A cylindrical specimen of aluminum having a diameter of (10...

A cylindrical specimen of aluminum having a diameter of (10 mm) and a gauge length of (60 mm) is pulled in tension. Use the load – elongation characteristics tabulated below to complete problems a through e. a. Plot the data as engineering stress versus engineering strain. b. Determine Young Modulus c. Determine the yield strength at a strain offset of 0.002. d. Determine maximum tensile strength of this alloy. e. What is the approximate ductility, in percent elongation? Load(P) (KN): 0, 6.32, 9.48, 15.8, 22.15, 22.11, 22.05, 25.27, 28.43, 25.27, 20.5 Elongation(?L) (mm): 0, 0.002, 0.9, 1.1, 1.98, 2.1, 2.383, 7.62, 11.91, 20.2, 21.9

Transcript

00:01 Hello everybody here in this problem we know that young's modulus y is equal to force p by area a divided by change in length that is del l by original length that is l so we can simply write it as force p by a into capital l by dell l.

00:28 Now p can be written as y a by l into dell l now if we put p versus elongation that is del l graph then it will become straight line passing through origin because this equation is equivalent to this equation y equals to mx and y equals to m x this equation is a straight line equation that passes through origin now m is the slope of the curve or slope of the straight line that means m equals to tan theta but theta is the angle the straight line makes with positive x axis so from this equation we can say if this is the slope and this is theta so simply m will become tan theta means this y or del y by del x here del x is del l and del y is del p can write m is equal to y a by l comparing the previous equation now here in this graph p is in kilo newton so kilo one kilo newton is equal to 10 to the above 3 newton and 1 millimeter millimeter is a unit of del l is equal to 10 to the power minus 3 meter so we have to find m by del y by del x that del y del y into 10 to the power 3 because we have converted it to newton and this one into 10 to the power minus 3.

02:33 So del y by del x what we get should be multiplied 6.

02:40 Now l is given by 60 millimeter or simply 0 .06 meter and d the diameter is given by 10 millimeter that is 10 into 10 to the power minus 3 meter or simply 10 to the power minus 2 meter now area is equal to pi d square by 4 some 78 .5 into 10 to the power minus 6 that much meter square m l by so simply we will get the value of y or youngs model now tensile strength maximum road by area from the data given we see that maximum load is 28 .4 3 that much kilo newton so simply we convert it to newton that is 28 .4 3 into 10 to the power 3 newton and area we found out as 78 .5 into 10 to the per minus 6 meters square.

04:15 We can easily find out tensile strength that is equal to 28 .43 into 10 to the power 3 divided by 78 .5 into 10 to the over minus 6.

04:35 That much newton per meter square.

04:39 So simply it will become 0 .36 into 10 to the power 9 newton per meter square.

04:47 Next is yield strength.

04:56 Ill strength is the stress where material changes from elastic to plastic behavior...

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