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luda_lava [24]
4 years ago
15

I need the answer, step by step

Engineering
1 answer:
Fynjy0 [20]4 years ago
6 0

Answer:

HFOUGTYSLDÑÑYHÑ.HDug hyfñi755+464654795

Explanation:

+10000000iq

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3 years ago
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A series circuit has 4 identical lamps. The potential difference of the energy source is 60V. The total resistance of the lamps
Alexxx [7]

Answer:

I=3A

Explanation:

From the question we are told that:

Number of lamps N=4

Potential difference V=60v

Total Resistance of the lamp is R= 20ohms

Generally the equation for Current I is mathematically given by

 I=\frac{V}{R}

 I=\frac{60}{20}

 I=3A

8 0
3 years ago
An inverted tee lintel is made of two 8" x 1/2" steel plates. Calculate the maximum bending stress in tension and compression wh
Kamila [148]

Answer:

hello your question lacks some information attached is the complete question

A) (i)maximum bending stress in tension = 0.287 * 10^6 Ib-in

    (ii) maximum bending stress in compression =  0.7413*10^6 Ib-in

B) (i)  The average shear stress at the neutral axis = 0.7904 *10 ^5 psi

    (ii)  Average shear stress at the web = 18.289 * 10^5 psi

    (iii) Average shear stress at the Flange = 1.143 *10^5 psi

Explanation:

First we calculate the centroid of the section,then we calculate the moment of inertia and maximum moment of the beam( find attached the calculation)

A) Calculate the maximum bending stress in tension and compression

lintel load = 10000 Ib

simple span = 6 ft

( (moment of inertia*Y)/ I ) = MAXIMUM BENDING STRESS

I = 53.54

i) The maximum bending stress (fb) in tension=

= \frac{M_{mm}Y }{I}  = \frac{6.48 * 10^6 * 2.375}{53.54} =  0.287 * 10^6 Ib-in

ii) The maximum bending stress (fb) in compression

= \frac{M_{mm}Y }{I} = \frac{6.48 *10^6*(8.5-2.375)}{53.54} = 0.7413*10^6 Ib-in

B) calculate the average shear stress at the neutral axis and the average shear stresses at the web and the flange

i) The average shear stress at the neutral axis

V = \frac{wL}{2} = \frac{1000*6*12}{2} = 3.6*10^5 Ib

Ay = 8 * 0.5 * (2.375 - 0.5 ) + 0.5 * (2.375 - \frac{0.5}{2} ) * \frac{(2.375 - (\frac{0.5}{2} ))}{2}

= 5.878 in^3

t = VQ / Ib  = ( 3.6*10^5 * 5.878 ) / (53.54 8 0.5) = 0.7904 *10 ^5 psi

ii) Average shear stress at the web ( value gotten from the shear stress at the flange )

t = 1.143 * 10^5 * (8 / 0.5 )  psi

  = 18.289 * 10^5 psi

iii) Average shear stress at the Flange

t = VQ / Ib = \frac{3.6*10^5 * 8*0.5*(2.375*(0.5/2))}{53.54 *0.5}

= 1.143 *10^5

4 0
4 years ago
If a structure can withstand seismic stress, what is it prepared for?.
mars1129 [50]

Answer:

Steel and wood

Explanation:

For a material to resist stress and vibration, it must have high ductility, which is the ability to undergo large deformations and tension. Modern buildings are often constructed with structural steel, a component that comes in a variety of shapes and allows buildings to bend without breaking.

4 0
2 years ago
A gasoline engine takes in air at 290 K, 90 kPa and then compresses it. The combustion adds 1000 kJ/kg to the air after which th
Inessa [10]

Answer:

attached below

Explanation:

4 0
4 years ago
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