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.
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:
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.
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 $f_u$:
- $f_y = 235 \ \text{MPa}$,
- $f_u = 360 \ \text{MPa}$
Strength of the member is determined by $N_{u,Rd}$ computed for ultimate strength with holes considered, since there is no reserve already.
$$ \underline{\underline{N_{Ed} = 419\ \text{kN}}} $$