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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:
Q = 8.845 DEGREE
Explanation:
given data:
combine Mass for 6 cylinder (M) =15 Kg/hr
mass of each cylinder (m) = 15/6 = 2.5 Kg/hr = 0.000694 Kg/ sec
Engine speed (N)= 1500rpm
Diameter of one nozzle hole ( d) = 200 micrometer = 0.0002 m
Discharge Coefficient (Cd) = 0.75
Pressure difference = 100 MPa
Density of fuel = 800 kg/m^3
velocity of fuel is
injected fuel volume (V) =Area of given Orifices × Fuel velocity × time of single injection × no of injection/sec
we know that p = m/ V
So
putting these value in volume equation and solve for Discharge
Q = 8.845 DEGREE