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Tensile Test of Steel Rod · Acedra
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Tensile Test of Steel Rod
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0:09
[Music] problem number one the following data were recorded during the tensile test of a 14 mm diameter mild steel rod the gauge length was 50 mm
0:20
so we have here the data of the load with the corresponding elongation or deformation so we are asked to plot the stress strain diagram and determine the following mechanical properties
0:34
first is the proportional limit [Music] second the modulus of elasticity third the yield point fourth the ultimate strength [Music] and the rupture strength
0:50
so what we are going to do is to determine the stress and the strain from the given data
1:00
so we have here from the given data so given the load and the corresponding elongation so we are going to compute for the stress and the strain for the corresponding load and elongation [Music]
1:15
so if we are going to recall the formula for the stress we have the stress is equal to force over area so given here is the force the load here is your P and the area can be computed from the given, if we go back on the given [Music] we have and the length of 50 mm so you must given that and we have diameter of 14 mm and the length equal to 50 mm so since given is the diameter then we can now compute for the area from the diameter
1:54
since this is equal to pi d square over 4 so the area is equal to pi
2:02
multiplied by the diameter which is 14 mm square over four then we would have the area equal to,
2:13
so we will not include pi in our computation so we have 14 squared over four and that will give us 49. so we have the area of 49 pi mm square so since we already have the value of the p given here and
2:35
the A here is the area so using the calculator and our equation would be force over area so replacing force by x and the area is 49 pi so the x that we enter here would be the value of the loads so here if x is zero we know that the stress would be zero and if x is six three ten so we have six thousand three hundred ten so we have here calc
3:11
six three ten and that is equal to forty point 99 so we would have here 40.99 megapascal and this is 40.99 mega pascal now if the load is 12 600 so 12 600 and that is equal to 81.85 megapascal
3:51
if this is 18 800 we have calc 18 800 and we have here 122.13 megapascal
4:12
[Music] for 25 100 so we have 25 100 and that is 163.05
6:08
calc zero two and that is equal to four times ten to the negative four
6:19
calc point zero three and that is
6:26
six exponent of negative four
6:32
calc point zero six and that would give us one point two exponent of negative
6:44
three calc point 163
6:54
and we have here 3.26 exponent of negative three
7:00
point four three and we have 8.66 exponent of negative 3. calc 1.25 and we have here 0.025 calc
7:21
2.5 and we have here 0.05 calc 4.5 and we have here 0.09 calc 7.5 and we have here 0.15
7:46
12.5 and we have here 0.25 calc 15.5 and we have here 0.31 calc 20 and we have here
8:06
0.4 so what if we are going to compute for the values starting from this point starting from this up to this point as you can see the elongation is constant with an increment of 0.01 so [Music]we use mode table that's seven and we input the equation so that's x the elongation over the length of 50 so and we start with 0 and we end at 0.06 with an increment of point zero one so when you hit equal then we have our answer
9:02
so for zero one that is two x one of negative four zero two that is four x one of the negative fours zero three that's six exponent of negative four and zero four that's eight exponent of negative four so zero five that's one exponent of negative three and we have here four zero six that's
9:27
one point two exponent of negative three okay since so if we have our answers for the strain since you already have the value of the stress and the strain and we are asked to find the proportional limit the modulus of elasticity yield point ultimate strength and the rupture strength [Music] I used here excel to graph the data for stress and strain diagram from the given data so next thing we do is we compute for the slope to see the behavior of our stress-strain diagram[Music]
10:16
stress so for this we have zero
10:23
this e here is the slope
10:27
at the proportional limit so [Music]
10:44
so we have
15:48
2 divided by 0.09 and that is equal to 42 to 2.44 441.74 divided by 0.15 and that is equal to
16:19
2944.98
16:29
383 point
16:33
383.27 divided by 0.25 and that is equal to [Music] 1533.08
16:53
440.44 divided by 31 and that is equal to
17:01
1420.7
17:11
422.25 divided by 0.4 and that is equal to 1055.63 [Music]
17:25
um if we are going to recheck we see here the trend of our slope but if we are going to recheck, the slope is much lower here 20 000 here so a checking the strain since we have
17:49
0.06 divided by the length of 50 and that is
18:02
and that 0.0012, we got another zero here. If we are going to recompute we have two four six point two divide point zero zero one two and that is equal to
18:20
two zero five one six six point sixty seven
18:35
we have the slope of in terms of thousand huh so we have the 204 203 205 then suddenly became 78
18:45
so we see here that the elongation is constant you know from this point until 0.06, meaning
19:02
this value here is the value of the proportional limit so that is
19:14
equal to 246.2 mega pascal and if we look at here the slope changed to 78 and this 78 here is
19:40
the value of your elastic limit
19:47
and the next one is the value of the yield point that is equal to 270.24
19:59
and the ultimate strength, looking at the stress data the highest the data is the 441 so this is the ultimate strength and the rupture strength is the load or the stress at failure so this is the rupture strength so this is the answer for a. 246.2
20:28
megapascal and the letter b since we are asked to compute for the modulus of elasticity
20:37
so modulus of elasticity is the slope at the proportional limit so this is now value modulus of elasticity and this is the value of the E since that is
20:55
205,166.67 the unit of this is mega pascal now if we are going to express this in gigapascal divide it by 1000 so we can say that modulus of elasticity E is equal to 205.2 GPa and this is the answer for the part b so part c is the yield point so this is the answer for the yield point this is the answer for the ultimate strength and this is the answer for the rupture strength
21:33
now I also solved the problem using excel
21:38
This is the computation using Excel [Music] If we look at the slope, we see that they are close in value that is 204, 203, 205
21:55
and here corresponding to this is the proportional limit so the stress at that point is 246.2 megapascal we have here the next point is the elastic limit that's 260 0.49 megapascal and the yield point is to 70.24 megapascal ultimate strength is 441.74 megapascal and the rupture strength of 399.51 megapascal now uh for the value of the
22:27
modulus of elasticity it is the slope at the proportional limit that is equal to 205 169.1 megapascal or equal to 205.2 GPa [Music]
22:44
so if you want to have access on this excel file you may check the description of this video and i posted there the link of this excel file [Music]
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