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Saturday, May 21

  1. page prob7.104 (team1) edited 7.104: the unpressurized clyindirical storage tank shown has a 5-mm wall thickness and is made of …
    7.104: the unpressurized clyindirical storage tank shown has a 5-mm wall thickness and is made of steel having a 420-MPA ultimate strength in tension determine the maximum height h to which it can be filled with water if a factoe of safety 4.0 (specific weight of water is 9810 N/M^3
    solution:
    t= 5 mm
    f.c.=4.0
    ultimate strengh=420 MPA
    f.c.=(ult/stress h)
    stress hoop=
    (420/4)=105 MPA
    volume of clyinder=(3.14/4)(7.5)^2*15=662.67 m^3
    weight of water= 9810*662.67=6500887.861 N
    Fh=stress(h)*h*2*t
    h=(650088.81/105*10^6*5*2)
    h=6.19 m

    (view changes)
    9:17 am
  2. page prob7.87 (team1) edited 7.87:the 36-mm diamater shaftis made of a grade of steel with a 250-MPA tensile yield stress. usin…
    7.87:the 36-mm diamater shaftis made of a grade of steel with a 250-MPA tensile yield stress. using the max- sheraing -stress critrion.determinethe magnitude of the torque Tfoe which yield occurs when P=200 KN
    {Image(001).jpg} solution:
    Tmax=0.5*yield stress
    =0.5*250
    =125 MPA
    Tmax=(T*C/J)
    stress(y)=(-p/A)=(-200*10^-3)/((3.14/4)(.036^2))
    stress(y)=-196.487 MPA
    C=(stressy/2)=-98.2435
    Txy^2=Tmax^2-(stressy-C)
    Txy^2=(125)^2-(89.2435)^2
    Txy=77.286 MPA
    T=708 N.m

    (view changes)
    8:49 am
  3. file Image(001).jpg uploaded
    8:35 am
  4. page prob7.85 (team1) edited 7.85: the 44-mm diamater shaft AB is made of a grade of steel for which the yield strength is stre…
    7.85: the 44-mm diamater shaft AB is made of a grade of steel for which the yield strength is stress yield= 250 MPA using the maximum-shearing -stress-criterion .determine the magnitude of the force P for yield occurs when T=1.5 KN.m.
    {Image.jpg} solution:
    stress(y)=0
    stress(x)=P/A
    Txy=(Tc/J)
    Txy=(1.5*1000*22*10^-3/0.5*3.14*(22*10^-3)^4)
    =89.68 MPA
    Tmax=0.5*stress yield=0.5*250 = 125 MPA
    (Tmax)^2=(Txy)^2+(stressx-c)
    (125)^2-(89.68)^2=0.5*stressx
    stress x=174.155 MPA
    P=stressx *A
    P=174.155*10^6*(3.14/4)(.044)^2
    P=264.8 KN

    (view changes)
    8:24 am
  5. file Image.jpg uploaded
    8:16 am

Saturday, May 14

  1. page prob7-49 (team1) edited 7-49 :solve problem 7.27 usnig mohr's circle {Image(218).jpg} solution: cy=sqrt(ac^2-ay^2) cy=s…
    7-49 :solve problem 7.27 usnig mohr's circle
    {Image(218).jpg} solution:
    cy=sqrt(ac^2-ay^2)
    cy=sqrt(75^2-20^2)
    =72.28 MPA
    oy=(2*cy)+ox
    oy=(2*72.28)+60
    oy=stress lareger value of Y=205 MPA

    (view changes)
    10:38 am
  2. file Image(218).jpg uploaded
    10:35 am
  3. page prob7-5 (team1) (deleted) edited
    10:24 am
  4. page prob7-37 (team1) edited 7-37 :solve prob 7-15 using mohr's circle {Image(217).jpg} solution: stress(X)= -60 MPA stres…
    7-37 :solve prob 7-15 using mohr's circle
    {Image(217).jpg}
    solution:
    stress(X)= -60 MPA
    stress(Y)= 90 MPA
    txy= 30 MPA
    stress(AVG)=C=(oy+ox)/2=(90-60)/2=15 MPA
    cy=oy-oc=90-15=75
    tan2$=(yk/cy)=(30/75)
    $=10.9
    R=sqrt(ky^2+cy^2) = 80.77
    first: anti clockwise (10 degrees)
    cos@=(cx'/R)
    cx'=80.77*cos 41.8
    cx'=60.21 MPA
    cx'=ox'+oc
    stress(X')=oX'=60.21-15
    =45.2 MPA
    second: with colckwise(25 degress):
    cos#=(cx'/R)
    cx'=80.77*cos28.4
    =71.049 MPA
    cx'=71.049
    ox'=cx'-oc
    =71.049-15
    =- 56.049 MPA

    (view changes)
    10:02 am

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