Skip to main content

Assignment in the Analysis of Trusses

Assignment - S.M.2,Unit-6
1.Find the internal forces in member AB,AC,BC by using Method of Joints.

2.Find the internal forces in all members of a pin jointed truss by using Method of Joints. P= 12kN

3.Find the internal forces in all members of a pin jointed truss by using Method of Joints.

4.Find the internal forces in BD, CD and CE members of a pin jointed truss by using Method of sections.
5.Find the internal forces in GF,GC,FC members of a pin jointed truss by using Method of sections.

Objective questions and answers 

1. How many equilibrium equations do we need to solve generally on each joint of a truss?
     a) 1    b) 2      c) 3       d) 4
2. If a member of a truss is in compression, then what will be the direction of force that it will apply to the joints?
   a) Outward     b) Inward     c) Depends on case 
   d) No force will be there
3.What should be ideally the first step to approach to a problem using method of joints?
   a) Draw fbd of each joint      b) Draw fbd of overall truss
   c) Identify zero force members 
   d) Determine external reaction forces
Answer: c
Explanation: Identifying zero force members should always be the first step to approach any truss problem as it eliminate a lot of variables and is fairly easy.
4. The free body diagram of which part of the section of the truss is made to make use of method of joints?
  a) Joints     b) Truss        c) The whole structure 
  d) The combination of joint and whole structure
5. For applying the method of joint at joints the forces need to coplanar.
  a) False     b) True
Answer: b
Explanation: The forces are coplanar that’s why they are there in the calculations in the method of joints. The forces are resolved in 2D. This makes the condition of the equilibrium fulfilled and hence the method of joints uses the forces in the same plane for its calculation.
6. For applying the method of joint at joints the forces need to be concurrent.
    a) False         b) True
Answer: b
Explanation: The forces are concurrent that’s why they are there in the calculations in the method of joints. The forces are resolved in 2D. This makes the condition of the equilibrium fulfilled by cancelling of various force members and hence the method of joints uses the concurrent forces for its calculation.
7. What is after taking the assumption of the direction of the force, the direction comes opposite?
a) The assumption made was wrong and the question can’t be solved further
b) The assumptions are not to be taken
c) The direction is in the opposite sense, and hence the direction is known to us
d) The direction will be already given to us, no need of assuming
Answer: c
Explanation: The direction is comes opposite, means that the assumption was wrong. But that doesn’t means that the question can’t be solved further. One needs to change the sign of the direction and the question is solved. Thus the calculations of the force direction.
8. If the whole truss is in equilibrium then all the joints which are connected to that truss is in equilibrium.This is known as:
a) Method of joints        b) Section method
c) Scalar field method   d) Vector equilibrium method
9. How many equilibrium equations are used in method of sections?
   a) 2     b) 4       c) 3      d) 5
10.In method of sections, what is the maximum no. of unknown members through which the imaginary section can pass?
 a) 1      b) 2       c) 3      d) 4
Labels:

Comments

Post a Comment

Popular posts from this blog

Relation between Modulus of Elasticity and Modulus of Rigidity

Modulus of Elasticity (E)   It is the ratio between Normal stress to Normal strain within the elastic limit. Elastic Modulus E = Normal stress/Normal strain E =  s/e Modulus of Rigidity (G)  It is the ratio between Shear stress to Shear strain within the elastic limit. Rigidity Modulus G = Shear stress/ Shear strain G =    Ƭ / ø   Relation between Modulus of Elasticity and Modulus of Rigidity: Consider a solid cube  PQRS is  subjected to a shearing force F.  Let  Ƭ     be the shear stress produced in the faces PQ and RS due to this shear force. The complementary shear stress consequently produced in the vertical faces PS and RQ is also equal to same and shown in figure as Ƭ   Due to the pure shearing force, the cube is deformed PQRS to PQR'S' . The point   S moved to S' and point R moved to R' as shown in fig.  The shear strain = The angle of distortion  ø           ...
5 comments
Read more

Shearing Stresses Distribution in Circular Section

Show that the shearing stress developed at the neutral axis of a beam with circular cross section is  τ max = (4/3)(F/πr2). Assume that the shearing stress is uniformly distributed across the neutral axis.  Solution : Let us consider the circular section of a beam as displayed in following figure. We have assumed one layer EF at a distance y1 from the neutral axis of the circular section of the beam Shear stress at a section will be given by following formula as mentioned here Where, F = Shear force (N) τ = Shear stress (N/mm2) A = Area of section, where shear stress is to be determined (mm2) ȳ = Distance of C.G of the area, where shear stress is to be determined, from neutral axis of the beam section (m) Q = A. ȳ = Moment of the whole shaded area about the neutral axis I = Moment of inertia of the given section about the neutral axis (mm4) For circular cross-section, Moment of inertia, I = ПR4/4 b = Width of the given section where shear stress is to be determined. Let us co...
1 comment
Read more

PORTAL METHOD and CANTILEVER METHOD

The behavior of a structure subjected to horizontal forces depends on its height to width ratio. The deformation in low-rise structures, where the height is smaller than its width, is characterized predominantly by shear deformations. In high rise building, where height is several times greater than its lateral dimensions, is dominated by bending action. To analyze the structures subjected to horizontal loading we have two methods. Portal method  and Cantilever method 1. PORTAL METHOD The portal method is an approximate analysis used for analysing building frames subjected to lateral loads such as Wind loads/ seismic forces.  Since shear deformations are dominant in low rise structures, the method makes simplifying assumptions regarding horizontal shear in columns.  Each bay of a structure is treated as a portal frame, and horizontal force is distributed equally among them. Assumptions in portal method   1. The points of inflection are located at the mid-height of e...
3 comments
Read more