Ansys Work

Q.1
a)

Section 1-3

∑ M B =0

5∗5∗400∗200−M =0
M =2000000 N . mm
∑ F y =0
 −5∗5∗400+ B y =0

B y =10000 N

∑ F x =0

∑ Bx =0

Section 1-2

0< I < 400

V =−5∗I∗t
When I =0 ; V =0
When I =200 ; V =−5000 N
When I =400; V =−10000 N
2
−5∗I ∗t
M=
2
When I =0 ; M =0
When I =200 ; M =−500000 N . mm
When I =400; M =−2000000 N . mm

b h3
I=
12
3
5∗100 4
I= =416666.667 mm
12
q 0 L3
δ= =4.571∗102 mm
8 EI
 M =F ()
I
2
=2∗106 N . mm

Shear force is the internal force on a member when the force is not applied at the axis.
Shearing force is the force divided by the cross-sectional area.

Bending moment is the force trying to rotate the member. Moment is the perpendicular
distance from the force to the axis multiplied by the force.

To better understand the two terms, consider the following 2 examples:

Get a piece of spaghetti & hold it in your hands with your thumbs touching each other.
Now push your right hand forward & pull your left hand backward. You just sheared the
spaghetti.

Get another piece of spaghetti & hold it in your hands with your thumbs as far apart as
possible but pointing at each other. Now rotate your right wrist clockwise & your left
wrist anti clockwise. You just broke the spaghetti in bending.

b) The Stress Function Method

An effective way of dealing with many two dimensional problems is to introduce a new
“unknown”, the Airy stress function  , an idea brought to us by George Airy in 1862.
The stresses are written in terms of this new function and a new differential equation is
obtained, one which can be solved more easily than Navier’s equations.

The Maxwell stress functions are defined by assuming that the Beltrami stress tensor is
restricted to be of the form;

The stress tensor which automatically obeys the equilibrium equation may now be
written as:
The solution to the elastostatic problem now consists of finding the three stress
functions which give a stress tensor which obeys the Beltrami–Michell compatibility
equations for stress. Substituting the expressions for the stress into the Beltrami-Michell
equations yields the expression of the elastostatic problem in terms of the stress
functions: These must also yield a stress tensor which obeys the specified boundary
conditions.

Consider a section x at a distance x from the fixed end A. The BM at this section is
given by
Mx = -W(L-x)
But bending moment at any section is given as
Equating the two values of bending moment we get,

Then integrating above equation,

--------------(1)

Integrating again we get

--------------(2)

Where C1 and C2 are the constants of integration, which is obtained from boundary
conditions, i.e., i) At x = 0, y = 0 ii) x = 0, dy / dx = 0

1. By substituting x = 0, y = 0
0 = 0 + 0 + 0 + C2
C2 = 0
2. By substituting x = 0, dy/dx = 0
0 = 0 + 0 + C1
C1 = 0
Then by substituting value of C1 in equation (1)

-------------(3)

Equation (3) is known as slope equation. We can find the slope at any point on the
cantilever by substituting the value of x. The slope and deflection are maximum at the
free end. These can be determined by substituting the values of C1 and C2 in equation
(2) we get
Equation (4) is known as deflection equation.
let
ϴ

= slope at end B i.e.,(dy/dx)

Y

=Deflection at the end B

a) Substituting ϴ

for dy/dx and x=L in equation (3), we get

Negative sign shows that tangent at B makes an angle in the anti-clockwise direction
with AB

Negative sign shows that tangent at B makes an angle in the anti-clockwise direction
with AB

b) Substituting Y

for Y and x=L in equation 4, we get

 Y B =1.219∗10−3 mm
2. Uniformly-loaded cantilever beams:

But bending moment at any section is given as

Equating the two values of bending moment we get,

Then integrating above equation,

-----------(1)

Integrating again we get

-----------(2)

Where C1 and C2 are the constants of integration, which is obtained from boundary
conditions, i.e., i) At x = 0, y = 0 ii) x=0, dy / dx = 0

1. By substituting x = 0, y = 0
2. By substituting x = 0, dy / dx = 0
Then by substituting value of C1 and C2 in equation (1) and (2), we get
-----------(4) deflection equation

From these equations the slope and deflection can be obtained at any sections.

To find the slope and deflection at point B, the value of x=L is substituted in these
equations.
let

= slope at free end B i.e.,(dy/dx) at b = ϴ

and Y

= Deflection at the free end B

From equation (3) we get slope at B as

From equation (4) we get deflection at B as

2
Y B =4.571∗10 mm
Then the deflection at any point x along the span of a uniformly loaded cantilevered
beam can be calculated using:

d)
1. Open ANSYS Mechanical and select Solution from the model tree.
2. Click Tools > Read Result Files and locate the .rst file containing the results.
3. Click User Defined Result and enter the SVAR to view. For example, type SVAR1 to
view the rupture status of the material.
4. Enter a Display Time.
5. Right-click User Defined Result in the model tree and select Evaluate All Results.
6. Switch to the Graph tab and click the Play button to monitor the SVAR throughout
the model.
Result Compression
Moment 2*10^6 N.mm 1.98*10^6 N.mm

Longitudinal force 5 N/mm^2 4.9 N/mm^2

Shear Stress 5 N/mm^2 4.87 N/mm^2

 Equivalent Stress 1.9874e5 MPa 1.786e4 MPa

Total Shear force 9772.7 N 100000 N

Total Bending Moment 2e6 2e6

Deformation 384.33 457.1

Minimum Stress -1.058^-9 -4800

Maximum 4800 -1.059^-9

Q.2
a) The disc with hole as shell elements is make in the work bench.
b) Therefore, the ultimate strength and breaking strength are the same. Typical brittle
materials like glass do not show any plastic deformation but fail while the deformation is
elastic. Hardness is the property of a material that enables it to resist plastic
deformation. A perfectly plastic material is such that it does not strain harden and is not
strain-rate sensitive. The strength is not a function of strains nor of strain rates. A
deformable material is said to be a perfectly plastic material when the following
conditions exist: 1. Hardness is a characteristic of a material, not a fundamental
physical property. Hardness is a measure of the resistance to localized plastic
deformation induced by either mechanical indentation or abrasion.
c) The ratio of the applied force divided by the unit area that has a tendency to elongate
or stretch the material. Tensile load, which tends to increase the length of a material, is
the opposite of compression, which tends to reduce the length of a material. Tensile
load is also called tensile stress. When the tensile force is applied then the radius of
circular is changed the intern shape.
d) The radial stress for a thick-walled cylinder is equal and opposite to the gauge
pressure on the inside surface, and zero on the outside surface. Radial stress is the
difference in wellbore pressure and pore pressure and acts along the radius of the
wellbore. Wellbore pressure and pore pressure both stem from fluid pressure acting
equally in all directions. The radial stress for a thick-walled cylinder is equal and
opposite to the gauge pressure on the inside surface, and zero on the outside surface .
The circumferential stress and longitudinal stresses are usually much larger for
pressure vessels, and so for thin-walled instances, radial stress is usually neglected.
e) Moment of force is represented as moment. Turning effect of force is called torque.
“Moment” is a concept of engineering and physics that refers to the tendency of a force
to move an object while torque is the tendency of a force to rotate an object in a pivot.
Force can make a body that is at rest to move. Force can stop a moving body or slow it
down. Force can accelerate the speed of a moving body. Force can change the
direction of a moving body along with its shape and size. The moment of a
vector results from the cross product of a position vector with a bound vector. A bound
vector is a vector that is bound to a point. A moment about O. This distance can be
calculated in various ways, e.g, using the unit vector u in the direction of FQ and/or the
angle θ between u and rQ/O . Moment of force is the rotational effect of a force, which is
equal to the force multiplied by the perpendicular distance between a pivot and the
force. The moment of a force is related to the size of the force and the size of the
perpendicular distance between the force and pivot. It can change the state of
movement of the body on which force is applied, i.e. it can move a stationary object or
stop a moving object. (ii) It can change the shape and size of an object.