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    Module 4 Activity No. 3

    Uploaded by

    John Mark Alves
    1) The document provides the details of an exercise problem involving calculating the minimum acceptable diameter of a round steel shaft transmitting 50 HP at 150 rpm and subjected to various loadings. 2) Shear, moment, and load diagrams are drawn and the minimum diameters are calculated for the shaft based on different material properties and load conditions. 3) Diameters ranging from 2.39 to 3.18 inches are obtained depending on the material type, load type, and allowable stresses.

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    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd

    Module 4 Activity No. 3

    Uploaded by

    John Mark Alves
    25 views7 pages
    1) The document provides the details of an exercise problem involving calculating the minimum acceptable diameter of a round steel shaft transmitting 50 HP at 150 rpm and subjected to various loadings. 2) Shear, moment, and load diagrams are drawn and the minimum diameters are calculated for the shaft based on different material properties and load conditions. 3) Diameters ranging from 2.39 to 3.18 inches are obtained depending on the material type, load type, and allowable stresses.

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    Module-4-Activity-No.-3

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    1) The document provides the details of an exercise problem involving calculating the minimum acceptable diameter of a round steel shaft transmitting 50 HP at 150 rpm and subjected to various loadings. 2) Shear, moment, and load diagrams are drawn and the minimum diameters are calculated for the shaft based on different material properties and load conditions. 3) Diameters ranging from 2.39 to 3.18 inches are obtained depending on the material type, load type, and allowable stresses.

    Copyright:

    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd
    Download as docx, pdf, or txt
    25 views7 pages

    Module 4 Activity No. 3

    Uploaded by

    John Mark Alves
    1) The document provides the details of an exercise problem involving calculating the minimum acceptable diameter of a round steel shaft transmitting 50 HP at 150 rpm and subjected to various loadings. 2) Shear, moment, and load diagrams are drawn and the minimum diameters are calculated for the shaft based on different material properties and load conditions. 3) Diameters ranging from 2.39 to 3.18 inches are obtained depending on the material type, load type, and allowable stresses.

    Copyright:

    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd
    Download as docx, pdf, or txt
    You are on page 1of 7

    Module 4: Activity No.

    3
    Exercises Problem on Shafts

    Name: Gatawa, Michael Paul T.


    Student No. 20180151648

    A round steel shaft transmits 50 HP at 150 rpm and subjected to loadings as shown.
    What minimum diameter is acceptable if the shaft weight is 10 lb/ft?
    a. For an allowable stress for flexure and shear are 6 ksi & 8 ksi respectively;
    b. For commercial shaft with keyway; gradually applied load;
    c. For AISI C1095 normalized steel; suddenly applied load with minor shock and
    d. For AISI C1030 as rolled steel; gradually applied load. Use ASME Code with
    keyway

    Draw the load, shear and moment diagrams, vertical and horizontal components.

    Solution:
    P=50 HP
    n=150 rpm
    lb
    W s=10 ¿
    ft

    nT
    P=
    63025
    150 ( T )
    50 HP=
    63025
    T =21008.33 lb .∈¿
    2000 cos (30) 2000 cos (30) 1650 cos( 45)

    22
    7 7 8
    22
    R1 R2

    For Vertical Moment:


    M R =0
    1
    36.67 lb
    0=( 2000 cos ( 30 ) ) ( 7 ) + ( 36.67 ) ( 22 ) + ( 2000 cos ( 30 ) ) ( 29 )−R 2 ( 36 )+ ( 1650 cos ( 45 ) ) ( 44 )
    R2=3180.46 lb

    M R 2=0
    0=( 1650 cos ( 45 ) ) ( 8 )−( 2000 cos ( 30 ) ) ( 7 ) −36.67 ( 14 )− ( 2000 cos ( 30 ) ) ( 29 ) + R1 ( 36 )
    R1=1487.04 lb

    *for shear
    v1 =1487.04
    v1 a =1487.04−10 ( 127 )=1481.21
    v a=1481.21−( 2000 cos ( 30 ) )=−250.85
    v ab =−250.85−10
    22
    12 ( )
    =−269.18

    v b=−269.18−( 2000 cos ( 30 ) )=−2001.23


    v b 2=−2001.23−10
    7
    12 ( )
    =−2007.06
    v 2=−2007.06+3180.46=1173.40
    v 2c =1173.40−10
    8
    12 ( )
    =1166.73

    v c =1166.73−( 1650 cos ( 45 ) )=0

    1487.04 1173.40 1166.73


    1481.21

    −250.85
    −269.18
    −2001.23 −2007.06

    *for moment
    1
    M a= ( 1487.04+ 1481.21) ( 7 ) =10388.88lb .∈¿
    2
    1
    M b=10388.88+ (−250.85−269.18 ) ( 22 )=4668.55 lb .∈¿
    2
    1
    M 2=4668.55+ (−2001.23−2007.06 ) ( 7 ) =−9360.47 lb.∈¿
    2
    1
    M c =−9360.47+ (1173.40+1166.73 )( 8 )=0
    2

    10388.88

    4668.55

    −9360.47

    For Horizontal Moment:

    2000 sin(30) 2000 sin(30) 1650 sin( 45)

    22
    7 7 8

    R1 R2

    M R =0
    1

    0=( 2000 sin ( 30 ) ) (7 ) + ( 2000 sin ( 30 ) ) ( 29 )−( R2 ) ( 36 ) +(1650 sin ( 45 ) )(44)


    R2=2426 lb

    M R =0
    2

    0=( 1650 sin ( 45 ) ) ( 8 ) −( 2000 sin (30 ) ) ( 7 )−( 2000 sin ( 30 ) ) ( 29 ) + R1 (36 )
    R1=740.73 lb
    *for shear
    v1 =740.73
    v a=740.73−( 2000sin ( 30 )) =−259.27
    v b=−259.27−( 2000 sin ( 30 ) )=−1259.27
    v 2=−1259.27+2426=1166.73
    v c =1166.73−( 1650 sin ( 45 ) )=0

    1166.73
    740.73

    −259.27

    −1259.27
    *for moment
    M a=( 740.73 ) ( 7 ) =5185.11lb .∈¿
    M b=5185.11−259.27 ( 22 )=−518.83 lb .∈¿
    M 2=−518.83−1259.27 (7 )=−9333.72lb.∈¿
    M c =−9333.72+1166.73 ( 8 ) =0
    5185.11

    −518.83

    M =√ ¿ ¿ −9333.72
    M a= √ ¿ ¿
    M a=11 610.95 lb.∈¿

    M b= √ ¿ ¿
    M b=4 695.29 lb.∈¿

    M 2=√ ¿ ¿
    M 2=13 218.80 lb.∈¿
    *use higher moment value
    M =13 218.80 lb.∈¿

    a.) For an allowable stress for flexure and shear are 6 ksi & 8 ksi respectively;
    τ d=8 000 psi
    σ d=6 000 psi
    K s =K m=1∗no load type given

    16
    τ d= 3
    ¿
    πD
    16
    8 000 psi= 3
    ¿
    πD
    D=2.51∈¿

    32
    σ d= ¿
    π D3
    32
    6 000 psi= 3
    ¿
    πD

    7
    D=3.18∈¿ 3
    16

    b.) For commercial shaft with keyway; gradually applied load;


    *with keyway
    τ d=6 000 psi
    σ d=12 000 psi
    K s =1; K m=1.5

    16
    τ d= ¿
    π D3
    16
    6 000 psi= 3
    ¿
    πD

    15
    D=2.91∈¿ 2
    16

    32
    σ d= ¿
    π D3
    32
    12 000 psi= 3
    ¿
    πD
    D=2.74∈¿

    c.) For AISI C1095 normalized steel; suddenly applied load with minor shock.
    K s =1.5 ; K m =2
    σ y =80 000 psi
    FS=3

    0.5 ( σ y ) 0.5 ( 80 000 )


    τ d= = =13 333.33 psi
    FS 3

    16
    τ d= 3
    ¿
    πD
    16
    13 333.33 psi= 3
    ¿
    πD

    15
    D=2.5=2
    16

    d.) For AISI C1030 as rolled steel; gradually applied load. Use ASME Code with
    keyway
    K s =1; K m=1.5

    τ d=0.3 ( σ y ) =0.13 (51 000 ) =15300 psi

    τ d=0.18 ( σ u )=0.18 ( 80 000 )=14 400 psi

    *use lower value, with keyway;


    τ d=14 400 ( 0.75 )

    16
    τ d= ¿
    π D3
    16
    14 400 ( 0.75 )= ¿
    π D3

    7
    D=2.39=2
    16

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