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JulijaS [17]
3 years ago
11

Problem 2

Engineering
1 answer:
mamaluj [8]3 years ago
8 0

Answer:

susmtqjqmjttqmjtqmjtmqutq

Explanation:

bakaf

fjgjgi5j6leny4mjtqjmu5tjmmwtjmjtj

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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
What are the assumptions made for air standard cycle analysis?
diamong [38]
D i took this hope it helps
4 0
3 years ago
We know that passengers can be either helpful or harmful to a driver. Describe a pro and a con of having passengers in your car.
anygoal [31]

Answer:

sub to pewdipie

6 0
3 years ago
Read 2 more answers
1. Declare a variable named num with an
FrozenT [24]

The code to

  • Declare a variable and increment with a while loop until the variable is not less than 5
  • Create a function that returns the product of two numbers
  • Uses the new keyword to create an array

were written in JavaScript and are found in the attached images

<h3>Declaring a variable</h3>

The first code declares a variable called num and gives it an initial value of 0. It then enters a while loop that lauches a message box (using <em>window.alert</em>) to print the message "<em>Keep going</em>" as long as num remains less than 5.

If nothing is done within the loop to increment num towards the value 5, the loop will go on endlessly notifying the user to "<em>Keep going</em>".

So, an increment of 1 was added to the loop body to increment the variable num. This makes sure the loop terminates.

<h3>Creating a function that returns the product of two numbers</h3>

Here, a function was created that receives two arguments (n1  and n2), then returns the product (n1 * n2)

<h3>Declaring an Array</h3>

This last code segment creates an array using the new keyword. The new keyword is generally used in constructing objects.

In this case the object constructed is an array having three strings;

  • my <em>first name</em>
  • my <em>nickname</em>, and
  • my <em>last name</em>

See another solved JavaScript problem here brainly.com/question/23610566

7 0
3 years ago
A compressed-air drill requires an air supply of 0.25 kg/s at gauge pressure of 650 kPa at the drill. The hose from the air comp
Klio2033 [76]

Answer:

L = 46.35 m

Explanation:

GIVEN DATA

\dot m  = 0.25 kg/s

D = 40 mm

P_1 = 690 kPa

P_2 = 650 kPa

T_1 = 40° = 313 K

head loss equation

[\frac{P_1}{\rho} +\alpha \frac{v_1^2}{2} +gz_1] -[\frac{P_2}{\rho} +\alpha \frac{v_2^2}{2} +gz_2] = h_l +h_m

whereh_l = \frac{ flv^2}{2D}

h_m minor loss

density is constant

v_1 = v_2

head is same so,z_1 = z_2

curvature is constant so\alpha = constant

neglecting minor losses

\frac{P_1}{\rho}  -\frac{P_2}{\rho} = \frac{ flv^2}{2D}

we know\dot m is given as= \rho VA

\rho =\frac{P_1}{RT_1}

\rho =\frac{690 *10^3}{287*313} = 7.68 kg/m3

therefore

v = \frac{\dot m}{\rho A}

V =\frac{0.25}{7.68 \frac{\pi}{4} *(40*10^{-3})^2}

V = 25.90 m/s

Re = \frac{\rho VD}{\mu}

for T = 40 Degree, \mu = 1.91*10^{-5}

Re =\frac{7.68*25.90*40*10^{-3}}{1.91*10^{-5}}

Re = 4.16*10^5 > 2300 therefore turbulent flow

for Re =4.16*10^5 , f = 0.0134

Therefore

\frac{P_1}{\rho}  -\frac{P_2}{\rho} = \frac{ flv^2}{2D}

L = \frac{(P_1-P_2) 2D}{\rho f v^2}

L =\frac{(690-650)*`10^3* 2*40*10^{-3}}{7.68*0.0134*25.90^2}

L = 46.35 m

5 0
3 years ago
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