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For the truss above known are the following: l= 1 + 0.02*n [m], h = 2 + 0.01*n [m], F_1 = 200 + 2*n [N], F_2= 500 + 6*n [N], F3 = 400 + 5*n [N], Young's modulus E = 2*10^11 [N/m2], allowable stress σ_a= 8*10^8 [N/m2] In the amounts above, n a natural number.


  1. Reaction forces at A and E .

  2. Tension forces in all members of the truss. Draw free body diagrams for all joints and members; indicate extension or compression for the member forces.

  3. Diameters of all the members considering they all have circular cross-sections. The diameters have to be selected from the following sequence of diameters (in mm): 2, 5, 8, 10, 12, 16, 20, 24, 30, 36, 40, 50, 60, 70.

  4. The axial deformation (Δl) of bars BC and FG.

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closed as too localized by dmckee Apr 20 '12 at 18:38

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Physics.SE is not a homework help site, and we have a strict policy on homework questions: we will answer well formed questions about the concepts your problem uses but we will not work you exercises for you. – dmckee Apr 20 '12 at 18:41
  1. Reaction forces can easily be obtained noting that internal forces cancel out (due to 3rd Newton law). So you have only five external forces ($F_1$, $F_2$, $F_3$, $N_\text{left}$ and $N_\text{right}$, which is perpendicular to ground). All you have to do is $\sum_i \vec{\tau}_i = 0$ around left support and then $\sum_i \vec{F}_i = 0$.

  2. Forces within members are strictly axial (along their axes) and no moment is transfered over joint. You therefore have only two forces per each member at its ends and these forces are equal and in the oposite directions. So you can draw $\sum_i \vec{F}_i = 0$ for each joint.

  3. Stress is $\sigma = \frac{F}{A}$, and this must be less than maximum allowable $\sigma = \frac{F}{A} \le \sigma_\text{max}$, meaning that $$A \ge \frac{F}{\sigma_\text{max}}.$$ Since you've obtained forces for each member in the previous step, you just have to put numbers in.

  4. You can get axial deformation from the expression

$$\frac{F}{A} = E \frac{\Delta l}{l}$$

This is actually the first time you need dimensions of members.

Note: All above calculations are made under the following assuptions:

  • weight of the members is much smaller than forces $F_1$, $F_2$ and $F_3$ - which is usually sane approximation in civil engineering for doing first approximation calulations;

  • all joints are perfect pin joints, so forces within members are strictly axial - usually joints are rigid joints (welded together), but in order to avoid problematic moments in the members of the construction, calculations are always done and construction designed as if they were all pin joints;

  • compression/enlongement of individual members is negligible - also a reasonable approximation.

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Yes, it is easy. How about tension forces with allowable stress – Aer Apr 20 '12 at 9:55
Wait it takes time to get there :) – Pygmalion Apr 20 '12 at 9:56
Gotta go to lectures now, so if there is no more questions, accept answer :) – Pygmalion Apr 20 '12 at 10:10
thank you! i got it – Aer Apr 20 '12 at 10:11
@Pygmalion: Try not to give full steps for a homework problem, just give concepts. – Manishearth Apr 20 '12 at 10:22

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