Machine-translated by Google. Questions are authored in English.
Top

Discussion

The maximum force F in kN that can be applied at C such that the axial stress in any of the truss members DOES NOT exceed 100 MPa is

Members AC and BC meet at joint C, where a horizontal force F acts to the right. Both are hinged to a wall on the left: AC rises at 45 degrees to the horizontal and BC falls at 60 degrees

  • A.8.17
  • B.11.15
  • C.14.14
  • D.22.30

Answer: B

The maximum force is about 11.15 kN.

From the joint equilibrium of the previous question, the member forces are proportional to F:

\[ F_{AC} = 0.897F,\quad F_{BC} = 0.732F \]

AC carries the larger force, so it reaches the limit first. A stress of 100 MPa on 100 mm2 is 10 kN:

\[ \begin{aligned} 0.897F &= 10\ \text{kN} \\ F &= 11.15\ \text{kN} \end{aligned} \]

Designing on member BC instead gives 13.66 kN, which would overstress AC. 8.17, 14.14 and 22.30 kN do not follow from the member forces.

No comment is present. Be the first to comment.
Loading…

Post your comment

Please log in or register to post a comment.