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Marta_Voda [28]
3 years ago
9

A cantilever beam of length L = 70 in is made from two side-by-side structural-steel channels of size 3 in weighing 5.0 lbf/ft.

The beam supports a uniform external distributed load of w = 6 lbf/in and a point load of F = 170 lbf at the end. Including the weight of the channels, use Castigliano's theorem to find the magnitude of deflection at A.
Engineering
1 answer:
natali 33 [55]3 years ago
6 0

Answer:

of 5 lb/ft and a concentrated service live load at midspan. .... length = 12 feet) to support a uniformly distributed load. Taking ... w 7..'{ 'f.- ~ s-·. 344 ft-kip. Fy : s-o ks I. 299 ft-kip. Li.. ::::- I 2.. }-t-. 150 ft-kip ..... The concrete and reinforcing steel properties are ... Neglecting beam self-weight . and based only on the ...... JI : Lf, 2. l.. ;VI.

Explanation:

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For the SR-latch below high levels of Set and Reset result in Q= 1 and 0, respectively. The next state is unknown when both inpu
dusya [7]

Answer:

hello your question lacks the required image attached to this answer is the image required

answer :  NOR1(q_) wave is complementary to NOR2(q)

Explanation:

Note ; NOR 2 will be addressed as q in the course of this solution while NOR 1 will be addressed as q_

Initial state is unknown i.e q = 0 and q_= 1

from the diagram the waveform reset and set

= from 0ns to 20ns reset=1 and set=0.from the truth table considering this given condition q=0 and q_bar=1 while  

from 30ns to 50ns reset=0 and set=1.from the truth table considering this condition q=1 and q_bar=1.so from 35ns also note there is a delay of 5 ns for the NOR gate hence the NOR 2 will be higher ( 1 )

From 50ns to 65ns both set and reset is 0.so NOR2(q)=0.

From 65 to 75 set=1 and reset=0,so our NOR 2(q)=1 checking from the truth table

also  from 75 to 90 set=1 and reset=1 , NOR2(q) is undefined "?" and is mentioned up to 95ns.

since q_ is a complement of q, then NOR1(q_) wave is complementary to NOR2(q)

3 0
2 years ago
Multiple Choice
ra1l [238]
I need more details to your question
4 0
2 years ago
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If the head loss in a 30 m of length of a 75-mm-diameter pipe is 7.6 m for a given flow rate of water, what is the total drag fo
Stolb23 [73]

Answer:

526.5 KN

Explanation:

The total head loss in a pipe is a sum of pressure head, kinetic energy head and potential energy head.

But the pipe is assumed to be horizontal and the velocity through the pipe is constant, Hence the head loss is just pressure head.

h = (P₁/ρg) - (P₂/ρg) = (P₁ - P₂)/ρg

where ρ = density of the fluid and g = acceleration due to gravity

h = ΔP/ρg

ΔP = ρgh = 1000 × 9.8 × 7.6 = 74480 Pa

Drag force over the length of the pipe = Dynamic pressure drop over the length of the pipe × Area of the pipe that the fluid is in contact with

Dynamic pressure drop over the length of the pipe = ΔP = 74480 Pa

Area of the pipe that the fluid is in contact with = 2πrL = 2π × (0.075/2) × 30 = 7.069 m²

Drag Force = 74480 × 7.069 = 526468.1 N = 526.5 KN

3 0
3 years ago
A rigid insulated tank is divided into 2 equal compartments by a thin rigid partition. One of the compartments contains air, ass
Illusion [34]
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A college student volunteers with the elderly in a hospice program and discovers her clients complain of dry skin. She has an id
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D

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