Answer:
hello the diagram attached to your question is missing attached below is the missing diagram
answer :
a) 48.11 MPa
b) - 55.55 MPa
Explanation:
First we consider the equilibrium moments about point A
∑ Ma = 0
( Fbd * 300cos30° ) + ( 24sin∅ * 450cos30° ) - ( 24cos∅ * 450sin30° ) = 0
therefore ;<em> Fbd = 36 ( cos ∅tan30° - sin∅ ) kN ----- ( 1 )</em>
A ) when ∅ = 0
Fbd = 20.7846 kN
link BD will be under tension when ∅ = 0, hence we will calculate the loading area using this equation
A = ( b - d ) t
b = 12 mm
d = 36 mm
t = 18
therefore loading area ( A ) = 432 mm^2
determine the maximum value of average normal stress in link BD using the relation below
бbd = 
= 20.7846 kN / 432 mm^2 = 48.11 MPa
b) when ∅ = 90°
Fbd = -36 kN
the negativity indicate that the loading direction is in contrast to the assumed direction of loading
There is compression in link BD
next we have to calculate the loading area using this equation ;
A = b * t
b = 36mm
t = 18mm
hence loading area = 36 * 18 = 648 mm^2
determine the maximum value of average normal stress in link BD using the relation below
бbd = = -36 kN / 648mm^2 = -55.55 MPa
Answer:
Functional Relationship
Explanation:
A functional relationship is only achieved through control and involves a specific change in one event (dependent variable) that can reliably be produced by specific manipulations of another event (independent variable), and the change in the dependent variable is unlikely to be the result of other extraneous factors (confounding variables
Answer:
The statement (a) In a non-deterministic FSM, a string is invalid if there is one path not leading to a final state is NOT true
Explanation:
A non-deterministic FSM, contrary to deterministic FSM which has only one possible thread of execution, has multiple threads and for the machine to be invalid, all threads should lead to a none accepting (final) state.
Answer:
\epsilon = 0.028*0.3 = 0.0084
Explanation:
\frac{P_1}{\rho} + \frac{v_1^2}{2g} +z_1 +h_p - h_l =\frac{P_2}{\rho} + \frac{v_2^2}{2g} +z_2
where P_1 = P_2 = 0
V1 AND V2 =0
Z1 =0
h_P = \frac{w_p}{\rho Q}
=\frac{40}{9.8*10^3*0.2} = 20.4 m
20.4 - (f [\frac{l}{d}] +kl) \frac{v_1^2}{2g} = 10
we know thaTV =\frac{Q}{A}
V = \frac{0.2}{\pi \frac{0.3^2}{4}} =2.82 m/sec
20.4 - (f \frac{60}{0.3} +14.5) \frac{2.82^2}{2*9.81} = 10
f = 0.0560
Re =\frac{\rho v D}{\mu}
Re =\frac{10^2*2.82*0.3}{1.12*10^{-3}} =7.53*10^5
fro Re = 7.53*10^5 and f = 0.0560
\frac{\epsilon}{D] = 0.028
\epsilon = 0.028*0.3 = 0.0084
