Equilibrium of Rigid Bodies
Loads
Learning Objectives:

Review basics of structural analysis include types of structures, types of loads.

Point loadsconcentrated forces exerted at point or location

Distributed loadsa force applied along a length or over an area. The distribution can be uniform or nonuniform.

Uniformly distributed Load (UDL):

Triangular Load:

The above figure represents how each type of load is represented (From top to bottom):
1. Point Load
2. UDL
3. Triangular Load
Types of Supports
Learning Objectives:

Understand different types of supports and the reactions offered by them.
Solved Example: 24101
A simply supported beam of span 'l' is subjected to a transverse point load 'W' acting at a distance of 'a' from the left support A. The reaction on the left support will be equal to: (VIZAG MT Mechanical 2015)
A. $\dfrac{W.a}{l}$
B. $\dfrac{W(la)}{l}$
C. $\dfrac{W}{2}$
D. $\dfrac{W.l}{a}$
Correct Answer: B
Conditions for Static Equilibrium
Learning Objectives:

Apply equilibrium conditions for a beam with various loading conditions, such as point load, uniformly distributed load, triangular load and eccentric/offset load.
\[\sum F_x = 0,\ \sum F_y = 0,\ \sum F_z = 0\] \[\sum M_x = 0,\ \sum M_y = 0,\ \sum M_z = 0\] If a problem is in 2D only then only three conditions will remain. \[\sum F_x = 0,\ \sum F_y = 0,\ \sum M = 0\]
Solved Example: 24301
Equilibrium occurs when:
A. All the forces acting on an object are balanced.
B. The sum of the +x forces on an object equals the sum of the x forces.
C. The net force on the object is zero.
D. All of the above.
Correct Answer: D
Solved Example: 24302
A simply supported beam is five meters in length. It carries a uniformly distributed load including its own weight of 300 N/m and a concentrated load of 100 N, 2 meters from the left end. Find the reactions if reaction A at the left end and reaction B at the right end.
A. $R_A$ = 810 N, $R_B$ = 700 N
B. $R_A$ = 820 N, $R_B$ = 690 N
C. $R_A$ = 830 N, $R_B$ = 680 N
D. $R_A$ = 840 N, $R_B$ = 670 N
\[V_{A}+V_{B}=100+300\times 5=1600\] Taking moment about point A, \[100\times 2+300\times 5\times 2.5=V_{B}\times 5\] \[V_{B}=790\ N \mathrm{,\ and\ }V_{A}=810\ N\]
Correct Answer: A
Free Body Diagram
Lamis Theorem
Learning Objectives:

Recount and apply 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 Example: 24501
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)
A. 0.63
B. 0.32
C. 1.26
D. 1.46
\[\dfrac {F}{\sin 105^\circ}=\dfrac {T_{2}}{\sin 120^\circ}=\dfrac {T_{1}}{\sin 135^\circ}\] \[\dfrac {T_{1}}{\sin 135^\circ}=\dfrac {F}{\sin 105^\circ}=\dfrac {1}{\sin 105^\circ}\] \[T_{1}=0.7320kN\] \[R_{NT1} = T_{1}\cos 30^\circ = 0.73205\times \cos 30^\circ = 0.634\ kN\]
Correct Answer: A
Solved Example: 24502
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)
A. 8.17
B. 11.15
C. 14.14
D. 22.30
\[\dfrac {F}{\sin 105^{\circ }}=\dfrac {T_{2}}{\sin 120^{0}}\] \[T_{2}=\dfrac {\sin 120^{0}}{\sin 135^{\circ }}\times F\] \[T_{1}=\left( 0.73205\right)F \] \[T_{2} > T_{1}\] \[\sigma =100\ MPa\] \[F=\sigma \times A_{1}\] \[F_{\max }=\sigma _{\max }\times A_{1}\] \[T_{2}=100\times 100\] \[0.8965F=100\times 100\] \[F_{1}=\dfrac {100\times 100}{0.8965} =11154.5\ N =11.15\ kN\]
Correct Answer: B
Solved Example: 24503
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.
A. 206 N
B. 150 N
C. 280 N
D. 240 N
\[\theta =\tan ^{1}\left( \dfrac {5}{20}\right) =14.03^\circ\] \[\alpha =\theta +90^\circ =104.03^\circ\] \[\beta =\alpha = 104.03^\circ\] \[\gamma =360\alpha \beta =151.93^\circ\] 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: 24504
According to Lami's theorem:
A. Three forces acting at a point will be in equilibrium
B. Three forces acting at a point can be represented by a triangle, each side being proportional to force
C. If three forces acting upon a particle are represented in magnitude and direction by the sides of a triangle, taken in order, they will be in equilibrium
D. If three forces acting at a point are in equilibrium, each force is proportional to the sine of the angle between the other two
Correct Answer: D
Principle of Virtual Work
Learning Objectives:

Apply principal of virtual work in statics by understanding virtual displacements and virtual work.
“If a system in equilibrium under the action of a set of forces is given a virtual displacement, the virtual work done by the forces will be zero.”
The terms used in this statement are defined as follows:

A virtual displacement \(\delta\)s is an imaginary infinitesimal variation of the coordinate given instantaneously. The virtual displacement must be compatible with the constraints of the system.

Virtual work \(\delta\)W is the work done by all the active forces in a virtual displacement. Since there is no significant change geometry associated with the virtual displacement, the force acting on the system are assumed to remain unchanged for the calculation of \(\delta\)W.
Solved Example: 24601
Using principle of virtual work, determine the force P which will keep the weightless bar AC in equilibrium. Take length AB as 2m and length AC as 8m. The bar makes an angle of 30$^\circ$ with horizontal. All the surfaces in contact are smooth. Refer figure.
A. 677 N
B. 1038 N
C. 1230 N
D. 1492 N
By using the principle of virtual work, \begin{align*} P\dfrac {d}{d\theta }\left( 8\cos \theta \right) +800\left( \dfrac {d}{d\theta }6\sin \theta \right) &=0\\ P\left( 8\sin\theta \right) +800\left( 6\cos \theta \right) &=0\\ 8P\sin \theta &=4800\cos \theta \\ P&=600\cot \theta \\ &=600 \cot 30\\ &=1038\ N \end{align*}Correct Answer: B
Solved Example: 24602
Virtual work refers to: (TNTRB ME 2017)
A. Virtual work done by Virtual forces
B. Virtual work done by Actual forces
C. Actual work done by Actual forces
D. Actual work done by Virtual forces
Correct Answer: B
Solved Example: 24603
The principle of virtual work states that, for a body to be in equilibrium, the virtual work should be: (CIL MT Civil 2020)
A. Any value between zero and one
B. Zero
C. Maximum
D. Minimum
Correct Answer: B
Solved Example: 24604
In virtual work equation some forces are neglected. Select the most appropriate answer from the following: (UPPSC AE Mechanical 2013 Paper I)
A. Reaction of a rough surface on a body which rolls on it without slipping.
B. Reaction of any smooth surface with which the body is in contact.
C. Reaction at a point or on an axis, fixed in space, around which a body is constrained to turn.
D. All of the above
Correct Answer: D