Question

An extruded, rectangular polypropylene strut of width 9 mm, depth 3 mm and length 80 mm is to be loaded from the ends in compression along its length with a creep load of 40 N for one year. [Creep curves for polypropylene are shown below. The buckling load is given by F = (2 p/ L)2 E I. For a rectangular beam the second moment of area is given by I = (w d3) /12.].

In order to improve buckling resistance, explain with reference to the buckling equation, what changes are possible in either material or beam design, indicating any likely constraints on such changes.

Discuss the manufacturing problems (in terms of extruder system design) which must be considered in producing such a beam.

Answer 1

Given that Buckling force

Fb = (2/ L) ² E I

Text book formula

lets work with the formulae in hand

Fb = 4² E I / L²

Resistance to buckling means More force of buckling

To have more force Fb, The numerator must be large as much as possible.

The only variable in numerator are E and I

Also To have More Fb the denominator should be as small as possible . But here i belive the length must be constant since we have to connect the beam between 2 points.

E = elastic modulous

I = moment of inertia of the crosection.

I - w * d³ / 12

Numerator values are the width and depth of beam , if we increase the depth considerably the I value is going to increase since d³.

So to improve the buckling resistance

1. Increase the CSA of the beam ( importantly width)

2. Change material for a higher elastic modulous.

Lets find the Buckling load for given values

w	9
d	3
I = (w* d^3)/12	20.25

Fb = 4² E I / L².

w	0.009	m
d	0.003	m
I = (w* d^3)/12	2.03E-11
E	1325	Mpa	1.33E+09	pa
length L	0.08
Fb	165.5086	N

Lets change the  

Changing the width from 9 to 12 increased the Fb to 220 N

w	12	mm	0.012	m
d	3	mm	0.003	m
I = (w* d^3)/12	2.7E-11
E	1325	Mpa	1.33E+09	pa
length L	0.08
Fb	220.6782	N

Increasing the depth to 5 mm changed the load Fb to 766 N

w	9	mm	0.009	m
d	5	mm	0.005	m
I = (w* d^3)/12	9.38E-11
E	1325	Mpa	1.33E+09	pa
length L	0.08
Fb	766.2437	N

There is a huge change .

The advantages of extrusion moulding:

1. High speed, high volume
2. Plastic products can be produced quickly and at a high volume, ensuring cost-efficiency and speed.
3. Low cost
4. Extrusion moulding is relatively inexpensive compared to other moulding processes. Leftover materials can be used, reducing waste and costs while extrusion machinery can operate continuously, allowing for a 24 hours-a-day manufacturing period.
5. Flexibility
6. Extrusion moulding allows considerable flexibility in the manufacturing of products with a consistent cross-section. A variation on the extrusion moulding process also allows for products that mix plastic attributes.
7. Post-extrusion alterations
8. The plastic materials remain warm after the process is complete, allowing for post-extrusion manipulations. Rollers, shoes and die can be used in order to alter the product while it is still warm.

The disadvantages of extrusion moulding:

1. Product limitations :This process places certain limits on the types of products that can be produced. Plastic soda bottles, for instance, cannot be produced as the shape narrows at one end, which one cannot achieve with standard extrusion moulding machinery. While alternatives, such as blow moulding exist, they require additional investment.
2. Size variances : As the plastic material exits the machinery, it frequently expands due to heat. It is difficult to predict the exact degree of expansion as the expansion is influenced by various factors in the process, meaning that manufacturers must often accept deviations from the standard product dimensions or tolerance.

Note that the main disadvantage of extrusion process that the size cannot be maintained especially for long columns

Also there will be warping which trigger and easy buckling when we increase the thickness beyond lomits.