Compare and contrast mechanical properties of plastics, metals and ceramics.
1 answer:
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Answer:
stress_ac = 5.333 MPa
shear stress_c = 1.763 MPa
Explanation:
Given:
- The missing figure is in the attachment.
- The dimensions of member AC = ( 6 x 25 ) mm x 2
- The diameter of the pin d = 19 mm
- Load at point A is P = 2 kN
Find:
- Find the axial stress in AE and the shear stress in pin C.
Solution:
- The stress in member AE can be calculated using component of force P along the member AE as follows:
stress_ac = P*cos(Q) / A_ae
Where, Angle Q: A_E_B and A_ac: cross sectional area of member AE.
cos(Q) = 4 / 5 ..... From figure ( trigonometry )
A_ae = 0.006*0.025*2 = 3*10^-4 m^2
Hence,
stress_ae = 2*(4/5) / 3*10^-4
stress_ae = 5.333 MPa
- The force at pin C can be evaluated by taking moments about C equal zero:
(M)_c = P*6 - F_eb*3
0 = P*6 - F_eb*3
F_eb = 0.5*P
- Sum of horizontal forces for member AC is zero:
P - F_eb - F_c = 0
F_c = 0.5*P
- The shear stress of double shear bolt is given by an expression:
shear stress = shear force / 2*A_pin
Where, The area of the pin C is:
A_pin = pi*d^2 / 4
A_pin = pi*0.019^2 / 4 = 2.8353*10^-4 m^2
Hence,
shear stress = 0.5*P / 2*A_pin
shear stress = 0.5*2 / 2*2.8353*10^-4
shear stress = 1.763 MPa
Answer:
A) n = 0.6143, c ≈ 640m/min
B) n = 0.6143 , c = 637.53m/min
Explanation:
using the given data
A) A plot of flank wear as a function of time and also A plot for tool when
Flank wear is 0.75 and cutting edge speed is 100m/min, Time of cutting edge is said to be 20.4 min also for cutting edge speed of 155m/min , time for cutting edge is 10 min
is attached below
calculate for the constant N from the second plot
note : the slope will be negative because cutting speed decreases as time of cutting increase
V1 = 100m/min , V2 = 155m/min, T1 = 20.4 min, T2 = 10 min
= - N =
therefore - N =
= - 0.6143
THEREFORE ( N ) = 0.6143
Determine for the constant C from the second plot as well
note : C is the intercept on the cutting speed axis in 1 min tool life
connecting the two points with a line and extend it to touch the cutting speed axis and measure the value at that point
hence C ≈ 640m/min
B) Calculate the values of N and C in the Taylor equation solving simultaneous equations
using the above cutting speed and time of cutting values we can find the constant N via Taylor tool life equation
Taylor tool life equation = vT = C ------------- equation 1
cutting speed = v = 100m/min and 155m/min
tool life = T = 20.4 min and 10 min
also constant n and c are obtained from the previous plot
back to taylor tool life equation = 100 * 20.4 = C
therefore C = (100)(20.4)^n ---------------- equation 2
also using the second values of v and T
taylor tool life equation = 155 * 10 = C
therefore C = ( 155 )(10)^n ----------------- equation 3
Equate equation 2 and equation 3 and solve simultaneously
(100)(20.4)^n = (155)(10)^n
To find N
take natural log of both sides of the equation
= In ((100)(20.4)^n) = In((155)(10)^n)
= In (100) + nIn(20.4) = In(155) + nIn(10)^n
= n(3.0155) - n (2.3026) = 5.043 - 4.605
= 0.7129 n = 0.438
therefore n = 0.6143
To find C
substitute 0.6143 for n in equation 2
C = (100)(20.4) ^ 0.6143
C = 637.53 m/min
Attached are the two plots for solution A
