The load on a bolt consists of an axial pull of 10 KN together with a transverse shear force of 5 KN. Find the diameter of bolt required according to I. Maximum principal stress theory; 2. Maximum shear stress theory; 3. Maximum principal strain theory, 4. Maximum strain energy theory, and 5 Maximum distortion energy theory. Take permissible tensile stress at elastic limit = 100 MPa and poisson's ratio = 0.3​

Explanation:

Find a minimal set of cycles that covers all vertices and edges of the circuit graph. For each cycle, define a "mesh" current, and write the Kirchhoff's Voltage Law (KVL) equation with respect to each of the edges in the cycle. Where an edge is part of more than one cycle, all current(s) defined for the edge will contribute to the voltage there.

This will give as many equations as there are mesh currents. Solve the resulting system of equations. The (signed) sum of the mesh currents through any edge is the current in that circuit branch.

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Example

Consider the attached circuit. It shows mesh currents I1, I2, and I3 in graph cycles with those numbers. The KVL equations are ...

  mesh 1: I1(R3 +R2 +R1) -I2·R1 -I3·R2 = Vi (the voltage across the current source)

  mesh 2: -I1·R1 +I2(r1 +1/(sC)) -I3(1/(sC)) = Vs

  mesh 3: -I1·R2 -I2(1/(sC)) +I3(R2 +sL +1/(sC)) = 0

You will note that the matrix of equation coefficients is symmetric.

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In this example, you will end with I1 as a function of Vi. If I1 is a given source value, that relation can be used to find Vi.

Answer:

[tex]D=0.41m[/tex]

Explanation:

From the question we are told that:

Discharge rate [tex]V_r=0.35 m3/s[/tex]

Distance [tex]d=4km[/tex]

Elevation of the pumping station [tex]h_p= 140 m[/tex]

Elevation of the Exit point [tex]h_e= 150 m[/tex]

Generally the Steady Flow Energy Equation SFEE is mathematically given by

[tex]h_p=h_e+h[/tex]

With

[tex]P_1-P_2[/tex]

And

[tex]V_1=V-2[/tex]

Therefore

[tex]h=140-150[/tex]

[tex]h=10[/tex]

Generally h is give as

[tex]h=\frac{0.5LV^2}{2gD}[/tex]

[tex]h=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]

Therefore

[tex]10=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]

[tex]D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}[/tex]

[tex]D=0.41m[/tex]

Answer:

Explanation:

From the given information:

weight of fiber [tex]w_f[/tex] = 3.0 g

weight of composite specimen [tex]w_c[/tex] = 4.0 g

specimen composite weight in water [tex]C_{wm}[/tex] = 2.0 g

specific gravity of fiber [tex]S_f[/tex] = 2.4

specific gravity of matrix [tex]S_m[/tex] = 1.3

The weight of the matrix = weight of the composite - the weight of fiber

⇒ (4.0 - 3.0) g

= 1.0 g

The theoretical density of the composite [tex]\rho_{ct}[/tex] can be determined by using the formula:

[tex]\dfrac{1}{\rho_{ct}} = \dfrac{w_f}{w_cS_f}+ \dfrac{w_m}{w_cS_m}[/tex]

[tex]\dfrac{1}{\rho_{ct}} = \dfrac{3.0}{(4.0 \times 2.4)}+ \dfrac{1.0}{(4.0\times 1.3)}[/tex]

[tex]\dfrac{1}{\rho_{ct}} = \dfrac{3.0}{9.6}+ \dfrac{1.0}{5.2}[/tex]

[tex]\dfrac{1}{\rho_{ct}} =0.505\\[/tex]

[tex]\rho_{ct} =\dfrac{1}{0.505}[/tex]

[tex]\mathbf{\rho_{ct} = 1.980 \ g/cm^3}[/tex]

The experimental density [tex]\rho _{ce}[/tex] is determined  by using the equation:

[tex]\rho _{ce} = \dfrac{w_f + w_c}{\dfrac{w_f }{S_f} + \dfrac{w_c }{S_m} }[/tex]

[tex]\rho _{ce} = \dfrac{3.0 + 4.0}{\dfrac{3.0 }{2.4} + \dfrac{4.0 }{1.3} }[/tex]

[tex]\rho _{ce} = \dfrac{3.0 + 4.0}{1.250 +3.077 }[/tex]

[tex]\mathbf{\rho _{ce} = 1.620 \ g/cm^3}[/tex]

The void fraction is: [tex]= \dfrac{\rho_{ct}-\rho_{ce}}{\rho_{ct}}[/tex]

[tex]= \dfrac{1.980-1.620}{1.980}[/tex]

= 0.1818

Answer:

5.16 M

Explanation:

Loads ; 1800N, 2100N, 2275N, 2200N

safety factor = 5

diameter of each column = 50 cm = 0.5 m

Elastic modulus = 10 MPa

Calculate the max height of structure

moment of inertia for a circular section ( I ) = πd^4 / 64

lets represent the required maximum height of the column as L

Applying Euler column theory

The bucker load of the column =  ( attached below )

attached below is the remaining solution

Answer:

f = 0.04042

Explanation:

temperature = 0°C = 273k

p = 600 Kpa

d = 40 millemeter

e = 10 m

change in  P = 235 N/m²

μ = 2m/s

R = 188.9 Nm/kgk

we solve this using this formula;

P = ρcos*R*T

we put in the values into this equation

600x10³ = ρcos * 188.9 * 273

600000 = ρcos51569.7

ρcos = 600000/51569.7

=11.63

from here we find the head loss due to friction

Δp/pg = feμ²/2D

235/11.63 = f*10*4/2*40x10⁻³

20.21 = 40f/0.08

20.21*0.08 = 40f

1.6168 = 40f

divide through by 40

f = 0.04042