The built-up shaft as shown in Figure 1 consists of a pipe AB and solid rod BC. The pipe has an inner diameter of 20 mm and outer diameter of 28 mm. The rod has a diameter of 12 mm. (i) Determine the resistive forces acting at points D and E if the stresses produced in the shaft at these points are 13.3 MPa (compressive) and 70.7 MPa (tensile) respectively. (ii) Draw free body diagrams and show the resistive forces on cross-sections at points D and E. (iii) Draw free body diagrams and show the stress distribution along cross-sections at points D and E.
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Answer:
Explanation:
Average Normal Stress at a point is defined as the axial force divided by cross section area.
sigma=F/A,
here,
Sigma= average normal stress ,
F = axial force ,
A= cross section area
At D,
Area Ad = pi [(28/2)2 - (20/2)]
= 301.44 mm2
Now,
sigma d = Fd/Ad
= 4×10^9 Pa / 301.44
= 13.3 MPa
At E,
Area Ae= pi (36)mm^2
= 113.04 mm^2
Now,
sigma e = Fe/Are
= 8×10^9 Pa/113.04
= 70.77 MPa.
Free body diagram
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Explanation:
Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation.
See more at .... https://brainly.in/question/17394351
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