Nothing Special   »   [go: up one dir, main page]

Elective-II: Finite Element Method: B.E. (Mechanical Engineering) Eighth Semester (C.B.S.)

Download as pdf or txt
Download as pdf or txt
You are on page 1of 4

B.E. (Mechanical Engineering) Eighth Semester (C.B.S.

)
Elective-II : Finite Element Method

P. Pages : 3 NJR/KS/18/4720
Time : Three Hours *0824* Max. Marks : 80
_____________________________________________________________________
Notes : 1. All questions carry marks as indicated.
2. Solve Question 1 OR Questions No. 2.
3. Solve Question 3 OR Questions No. 4.
4. Solve Question 5 OR Questions No. 6.
5. Solve Question 7 OR Questions No. 8.
6. Due credit will be given to neatness and adequate dimensions.
7. Assume suitable data whenever necessary.
8. Illustrate your answers whenever necessary with the help of neat sketches.
9. Use of non programmable calculator is permitted.

1. a) The composite element is shown in fig.1 (a) Determine the nodal displacement and 14
stresses in all element.

50 mm  Steel
240 mm 

60 mm  Brass 80 mm 
P1 P2

Steel
50 mm  Steel

400 mm 300 mm 300 mm

Esteel = 200 GPa E Br =100 GPa


P1 = 15 kN P2 = 10 kN
Fig. 1 (a)

b) Explain in brief the plane stress and plane strain analysis alongwith assumptions and 6
application.

2. a) Explain in brief the various types of elements used in FEM with their salient features. 6

b) Determine the angle of twist in degree at the step and maximum shear stress at section and 14
reaction at wall, for the steeped circular bar subjected to torque as shown in fig. 2 (b)
G = 77GPa.
5 kN.m

75 mm 
50 mm 

400 mm 400 mm

Fig. 2 (b)

NJR/KS/18/4720 1 P.T.O
3. The three bar truss supported by a spring is shown in Fig. (3). The cross sectional area of 20
3 2
all member, A = 1000 mm and Elastic constant E = 200  10 N / mm . Determine the
2

nodal displacement and stresses in all three member the truss members are subjected to
uniform temperature rise of 80ºC along with applied point load as shown.
Take  = 12  10−6 / º C Length of all member

L1 = L 2 = L3 = 5 meter
2  = 12  10 −6 / º C
E = 200  10 3 N / mm 2
o
30

40o 25 kN
4 10 kN
3
Fig. 3 K = 3000 kN/m

5 Fig. [3]

4. a) For a beam with UDL as shown in fig. (4) Determine: 15


i) The slope at 2 and 3.
ii) The vertical displacement [deflection] at the midpoint of the distributed load.
25 kN/mtr

2 mtr 2 mtr
E = 200GPa ; I = 10  106 mm4

b) Write a short note on plane frame analysis. 5

5. A two dimensional triangular plate is subjected to point load as shown in fig. (5). 20
Determine the nodal displacements and element stresses using one element model.
10 kN
750 mm

5 kN
o
90
400 mm E = 200GPa
(Poission ratio)  = 0.30
thickness = t = 15 mm
Fig. (5)

NJR/KS/18/4720 2
6. a) A structure shown in fig. 6 (a) ; It consists of a rigid bar of negligible mass, pined at one 15
end is supported by a steel rod and aluminium rod. A load P = 30kN is applied as shown.
Determine the deflection and stresses in the steel rod and aluminium rod.
Steel
Aluminium
A = 1200 mm2
E = 200 GPa A = 900 mm2
St E = 70 GPa
L = 4.5 meter AL
L = 3 meter

30 kN
2m 3m 1m
Fig. 6 (a)

b) Explain in brief Isoparametric, subparametric and superparametric element. 5

7. a) The one dimensional composite bar & shown in fig. 7 (a). Determine the interface 10
temperature.
Element – I Element – II Element – III
K xx 200 W/mºC 100 W/mºC 50 W/mºC
Length 2m 1m 0.5m
2
The Area is 0.15m . The left end has a constant temperature of 100ºC and the right end
has a constant temperature of 300ºC.

100ºC 300ºC
I II III

2m 1m 0.5 m
Fig. 7 (a)
b) Enlist various software [commercial] used for finite element analysis of solid mechanics. 10
Explain any one with procedure to solve one-dimensional problem. [From pre-processing
to Post processing stage]

8. a) The axisymmetric element is shown in fig. 8 (a). Determine the element stresses. The 15
nodal displacements for each node are
u1 = 0.001mm w1 = 0.0015mm
u 2 = 0.004 mm w 2 = 0.006 mm
u3 = 0 w3 = 0
Elastic constant E = 2  105 N / mm2
Poission ratio  = 0.25
[30, 30]
3

All coordinates
z
are in mm

r 1 2
[10, 0] [30, 0]
Fig. 8 (a)
b) Explain in brief the Lumped mass matrix and consistent mass matrix used for dynamic 5
analysis.
**********

NJR/KS/18/4720 3 P.T.O
NJR/KS/18/4720 4

You might also like