Lamis Theorem

  • Recount and apply Lami’s theorem.

Lami's theorem

If three rigid bodies are in static equilibrium, then the forces are directly proportional to the sine ratio of the angle between the remaining two forces.

\[\frac{F_1}{\sin\ \alpha} = \frac{F_2}{\sin\ \beta} = \frac{F_3}{\sin\ \gamma}\]

Solved Examples

Solved Example:

24-5-01

Two steel truss members, AC and BC, each having cross sectional area of 100 mm$^2$, are subjected to a horizontal force F as shown in figure. All the joints are hinged. If F = 1 kN, the magnitude of the vertical reaction force developed at the point B in kN is: (GATE ME 2012)

24.5-01



Solution:
\[\dfrac {F}{\sin 105^\circ}=\dfrac {T_{2}}{\sin 120^\circ}=\dfrac {T_{1}}{\sin 135^\circ}\] \[\dfrac {T_{1}}{\sin 135^\circ}=\dfrac {F}{\sin 105^\circ}=\dfrac {1}{\sin 105^\circ}\] \[T_{1}=0.7320kN\] \[R_{NT1} = T_{1}\cos 30^\circ = 0.73205\times \cos 30^\circ = 0.634\ kN\]

Correct Answer: A

Solved Example:

24-5-02

The maximum force F is kN that can be applied at C such that the axial stress in any of the truss members DOES NOT exceed 100 MPa is: (GATE ME 2012)

Solution:
\[\dfrac {F}{\sin 105^{\circ }}=\dfrac {T_{2}}{\sin 120^{0}}\] \[T_{2}=\dfrac {\sin 120^{0}}{\sin 135^{\circ }}\times F\] \[T_{1}=\left( 0.73205\right)F \] \[T_{2} > T_{1}\] \[\sigma =100\ MPa\] \[F=\sigma \times A_{1}\] \[F_{\max }=\sigma _{\max }\times A_{1}\] \[T_{2}=100\times 100\] \[0.8965F=100\times 100\] \[F_{1}=\dfrac {100\times 100}{0.8965} =11154.5\ N =11.15\ kN\]

Correct Answer: B

Solved Example:

24-5-03

A rope is stretched between two rigid walls 40 meters apart. At the midpoint, a load of 100 N was placed that caused it to sag 5 meters. Compute the approximate tension in the rope.

Solution:
\[\theta =\tan ^{-1}\left( \dfrac {5}{20}\right) =14.03^\circ\] \[\alpha =\theta +90^\circ =104.03^\circ\] \[\beta =\alpha = 104.03^\circ\] \[\gamma =360-\alpha -\beta =151.93^\circ\] Using Lami's theorem, \[\dfrac {F_{1}}{\sin \alpha }=\dfrac {F_{2}}{\sin \beta }=\dfrac {F_{3}}{\sin \gamma }\] \[\dfrac {F_{1}}{\sin 104.03^\circ}=\dfrac {100}{\sin 151.93^\circ}\] \[F_{1}=206.16\ N\]

Correct Answer: A

Solved Example:

24-5-04

According to Lami's theorem:

Correct Answer: D