Compression test on MS and CI - Guide PDF

Title Compression test on MS and CI - Guide
Course Engineering Mechanics
Institution The Maharaja Sayajirao University of Baroda
Pages 6
File Size 261.8 KB
File Type PDF
Total Downloads 44
Total Views 149

Summary

Compression test on MS and CI - Practical Guide...


Description

Materials and Structures Laboratory





   To study behavior of mild steel and cast iron under uniaxial compression.  Universal Testing Machine (UTM), Mild Steel and Cast Iron specimen, Vernier Callipers, etc.  In this test, a prepared specimen is subjected to a gradually increasing uniaxial compressive load until failure occurs. In order to uniform stress distribution, circular cross section is preferred. In addition, it is important to confine the specimen height such that effect of bending and friction at the specimen ends are avoided.

As per ASTM E 9, a specimen can be categorized as short, medium and long depending on H/D ratio where H is the height and D is the diameter of the specimen.  !"#

 $% &'  (  H/D < 1.0

2.0 < H/D > 3.0

3.0 < H/D >10.0

If the length of the specimen is too long, there is increasing tendency towards bending of the specimen as shown in Fig. 1. Also if the specimen is too short, effect of frictional resistance at the ends is a concern. Moreover, the diagonal plane along which failure would take place might intersect the base, apparently increasing the strength, Fig. 2. Considering these two factors, H/D ratio between 2 and 3 is recommended for specimens to be used in a compression test.

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Materials and Structures Laboratory

Fig.1 Long Specimen

Fig.2 Short Specimen

   Measure initial diameter and height of cast iron and mild steel specimen. Insert one of the specimens in the UTM for testing. Make sure that the specimen is aligned properly with the plunger in order to get uniaxial compression. Apply compressive load at an increment of 5 BT and note down height, top diameter, diameter at the middle and bottom diameter of the specimen after each increment of load. For cast iron, record the load at which the specimen breaks. Mild steel specimen will not break but will show excessive deformation. Application of load on mild steel specimen should be stopped when change in height is approximately between 15 to 20 % of the original height. After completion of test, calculate ultimate and permissible compressive strength of cast iron using factor of safety equal to 3. Also plot graphs of load versus height and load versus diameter for both specimens and comment on the results.

)*  For cast iron specimen 1. Initial height = ______ mm 2. Initial Diameter = _____mm

For mild steel specimen 1. Initial height = ______ mm 2. Initial Diameter = _____mm  14

Materials and Structures Laboratory

)*)

No. Load Load

Mild Steel Cast Iron Height Diameter (mm) Height Diameter (mm) (BT) (N) (mm) Top Middle Bottom (mm) Top Middle Bottom

 For cast iron specimen Resisting Area =

σ ult 

Pult Re sisting Area

 σ per 

σ ult  Factor of safety

      15

Materials and Structures Laboratory

Following care should be taken in a compression test. 1. As far as possible, the applied load should be axial or truly concentric in order to avoid effect of bending stress in the result. 2. Length of specimen should be short enough that under axial

compression,

it should

not buckle. 3. Specimen should be long enough so that friction between the bearing plates and end surfaces of specimen is minimum

+  Mild steel being a ductile material, the specimen will not rupture but will show excessive deformation.

For brittle materials, specimen ruptures along a diagonal plane or with a cone or pyramidal shaped fracture as shown in Fig. 3.

For a material whose resistance to failure is due to internal friction as well as cohesion, the angle of rupture θ is a function of angle of internal friction Φ. For CI, failure occurs due to shear stress and the failure plane is approximately at θ = 45 + Φ/2 to the major principal plane as shown in Fig. 4.



 

 Fig. 3 Specimen

Fig.4 Mohr’s Circle Diagram

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Materials and Structures Laboratory















 !! 











AFTER TEST GRAPHS:  For Mild Steel Specimen: Height v/s Load Diameter v/s Load

 For Cast Iron Specimen: Height v/s Load Diameter v/s Load 17

Materials and Structures Laboratory

RESULT:

CONCLUSION:

18...


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