The circular shaft made of 6061-T6 aluminum (E = 10 × 10³ksi, yielding strength oysensile 37 ksi, oyear = 19 ksi) has a diameter of 0.6 in. If on a cross-section of the shaft, it is subjected to the internal resultant loads as shown, (a) calculate the stress components that act at point A due to each force/moment/torsion, (b) determine the combined state of stress at point A, (c) determine the combined state of stress at point B, (d) determine which point has the more (a) stress component at A severe stress state and calculate the principal stresses and maximum shear due to F, due to V, due to T, due stress at that point, and (e) calculate the safety factor based on the to My, due to Mz maximum normal stress yielding theory and (f) based on the maximum (b) at point A shear stress yielding theory. Oz = ,Jy = ,txy = %3! (c) at point B Ox = (d) o = (e) S.F. = (f) S.F. = Jy = ,02 = ,Txy = ,Tmax v300 My-200 bin T-250 in

Mechanics of Materials (MindTap Course List)
9th Edition
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Author:Barry J. Goodno, James M. Gere
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Chapter3: Torsion
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The circular shaft made of 6061-T6 aluminum (E = 10 × 10³ksi, yielding strength ơy,ensite
37 ksi, oyear = 19 ksi) has a diameter of 0.6 in. If on a cross-section of the shaft, it is subjected to the internal
resultant loads as shown, (a) calculate the stress components that act at point A due to each force/moment/torsion,
(b) determine the combined state of stress at point A, (c) determine the
combined state of stress at point B, (d) determine which point has the more (a) stress component at A
severe stress state and calculate the principal stresses and maximum shear due to F, due to V, due to T, due
stress at that point, and (e) calculate the safety factor based on the to My, due to Mz
maximum normal stress yielding theory and (f) based on the maximum (b) at point A
shear stress yielding theory.
Ox = ,Jy = ,txy =
(c) at point B
Ty =
,02 =
Txy =
, Tmax
v300
( d) σ1-
(e) S.F. =
(f) S.F. =
My-200 bin
T-250 bin
F500 b
Transcribed Image Text:The circular shaft made of 6061-T6 aluminum (E = 10 × 10³ksi, yielding strength ơy,ensite 37 ksi, oyear = 19 ksi) has a diameter of 0.6 in. If on a cross-section of the shaft, it is subjected to the internal resultant loads as shown, (a) calculate the stress components that act at point A due to each force/moment/torsion, (b) determine the combined state of stress at point A, (c) determine the combined state of stress at point B, (d) determine which point has the more (a) stress component at A severe stress state and calculate the principal stresses and maximum shear due to F, due to V, due to T, due stress at that point, and (e) calculate the safety factor based on the to My, due to Mz maximum normal stress yielding theory and (f) based on the maximum (b) at point A shear stress yielding theory. Ox = ,Jy = ,txy = (c) at point B Ty = ,02 = Txy = , Tmax v300 ( d) σ1- (e) S.F. = (f) S.F. = My-200 bin T-250 bin F500 b
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