 - 18 \times 10 = 1620 \ \text{mm}^2 \nonumber \\ A_{net2} = A - 2 d_0 t = 1800 - 2 \times 18 \times 10 = 1440 \ \text{mm}^2 \nonumber \\ A_{net3} = A - 2 d_0 t + \frac{40^2 \times 10}{4 \times 60} = 1507 \ \text{mm}^2 \nonumber \\ A_{net4} = A - 3 d_0 t + 2 \times \frac{40^2 \times 10}{4 \times 60} = {\underline{1394 \ \text{mm}^2}} \nonumber \end{align} \right\} \implies \underline{\underline{A_{net} = 1394 \ \text{mm}^2}}

#### Example (design tension resistance)

Find the strength (design tension resistance $N_{t,Rd}$) of the above plate. The basic material is steel S235.

Nominal values of yield strength $f_y$, ultimate tensile strength $f_u$:

• $f_y = 235 \ \text{MPa}$,
• $f_u = 360 \ \text{MPa}$
\begin{align} A_g &= 2400\ \text{mm}^2 \nonumber \\ A_{net} &= 1680\ \text{mm}^2 \nonumber \\ N_{pl,Rd} &= \frac{A f_y}{\gamma_{M0}} = \frac{2400\times10^{-6} \times 235\times10^{6}}{1.15} = 490 \ \text{kN} \nonumber \\ N_{u, Rd} &= \frac{0.9 \times A_{net} \times f_u}{\gamma_{M2}} = \frac{0.9 \times 1680 \times 360}{1.3} = 419 \ \text{kN} \nonumber \end{align}

Strength of the member is determined by $N_{u,Rd}$ computed for ultimate strength with holes considered, since there is no reserve already.

$$\underline{\underline{N_{Ed} = 419\ \text{kN}}}$$

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