QUESTION
- (a). With the aid of diagrams, explain the modulus of elasticity and modulus of rigidity, and use examples to show these parameters may vary for different materials.
(b). In a tensile test conducted on a sample of the material, a full range of stress-strain curve can be measured until the sample breaks. Draw a diagram to show such a typical relation between strain and stress, and explain properly for such key concepts as elastic range, plastic range, yield stress, ultimate tensile stress.
(c). A square metal column is to support a load of 1056 Tonnes and it must not compress more than 0.25 mm. The modulus of elasticity is 225 Gpa, the column is 1.8 m long. Calculate the cross section area and the width of the column.
(d). A punch must cut a hole 45 mm diameter in a sheet of steel 3.2 mm thick. The ultimate shear stress is 80 Mpa. Determine the minimum force that is required to cut the hole.
- Shear force diagram (SFD) and bending moment diagram (BMD) are a graph of shear force / bending moment plotted against the horizontal axis , indicating the shear force / bending moment withstood by the beam section along the length of the beam.
(a) Draw an appropriate shear force diagram and bending moment diagram for each of the following cases, where necessary calculations and expressions may be needed to show the details of the work.
(b) With the bending moment distribution, determine the location of the maximum bending moment and its value in each of the following cases.
(c) If the beams used in these examples have a circular cross-section, with diameter D = 35 mm, calculate the maximum stress produced by the maximum bending moment. The modulus of elasticity of the material is given E = 120 Mpa.
(1). Single point loading and uniform loading on a cantilever beam
(2). Double point loading and uniform loading on a cantilever beam
(3). Point loading and uniform loading on a simply supported beam
- A mass of 6 kg lies on a smooth horizontal table, and is attached by a spring to a fixed point, as shown in the figure below. The spring has a stiffness of 450 N/m.
(a). Determine the natural frequency of this system.
(b). If the mass is given a displacement, x = 0.65 m, to the left, and is then released, by neglecting the effect of friction, determine the equation of motion in this case
(c). With the aid of graphical package, i.e., Excel, Derive, Matlab etc., plot curves of displacement, velocity and acceleration against time based on the equation obtained in question (b).
(d). Calculate displacement, velocity and acceleration at time t = 0.12 seconds, and compare with the plots obtained in question (c).
- A cam is required to a lift of 30 mm for a rotation range of 60o. The follower then dwells at this level for a further 90o and then falls over the next 60o. Finally the rest of rotation of the cam is applied to complete a single cycle. The base circle of the cam is 120 mm diameter. The velocity of the lift and fall is kept constant for both rising and falling periods.
(a). Design a cam profile to satisfy the requirements.
(b). If the speed of the rotation is 12 rev/min determine the velocity of the follower.
(c). Draw the displacement, velocity and acceleration diagrams for the follower.
Necessary calculations and descriptions are expected showing the details of the design.
- A roller 0.325 m diameter rolls down a slope starting from rest. It takes 10 seconds make 6 complete rotations along the sloping surface accelerating uniformly as it moves. Calculate the following:
(1). The angular velocity at the end.
(2). The linear velocity at the end.
(3). The angular acceleration.
(4). The linear acceleration.
(5). The distance travelled.
- Mechanical failure is one of the most important factors causing component/structure/product no longer fit for its intended purpose. Use examples to explain at least three modes of mechanical failure, and discuss major causes to each mode.
(The modes of mechanical failure may include ductile, brittle, buckling, creep, fatigue, corrosion, impact, wear, thermal shock and degradation etc.)
- One of the most important alloys are iron and carbon steel, where the iron-carbon phase diagram describes the iron-carbon system of alloys in relation to the carbon percentage, and discloses the phases compositions and their transformations occurring with the alloys during their cooling or heating.
In this task, describe the iron-carbon phase diagram for the percentage by weight of carbon up to 3%, where appropriate explanation and drawings shall be provided. Then classify the steels with carbon that fall between the extremes of pure iron and cast iron by showing their carbon content and typical applications.
SOLUTION
1.Modulus of elasticity: All the objects are deformable to certain extent, when acted by a force upon it. This force, the external force, which acts on the object across a cross-section, is defined as Stress. There is certain finite deformation in the shape and the size of the object, which is called as Strain. The strain produced in the object is proportional to the stress acted upon it, unless the force or the stress is within the limit of elasticity. The proportionality depends on the nature of the material and the force applied. The constant of proportionality is called the Modulus of Elasticity and is given by the following ratio.
In the above figure the long bar is stretched by a force F to .So, as explained above the stress is F and strain is .Hence the Modulus of elasticity is given by
The Modulus of rigidity:The modulus of rigidity is defined as the ratio of stress by strain, in this case produced by a tangential force acting on one surface of the body, when the opposite surface of the body is fixed by the another force.
I the above figure the rectangular block is being acted upon by a tangential force F when the other side has been kept fixed. Thus a deformation is produced as as shown above.
There are different daily life examples which tell us that these constants vary with different materials. The rubber can be easily stretched as compared to the metals.
b)
The following figure shows the Stress-strain curve for an elastic solid.
As it can be seen in the above graph the solid object has a linear stress-strain curve until the elastic point is reached. At this point if the object is left by the external force, it may return back to its initial state, but beyond this point the curve is concave downward .The object would show cetin deformation even after it is left by the force. In this range too, if the stress continues beyond the Breaking point the object may break under pressure. This point is called Breaking Point. There is another term called yield stress, which is the stress at which the elastic material begins to deform plastically
c) Clearly as given the force acting is equal to the weight.
M=1056 tonnes = kg
W= kg=1.03
=0.25
E=225 Gpa =225 pa.
E= stress/strain
225 pa.=
So, A=30.33
Width=0.182 m=182 mm
d) Given Shear stress =80Mpa
Area of the cut=0.0127
So, the minimum force would be F=80=1.016 N
2. Lets tabulate the SF and the BD for the different diagrams.
a) As the figure shown in the questions already has the SFD we have noted down the bending moments in the table at different points. If we take the axis along the rod, the direction of the moment would be towards the reader.
| Points | Shear Force | Bending Moment |
| X1(=0.85m) | 52 N | 44.2 Nm |
| X2=(1.71m) | 52 N | 88.92 Nm |
| X3=(2.56m) | 52 N | 133.12 Nm |
| X4=(3.41m) | 52 N | 163.28 Nm |
| X5=(4.26m) | 52 N | 221.52 Nm |
| X6=(5.11m) | 52 N | 265.72 Nm |
| X7=(6m) | 195 N | 1170 Nm (Maximum BM) |
b)
| Points(in meters) | Shear force(in the units of N) | Bending Moment(in Nm) |
| X1(1.42) | 50 | 71 |
| X2(2.85) | 50 | 142.5 |
| X3(4.27) | 50 | 213.5 |
| X4(5.69) | 50 | 284.5 |
| X5(6) | 365 | 2190(Maximum BM) |
| X6(7.12) | 50 | 356 |
| X7(8.53) | 50 | 426.5 |
| X8(10) | 305 | 305 |
c)
| Points(in meters) | Shear force(in the units of N) | Bending Moment(in Nm) |
| X1(1.5) | 40 | 600 |
| X2(3) | 255 | 765 |
| X3(4.5) | 40 | 180 |
| X4(6.0) | 40 | 240 |
| X5(6.5) | 400 | 2600(Maximum BM) |
| X6(7.5) | 40 | 300 |
| X7(9) | 40 | 360 |
2(c). The shear stress is given by Stress=
Since the A is common in each which is equal to
In case 1: Maximum Shear Stress=195/=Mpa
In case 2: Maximum Shear Stress=365/ =379.4Mpa
In case 3: Maximum Shear Stress=400/ =415 Mpa
3.
a) The natural frequency of the system would be =
So,f=0.159 x 8.66=1.376 Hz
b) We equate the Kinetic and the potential energy of the body at x:
Where, A=0.65 m the maximum displacement.
Hence, V==8.66) m
c) The velocity is related to the position as
V= ,clearly it is a parabolic curve
Please note the vertical axis is the velocity and the horizontal axis is the position.
Similarly, the acceleration curve is given by :
a=So, this is the linear curve:
KD18
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