Compression Lab report PDF

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Summary

experiment on compression for different ductile and brittle materials...

Description

The American University in Cairo

ENGR 2112

Strength and Testing of Materials – 229

Section #84

Impact Lab Report

Ahmad Ali 900170037 26th, September 2019

Dr. Mohamed Salama

Abstract: During this experiment, a compression test is done on cast iron, lead and wood. The compression test for the cast iron is used to draw the engineering stress- strain graph for the cast iron which from we were able to calculate the mechanical properties for the cast iron and compare it with the results in case of tension. Furthermore, we did the test 2 times on lead where the first time had lead covered with grease and the second time had lead without grease. The purpose of doing this is to study how grease can affect the malleability of lead. Finally, we also did the test 2 times on wooden cubes where the first time the fibers was parallel to the direction of force and the second time the fibers were perpendicular. From the results, we were able to conclude how the direction of force can affect the wood mechanical properties. Finally, the fracture of the five trials is examined and a conclusion is written about it. The results have shown that grease reduce the malleability of lead and that the timber can withstand larger forces if the fibers are parallel to the direction of the force applied.

Table of Contents: Introduction---Objective---Theory---Description of the machine and Specimen---Procedure---Results and calculations---Conclusions and recommendations----

List of Figures: Figure 1 - The Universal Testing Machine---Figure 2 – Specimens before fracture---Figure 3 – Specimens after fraction ----

Nomenclature:

Unit

Symbol

Description

D0

mm

Initial diameter

Df

mm

Final diameter

L0

mm

Initial length

Lf

mm

Final length

DL

mm

Change in length

A0

mm2

Initial area

Af

mm2

Final area

P

KN

Applied Load

s

MPa

Stress

e

Unitless

Strain

sE

MPa

Elastic Strength

sY

MPa

Yield Strength

sU

MPa

Ultimate Strength

sf

MPa

Fracture Strength

εE

Unitless

Strain at yield

εf

Unitless

Strain at fracture

E

MPa

Modulus of Elasticity

R

J/m3

Modulus of Resilience

T

J/m3

Modulus of Toughness

Introduction: A compression test is a method for determining the behavior of materials under a compressive load. Compression tests are conducted by loading the test specimen between two plates, and then applying a force to the specimen by moving the crossheads together. The compression test is used to determine elastic limit, proportional limit, yield point, yield strength, and (for some materials) compressive strength, which is the maximum compressive strength a material can withstand.

Objectives: 1. Neatly plot the Engineering Stress vs. Engineering Strain Diagram for cast iron. 2. Determine the following Mechanical Properties for cast iron: a. Elastic Strength; b. Yield Strength; c. Compressive Strength (σmax); d. Modulus of Elasticity; e. Modulus of Toughness; f. Malleability. 3. Determine the following Mechanical Properties for the other Specimens: a. Compressive Strength (σmax); b. Malleability. (not for wood) 4. Compare the malleability of the lead specimen with and without grease. 5. Compare the behavior of the wood specimen when the load is applied parallel to its fibers. 6. Examine the Surface of Fracture for the tested Specimens (cast iron, lead with grease, lead without grease, wood with parallel load, and wood with perpendicular load) and Plot a schematic for their shape.

Theory:

This test is performed to determine the strength of a material under compression. Generally, the compression test is carried out to know either simple compression characteristics of material or column action of structural members. It has been observed that for varying heights of members, keeping the cross sectional and the load applied constant, there is an increased tendency towards bending of a member. Members under compression usually bend along the minor axis, i.e, along least lateral dimension. Below are the used formulas for calculating the required values.

Young’s modulus=slope of stress vs strain graph Ultimate compressive strength= Force (N) just before rupture/ (original c/s area) Percentage reduction in length= (initial length -final length) *100/initial length  = P/A Where: P is the applied load A is the cross-sectional area

Description of Machine and Specimen:

The cast-iron specimen used in the experiment had a cylindrical shape with an initial diameter of 1 cm and an initial length of 2.2 cm. The second specimen was lead with grease with an initial diameter of 1.17 cm and an initial length of 2 cm. The third specimen was lead without grease that had an initial diameter of 1.17 cm and an initial length of 2 cm. The wood specimens were two specimens. The first had perpendicular tissues which had an initial length of 4.53 cm and an initial height of 4.46 cm. The second wooden specimen, which had parallel tissues, had an initial length of 4.43 cm and an initial height of 4.62 cm. The universal testing machine had 2 plates that would compress the material until fracture applies.

The universal testing machine.

The universal testing machine. The lead with and without grease specimen

The cast- iron specimen The wooden specimens

Procedure: 1. Dimension of test piece is measured at three different places along its height/length to determine the average cross-section area. 2. Ends of the specimen should be plane. For that the ends are tested on a bearing plate. 3. The specimen is placed centrally between the two Compression plates, such that the center of moving head is vertically above the center of the specimen. 4. Load is applied on the specimen by moving the movable head. 5. The load and corresponding contraction are measured at different intervals. 6. Load is applied until the specimen fails.

Conclusion and recommendations: After examining the material in the universal testing machine, load and extension data were derived to calculate the following stress-strain graph.

The dimension used in the calculations for cast-iron, lead with and without grease.

stress

cast-iron stress vs. strain

15 14 13 12 11 10

9

8

7

6

5

4

3

2

90 80 70 60 50 40 30 20 10 0 1

strain stress

strain

Elastic strength: load/Area = 31 Yield strength: 38 Compressive strength (σmax): 82 Mod of elasticity: stress/strain = 18/3= 6 Mod of toughness: 0.5(yield+ultimate)*strain at fracture = 0.5*120*2.8=168 Malleability: %area decreased= (A-A.)/A.*100= 0.785-1.862/1.862=57% length increased= (L-L.)/L.*100= (1.21-2.2)/2.2*100= 45%

The cast-iron after deformation.

The cast-iron specimen fails after applying loads to it and it takes the shape of a triangle due to the slipping of the material so it forms a deformation along its fracture.

Lead with and without grease after deformation.

The lead specimen without grease was completely deformed and the buckling was very clear and obvious. On the other hand, the lead specimen with grease was ununiformly deformed.

Wood specimens after fracture and its fracture shape.

The test showed that wood seems to be stronger when the load is applied parallel to its fibers. The reason orientation of fibers affects the performance of material in wood is because wood is a heterogeneous material. However, it also eventually fractured for both orientation but the timing of course differed. This shows how the orientation of the samples with relevance to the fibers affect the performance of the material under compressive forces. Below are the dimensions of the wood specimens before and after applying the test.

Lead with grease Compressive Strength (σmax): 600/1.075= 558.07 Pa Malleability for lead with grease: %area decreased= (A-A.)/A.*100=(0.841.075)/1.075*100= 21.86% %length increased = (L-L.)/L.*100= (1.14-2)/1.14*100= 75.43% Malleability for lead without grease: %area decreased= (A-A.)/A.*100= (2.0351.075)/1.075*100= 89.3% %length increased= (L-L.)/L.*100= (1.14-2)/2*100= 43% Lead without grease Compressive Strength (σmax); 600/0.01075= 558.07 KPa Wood with parallel tissues Compressive Strength (σmax); 98/0.2020= 485 KPa Wood with perpendicular tissues Compressive Strength (σmax); 8.7/0.20466=42.5 KPa

Conclusion and recommendation: This test, just like any other test performed in the lab, can have room for some error. The error might arise from the fact that measurements were manually taken and read inside the lab using the Vernier caliper. Also, the environment in which the experiment is performed could somehow affect the results. Because there are some sources of error in this experiment, there definitely were some differences in the calculations done here for the mechanical values of materials and the standard ASTM ones that show different values for the materials. For example, cast-iron is shown to have yield strength of 230 MPa which is obviously different than our obtained results that were discussed earlier in the results section.

References: Philipfigari, & Instructables. (2017, October 10). Steps to Analyzing a Material's Properties From Its Stress/Strain Curve. Retrieved from https://www.instructables.com/id/Steps-to-Analyzing-a-Materials-Propertiesfrom-its/...