Unit 2
Unit 2
Moment of Inertia
Question 1) Determine the forces in all members of truss by joint method
Answer:
Consider FAB of Truss, Applying conditions of equilibrium,
= 0
RHA + 30 Kn
= 0
RvA + RB =0
Taking moment at point A,
= 0
-(RB* 3) - (30*3) = 0
RB = -30 KN
RB =30 KN RVA = 30 KN
Consider Joint c, Assuming forces in member AC & BC to be Tensile,
Applying conditions of equilibrium,
= 0
-30 + FCB cos 45 =0
FCB =30/cos 45 = 42.42 KN (T)
- 
= 0
- FcA – FCB sin 45=o
- F cA -42.42 sin 45 =0 
-FCA – 30= 0
-FCA = -30 KN
Consider Joint c, Assuming forces in member AC & BC to be
Tensile, Applying conditions of equilibrium,
Fx = 0 30+FAB =0
FAB = -30KN
Question 2) Determine the forces in each member of the plane truss as shown in fig. In terms pf the external loading and state if the members are in tension or compression. Use 0+ 30 deg, L = 2 m and p =100N.
Diagram
Answer:
Consider FBD of Truss,
For equilibrium, 
Fx =0
RHA+100 =0
RHA =-100 KN
Fy =0 RVA = RD – 100 ….. 1
MA = 0 -------Taking moment @ A
- (Roxz) + (100*1) + (100*1.732) =0
- - 2 Rd + 100 + 1.7320 =0
- -2 Rd + 100 + 173.2 =0
- RD =136.66 N ()
From eqn (1)
RVA = 100 – 136.6
RVA= 36.6N
RVA = 36.6 N
Consider Joint D, for equilibrium,
Fx =0
-FAD + FCD cos 60 =0 – (11)
Fy = 0
136.6 +FCD sin 60 +0 
FCD = -157.73 N ©
From eqn (11), Fad =- 78. 87 N (c)
Consider point A, for the equilibrium of point A,
Fx = 0
- 100 + FAD + FAccos + 30 FAb cos60 = 0
-100 + (-78.87) +FAC cos 30 +FAb cos60 =0
FAccos 30 + FAB cos60 + 178. 87 (3)
fx =0
-36.6 + FAC sin 30 + FAB sin 60 =0
FAC sin 30 + FAb sin 60 = 36.6
Solving eqn (3) and (4)
FAC = 273.21 N (T)
FAB = -115.47 N
FAB= 115.47 N ©
Consider point B, for the equilibrium of point,
Fx =0
-FAB cos60 + FBC =0
-[9-115.47) cos60]+ FBC =0
FBC = -57.73 N
FBC =57.73N
Member | AB | BC | CD | AD | AC |
Force | 115.47N | 57.73N | 157.73N | 78.87N | 273.21N |
Nature | c | c | c | c | T |
Question 3) Determine the axial forces in each member of the plane truss as shown in figure.
Answer:
Consider FBD of Truss,
For the equilibrium of Truss, 
fx =0
RHA + 10 =0
RHA = -10 KN
RHA =10KN (
Resolving forces vertically,
fy =o
RVA + RD -15 =0
RVA + RD = 15 …… (1)
Taking moment about point A,
Fy =0
(10*3) – 3 Rd =0
30+ 3 Rd =0
RD= 10 KN 
RVA = 5KN (
)
Now Consider joint B, FBD of joint B is shown below.
Assumbing forces developed in all members to be Termile,
For the equilibrium of joint We have
fx =0
10+FBC =0 
FBC = -10 KN
FBC = 10KN (c)
fy =0
FAB = -15 KN
FAB = 15 KN (c)
Now consider joint c,
For the equilibrium of joint,
fx =0
-FBC – FAC cos 45 =0
- (-10) – FAC cos 45=0
10= FAC cos 45
FAC = 10/cos 45 
FAc = 14.14 KN (T)
fy =0
-Fac sin 45 – FCD=0
- 14.14 sin45 = FCD
FCD = -10 KN (c)
FCD= 10 KN (c)
Consider joint D,
By observation,
FAD=0
Sr .No | Member | Force | Nature |
1 | AB | 15 KN | c |
2 | BC | 10KN | c |
3 | CD | 10KN | c |
4 | DA | 0 | - |
5 | AC | 14.14 | T |
Question 4) Member AB & BC can support a maximum compressive force of 800 N & members AD, DC, BD can support a max. Tensile. Force of 2000N Determine the greatest land p that
Answer :
Consider following geometry of the figure.
Tan
1 = ¼
14.04
2 = 
=45
Consider
fx =0 RHA = 0
∈Fy =o RVA + Rc = p………(1)
∈ma =0
Hp- 8 Rc =0
Rc =P/2 N (
) RvA = (P/2) N (
)
Consider point A C assuming all forces as Tensile)
∈fx =0
FAD cos 14.04+FAB cos 45 =0… (2)
∈fy = 0 p/2= Fad sin 14.04 + FAB sin 45 =0….. (3)
FAB = 2743.9 N > 800 N ( Not Allowed)
Let FAD = 2000 N (T) Then from eqn (2) & (3)
FAB =-274.9 N > 800 ( Not allowed)
Let FAB = .800 N ©, Then put this eqn (2) (3) we get
FAB = 583 N < 2000 N (Allowed) From Eqn (2) P= 848.9 N
Question 5) Explain Equation of equilibrium?
Answer:
- A body is said to be in equilibrium if the resultant of all forces acting on it is zero.
- When the resultant of all the forces acting on the particle is zero, then, the particle is in equilibrium.
- If the resultant force acting on a particle is zero (or if particle is in equilibrium state) the, particle will remain at rest condition if it is originally at rest; or particle will move with constant speed in a straight line if it is originally in motion.
Condition for Equilibrium: R = 0 { 
}
∑M = 0
Question 6)Explain type of truss?
Answer:
TYPES OF TRUSSES
- Simple truss – indicates a single triangular truss. These trusses are most often used as the roof trusses.
- Planar truss – as the name implies it is a two dimensional truss. If all the members and the nodes are in a planar surface, then this truss is a planar truss.
- Space frame truss – Contrast to planar truss, the members and the nodes are located in the three dimensional space. Electrical and telecom towers are the one of the simplest example that we are seeing in the day to day life.
Question 7) Explain Assumptions in Truss Analysis.
Answer:
1. Members are connected at their ends by frictionless pins.
2. Members are straight and, therefore, are subjected only to axial forces.
3. Members’ deformation under loads are negligible and of insignificant magnitude to cause appreciable changes in the geometry of the structure.
4. Loads are applied only at the joints due to the arrangement of members.
Question 8) Explain method of joint.
Answer:
Method of joint: This method involves isolating each joint of the truss and considering the equilibrium of the joint when determining the member axial force. Two equations used in determining the member axial forces are ∑ Fx = 0 and ∑ Fy = 0. Joints are isolated consecutively for analysis based on the principle that the number of the unknown member axial forces should never be more than two in the joint under consideration in a plane trust.
Question 9) Explain truss.
Answer 9) Truss:-
A Rigid structure formed by connecting various two force members to each other by using pin joint..
Plane truss: When all member of the truss lies in one plane, Then truss is known as plane truss
Rigid truss:- A truss which do not collapse when external Load is applied on it.
Simple truss:-The structure formed by basic triangle made by Connecting various members are called simple truss.
Question 10) Explain method of section.
Answer:
Method of section: This method entails passing an imaginary section through the truss to divide it into two sections. The member forces are determined by considering the equilibrium of the part of the truss on either side of the section. This method is advantageous when the axial forces in specific members are required in a truss with several members.
Analysis of Trusses by Method of Section
Sometimes, determining the axial force in specific members of a truss system by the method of joint can be very involving and cumbersome, especially when the system consists of several members. In such instances, using the method of section can be timesaving and, thus, preferable. This method involves passing an imaginary section through the truss so that it divides the system into two parts and cuts through members whose axial forces are desired. Member axial forces are then determined using the conditions of equilibrium. The detailed procedure for analysis by this method is presented below.
Procedure for Analysis of Trusses by Method of Section
•Check the stability and determinacy of the structure. If the truss is stable and determinate, then proceed to the next step.
•Determine the support reactions in the truss.
•Make an imaginary cut through the structure so that it includes the members whose axial forces are desired. The imaginary cut divides the truss into two parts.
•Apply forces to each part of the truss to keep it in equilibrium.
•Select either part of the truss for the determination of member forces.
•Apply the conditions of equilibrium to determine the member axial forces.
