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gtnhenbr [62]
3 years ago
13

Calculate the reactions at 4 ends (supports) of this bookshelf. Assume that the weight of each book is approximately 1 lb. The w

idth of shelf is 10inches and its length is 40 inches. (Hint: Model the shelf as a beam with distributed load).

Engineering
1 answer:
hoa [83]3 years ago
8 0

Answer:

R = 4.75 lb  (↑)

Explanation:

Number of books = n = 19

Weight of each book = W = 1 lb

Length of the bookshelf = L = 40 inches

We can get the value of the distributed load as follows

q = n*W/L = 19*1 lb/ 40 inches = 0.475 lb/in

then the reactions at 4 ends (supports) of the bookshelf are

R = (q/2)/2 = 4.75 lb

We can see the bookshelf in the pic.

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What motivated software engineers to move from the waterfall model to the incremental or spiral model
scoray [572]

Answer:

1. They needed to develop multiple components in software programs.

2. The ability to overlap the development to be more evolutionary in nature.

3. The need to be more risk-averse or the unwillingness to take risks led to the use of a spiral model.

Explanation:

Software development life cycle (SDLC) can be defined as a strategic process or methodology that defines the key steps or stages for creating and implementing high quality software applications.

In SDLC, a waterfall model can be defined as a process which involves sequentially breaking the software development into linear phases. Thus, the development phase takes a downward flow like a waterfall and as such each phase must be completed before starting another without any overlap in the process.

An incremental model refers to the process in which the requirements or criteria of the software development is divided into many standalone modules until the program is completed.

Also, a spiral model can be defined as an evolutionary SDLC that is risk-driven in nature and typically comprises of both an iterative and a waterfall model. Spiral model of SDLC consist of these phases; planning, risk analysis, engineering and evaluation.

<em>What motivated software engineers to move from the waterfall model to the incremental or spiral model is actually due to the following fact;</em>

  • They needed to develop multiple components in software programs.
  • The ability to overlap the development to be more evolutionary in nature.
  • The need to be more risk-averse or the unwillingness to take risks led to the use of a spiral model.
6 0
3 years ago
Which of the following is typically wom when working in an atmosphere containing dust?
alukav5142 [94]

Answer:

Either D or C

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3 0
3 years ago
Where does Henrietta think William might be?
umka2103 [35]

Answer:

The cemetery

Explanation:

5 0
3 years ago
Read 2 more answers
What is JIT and why would you advocate it to reduce inventory?
bazaltina [42]

Explanation:

Just in time inventory is a system of managing inventory. It is a method to reduce waste, lower cost and increase profit. According to it the raw material orders are directly aligned with the production schedules. In simple words every component of a unit is arriving just in time to be used.

So JIT would be advocated to reduce inventory.

7 0
3 years ago
An aluminum alloy [E = 73 GPa] cylinder (2) is held snugly between a rigid plate and a foundation by two steel bolts (1) [E = 18
V125BC [204]

Answer:

\sigma_A = 58.43 N/mm^2

Explanation:

Given data:

length of Steel bolt L_1 = 335 mm

Length of aluminium cylinder L_2 = 275 mm

Pitch of bolt p = 1mm

Modulus of elasticity of steel E = 215 GPa

Modulus of elasticity of aluminium =  74 GP

Area of bolt = \frac{\pi}{4} 14^2 =  153.93 mm^2

Area of cylinder  = 2300 mm^2

n =1

By equilibrium

\sum F_y = 0

P_A -2P_S = 0

P_A =2P_S

By the compatibility

\delta _s + \delta_A = nP

Displacement in steel is \delta_s = \frac{P_sL_s}{E_sA_s}

Displacement in Aluminium is \delta_A = \frac{P_AL_A}{E_AA_A}

from compatibility equation we have

\frac{P_sL_s}{E_sA_s} +  \frac{P_AL_A}{E_AA_A} = nP

\frac{P_s\times 335}{180\times 10^3\times 153.93} +  \frac{P_A\times 275}{73\times 10^3\times 2300} = 1\times 1

1.20\times 10^[-5} P_s  + 1.44\times 10^{-6}P_A = 1

substituteP_A =2P_S

1.20\times 10^[-5} P_s  + 1.44\times 10^{-6} (2\times P_s) = 1

1.488\times 10^{-5} P_s = 1

P_s = 67204.30 N

P_A = 134,408.60 N

Stress in Aluminium \sigma  = \frac{P_A}{A_A}

                                               = \frac{134,408.60}{2300}

                             \sigma_A = 58.43 N/mm^2

8 0
3 years ago
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