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hodyreva [135]
3 years ago
7

A fully braced structural member in a building is subjected to several different loads, including roof loads of D = 5 k and L_r

= 7 k; floor loads of D = 6 k and occupancy L = 15 k; S = 18 k; and W = 17 k and E = 12 k (resulting from overturning forces on the lateral-force-resisting system). Consider the ASD load combinations described in Sec. 2.17 and assume the wind and seismic loads act in the same direction as the gravity toads. Find: The critical ASD combination of loads.

Engineering
1 answer:
allochka39001 [22]3 years ago
7 0

Answer: 37.4K

Explanation:

See attachment

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A plumbed eyewash station is portable.
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14. An engine is brought into the shop with a
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B. To accurately measure spark advance, use a timing light that incorporates an

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2 years ago
A transformer has 300,000 windings in its primary coil and uses 12,000V AC input. (4 points) How many windings would be needed t
viva [34]

Answer:

  2750

Explanation:

The number of windings and the voltage are proportional.

__

Let n represent the number of windings to produce 110 Vac. Then the proportion is ...

  n/110 = 300,000/12,000

  n = 110(300/12) = 2750 . . . . multiply by 110

2750 windings would be needed to produce 110 Vac at the output.

7 0
1 year ago
The aluminum rod (E1 = 68 GPa) is reinforced with the firmly bonded steel tube (E2 = 201 GPa). The diameter of the aluminum rod
Vsevolod [243]

Answer:

Explanation:

From the information given:

E_1 = 68 \ GPa \\ \\ E_2 = 201 \ GPa  \\ \\ d = 25 \ mm \  \\ \\ D = 45 \ mm \ \\ \\ L   = 761 \ mm  \\ \\ P = -88 kN

The total load is distributed across both the rod and tube:

P = P_1+P_2 --- (1)

Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.

\delta_1=\delta_2

\dfrac{P_1L}{A_1E_1}= \dfrac{P_2L}{A_2E_2}

\dfrac{P_1 \times 0.761}{(\dfrac{\pi}{4}\times .0025^2 ) \times 68\times 10^4}= \dfrac{P_2\times 0.761}{(\dfrac{\pi}{4}\times (0.045^2-0.025^2))\times 201 \times 10^9}

P_1(2.27984775\times 10^{-8}) = P_2(3.44326686\times 10^{-9})

P_2 = \dfrac{ (2.27984775\times 10^{-8}) P_1}{(3.44326686\times 10^{-9})}

P_2 = 6.6212 \ P_1

Replace P_2 into equation (1)

P= P_1 + 6.6212 \ P_1\\ \\ P= 7.6212\ P_1 \\ \\  -88 = 7.6212 \ P_1  \\ \\ P_1 = \dfrac{-88}{7.6212} \\ \\  P_1 = -11.547 \ kN

Finally, to determine the normal stress in aluminum rod:

\sigma _1 = \dfrac{P_1}{A_1} \\ \\  \sigma _1 = \dfrac{-11.547 \times 10^3}{\dfrac{\pi}{4} \times 25^2}

\sigma_1 = - 23.523 \ MPa}

Thus, the normal stress = 23.523 MPa in compression.

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3 years ago
What type of building project is most like an enhanced skills occupation?
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