Question Solved1 AnswerPlease no copy paste of other answers. Please explain throughly.Thanks 3. The following figure shows a shaft mounted in bearings at A and D and having pulleys at B and C. The forces shown acting on the pulley surfaces represent the belt tensions. The shaft is made of AISI 1010 Hot Rolled steel (shaft diameter:1.5in). (a) Calculate the maximum moments at points B and C respectively and find the point of a maximum moment (B or C). (b) Find maximum stresses (tensile, compressive, and shear stresses) at the point. 8-in D. . . 400 lbf 4 50 lbf ✓ 5016 122.5lbf 460 lbf D 160 bi D ī 10-in D. 9 in z B gin 14 coin

3BU89O The Asker · Mechanical Engineering

Please no copy paste of other answers. Please explain throughly. Thanks

Transcribed Image Text: 3. The following figure shows a shaft mounted in bearings at A and D and having pulleys at B and C. The forces shown acting on the pulley surfaces represent the belt tensions. The shaft is made of AISI 1010 Hot Rolled steel (shaft diameter:1.5in). (a) Calculate the maximum moments at points B and C respectively and find the point of a maximum moment (B or C). (b) Find maximum stresses (tensile, compressive, and shear stresses) at the point. 8-in D. . . 400 lbf 4 50 lbf ✓ 5016 122.5lbf 460 lbf D 160 bi D ī 10-in D. 9 in z B gin 14 coin
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Transcribed Image Text: 3. The following figure shows a shaft mounted in bearings at A and D and having pulleys at B and C. The forces shown acting on the pulley surfaces represent the belt tensions. The shaft is made of AISI 1010 Hot Rolled steel (shaft diameter:1.5in). (a) Calculate the maximum moments at points B and C respectively and find the point of a maximum moment (B or C). (b) Find maximum stresses (tensile, compressive, and shear stresses) at the point. 8-in D. . . 400 lbf 4 50 lbf ✓ 5016 122.5lbf 460 lbf D 160 bi D ī 10-in D. 9 in z B gin 14 coin
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(a)M_(B)=3063.52 l lbfinM_(C)=3194.32lbf fin&#160; &#160; &#160;:.M_(max)=M_(c)C is the point of maximum moment.&#160;(b)sigma_(1)=10.31Kppsisigma_(2)=-10.31Kpsitau_(1)=5.49Kpsi&#160;The detailed solution is given below:Given: rarr{:[d_(B)=10in],[d_(C)=sin]:}∣d_("shaft ")=1.5sin#(x-y" planl ")/(y)Put above value in (1), we getput above value in (2), we get(R_(A)|_(Z)=482.5-321.67=160.83lbf:}(a)Bending moment:-{:[" \# In "x" cyplane "],[longrightarrowM_(B)=(R_(A))_(y)xx9=300lbfxx9in=2700lbf"-in "],[rarrM_(C)=(R_(D))_(y)xx9=150lbfxx9in=1350lbf"-in "],[" \# In "x-z" planl "],[rarrM_(B)=[R_(A)|_(Z)xx9=160.83lbfxx" in "=1447.47lbf-in:}],[rarrquadM_(C)=(R_(D))_(Z)xx9=321.67lbfxx9_(in)=2895.03lbf"-in "]:}Resultant B.M{:[rarrM_(B)=sqrt((2700)^(2)+(1447.47)^(2))=3063.52lbf-in],[rarrM_(C)=sqrt((1350)^(2)+(2895.03)^(2 ... See the full answer