Answer:
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Explanation:
Answer:
B A and C
Explanation:
Given:
Specimen σ
σ
A +450 -150
B +300 -300
C +500 -200
Solution:
Compute the mean stress
σ
= (σ + σ)/2
σ
= (450 + (-150)) / 2
= (450 - 150) / 2
= 300/2
σ = 150 MPa
σ
= (300 + (-300))/2
= (300 - 300) / 2
= 0/2
σ = 0 MPa
σ
= (500 + (-200))/2
= (500 - 200) / 2
= 300/2
σ = 150 MPa
Compute stress amplitude:
σ
= (σ - σ)/2
σ
= (450 - (-150)) / 2
= (450 + 150) / 2
= 600/2
σ = 300 MPa
σ
= (300- (-300)) / 2
= (300 + 300) / 2
= 600/2
σ = 300 MPa
σ
= (500 - (-200))/2
= (500 + 200) / 2
= 700 / 2
σ = 350 MPa
From the above results it is concluded that the longest fatigue lifetime is of specimen B because it has the minimum mean stress.
Next, the specimen A has the fatigue lifetime which is shorter than B but longer than specimen C.
In the last comes specimen C which has the shortest fatigue lifetime because it has the higher mean stress and highest stress amplitude.
Explanation:
Production planning: Planning is ideal so that there are the right resources, at the right time and in the right quantity that can meet the production needs of a period.
Strategies: The strategic development of production is the area that will assist in organizational competitiveness and in meeting consumer demand and needs.
Product and service design: Development of new products and services and their improvement, innovations and greater benefits
Production systems: Study of physical arrangements so that production takes place effectively according to the ideal layout for each type of product or service.
Production capacity planning: Analysis of the short, medium and long term related to production, and identification if necessary to obtain more resources, increase in staff, machinery, etc., to meet present and future demands.
<em>Each area of knowledge acquired will assist in the development of a professional career, as technical knowledge is essential in decision-making, provision, problem solving, the development of new ideas and innovation.</em>
