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3 years ago

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A rigid bar ABCD is pinned at A and supported by two steel rods connected at B and C, as shown. There is no strain in the vertic

al rods before load P is applied. After load P is applied, the normal strain in rod (2) is 800 με. Determine : (ajthe axial normal strain in rod (1). b)the axial normal strain in rod (1) if there is a 1 mm gap in the connection between the rigid bar and rod (2) before the load is applied. 2.7 m 1.5 m Rigid bar 2.5 m

1 answer:

mylen [45]3 years ago

4 0

Answer:

See attached picture.

Explanation:

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3 years ago

Bananas are to be cooled from 24°C to 13°C at a rate of 215 kg/h by a refrigeration system. The power input to the refrigerator

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Answer:

a) $\dot Q_{L} = 132.046\,\frac{kJ}{min}$ , b) $COP_{R} = 1.467$

Explanation:

a) The heat rejected by the bananas is:

$\dot Q_{L} = \dot m \cdot c\cdot (T_{f}-T_{o})$

$\dot Q_{L} = \left(215\,\frac{kg}{h}\right)\cdot \left(\frac{1\,h}{60\,min} \right)\cdot \left(3.35\,\frac{kJ}{kg\cdot ^{\textdegree}C} \right)\cdot (24^{\textdegree}C-13\,^{\textdegree}C)$

$\dot Q_{L} = 132.046\,\frac{kJ}{min}$

b) The coefficient of performance for the refrigeration system is:

$COP_{R} = \frac{\dot Q_{L}}{\dot W}$

$COP_{R} = \frac{\left(132.046\,\frac{kJ}{min}\right)\cdot \left(\frac{1\,min}{60\,s} \right)}{1.5\,kW}$

$COP_{R} = 1.467$

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3 years ago

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a

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Answer:

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3 years ago

A beam of span L meters simply supported by the ends, carries a central load W. The beam section is shown in figure. If the maxi

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Answer:

W = 11,416.6879 N

L ≈ 64.417 cm

Explanation:

The maximum shear stress, $\tau_{max}$ , is given by the following formula;

$\tau_{max} = \dfrac{W}{8 \cdot I_c \cdot t_w} \times \left (b\cdot h^2 - b\cdot h_w^2 + t_w \cdot h^2_w \right )$

$t_w$ = 1 cm = 0.01

h = 29 cm = 0.29 m

$h_w$ = 25 cm = 0.25 m

b = 15 cm = 0.15 m

$I_c$ = The centroidal moment of inertia

$I_c = \dfrac{1}{12} \cdot \left (b \cdot h^3 - b \cdot h_w^3 + t_w \cdot h_w^3 \right )$

$I_c$ = 1/12*(0.15*0.29^3 - 0.15*0.25^3 + 0.01*0.25^3) = 1.2257083 × 10⁻⁴ m⁴

Substituting the known values gives;

$I_c = \dfrac{1}{12} \cdot \left (0.15 \times 0.29^3 - 0.15 \times 0.25^3 + 0.01 \times 0.25^3 \right ) = 1.2257083\bar 3 \times 10^{-4}$

$I_c$ = 1.2257083 $\bar 3$ × 10⁻⁴ m⁴

From which we have;

$4,500,000 = \dfrac{W}{8 \times 1.225708\bar 3 \times 10 ^{-4}\times 0.01} \times \left (0.15 \times 0.29^2 - 0.15 \times 0.25^2 + 0.01 \times 0.25^2 \right )$

Which gives;

W = 11,416.6879 N

$\sigma _{b.max} = \dfrac{M_c}{I_c}$

$\sigma _{b.max}$ = 1500 N/cm² = 15,000,000 N/m²

$M_c$ = 15,000,000 × 1.2257083 × 10⁻⁴ ≈ 1838.56245 N·m²

From Which we have;

$M_{max} = \dfrac{W \cdot L}{4}$

$L = \dfrac{4 \cdot M_{max}}{W} = \dfrac{4 \times 1838.5625}{11,416.6879} \approx 0.64417$

L ≈ 0.64417 m ≈ 64.417 cm.

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3 years ago

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andrezito [222]

Why not simply include each step's setup instructions within it? is it due to our desire to maintain DRY (Don't Repeat Yourself) code?

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Program testing is known as "unit testing" involves testing individual software components. When developing an application, unit testing is done on the software product. An individual component could be a technique or a specific function.

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The core of the testing process consists of unit testing and functional testing. The primary distinction between the two is that during the development cycle, the developer conducts unit testing. The tester does functional testing at the system testing level.

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