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Thursday, June 9

Saturday, June 4

  1. page (8_53) team 5 edited ... τB=VxQyIyt+NyQxIxt=9.47MPa {https://blogger.googleusercontent.com/tracker/1972821090574092599…
    ...
    τB=VxQyIyt+NyQxIxt=9.47MPa
    {https://blogger.googleusercontent.com/tracker/1972821090574092599-260382311407418906?l=aero103.blogspot.com}
    
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    12:50 pm
  2. page (8_17)team 5 edited The following was originally published June 4, 2011 at http://aero103.blogspot.com/2011/06/853-tea…
    The following was originally published June 4, 2011 at http://aero103.blogspot.com/2011/06/853-team-5.html
    (8-17)
    Determine the smallest allowable diameter of the solid shaft ABCD knowing that τall=60 MPa and that the radius of disk B is r= 80 mm
    {https://viewer.zoho.com/getimage.do?genurl=a938bc18f691d6badacf758e2a824cf40&image=a938bc18f691d6badacf758e2a824cf40_001.png}
    M=0
    T-Pr=0
    P=7.5*103 N
    RA=Rc=12P=3.75*103 N
    MB=3.75*103150*10-3=562.5 N.M
    Critical section at point B
    JC=π2C3=2πM2+T2τall=2π562.52+600260*106=8.726*10-6 m3
    C=20.58*10-3m D=2C=41.2*10-3m
    {https://blogger.googleusercontent.com/tracker/1972821090574092599-260382311407418906?l=aero103.blogspot.com}

    (view changes)
    12:33 pm
  3. page (8_7)team5 edited The following was originally published June 4, 2011 at http://aero103.blogspot.com/2011/06/853-tea…
    The following was originally published June 4, 2011 at http://aero103.blogspot.com/2011/06/853-team-5.html
    {https://viewer.zoho.com/getimage.do?genurl=r7deee2ee79244faccf13ba23f91903a9&image=r7deee2ee79244faccf13ba23f91903a9_001.png}
    {https://viewer.zoho.com/getimage.do?genurl=r7deee2ee79244faccf13ba23f91903a9&image=r7deee2ee79244faccf13ba23f91903a9_002.png}
    RA=375 KN RB=375 KN
    Vmax=375 KN Mmax=450 KN
    Smin=Mmaxσall=450*1000169*106=2.8125*10-3m3
    Use: W 690*125
    σm=Mmaxsx=450*1033510*10-6=128.2 MPa
    τmax=VmaxAw=375*103678*10-3(11.7*10-3)=47.3 MPa
    At point (C)
    V=125 KN
    τw=VAw=125*103678*10-3(117*10-3)
    σb=0.5d-tf0.5dσm=122 MPa
    R=σb22+ τw2=63 MPa
    σmax=σb2+R=124 MPa
    {https://blogger.googleusercontent.com/tracker/1972821090574092599-260382311407418906?l=aero103.blogspot.com}

    (view changes)
    12:02 pm
  4. page (8_53) team 5 edited ... τB=VxQyIyt+NyQxIxt=9.47MPa {https://blogger.googleusercontent.com/tracker/1972821090574092599…
    ...
    τB=VxQyIyt+NyQxIxt=9.47MPa
    {https://blogger.googleusercontent.com/tracker/1972821090574092599-260382311407418906?l=aero103.blogspot.com}
    
    (view changes)
    11:51 am
  5. page (8_53) team 5 edited The following was originally published June 4, 2011 at http://aero103.blogspot.com/2011/06/853-tea…
    The following was originally published June 4, 2011 at http://aero103.blogspot.com/2011/06/853-team-5.html
    {https://viewer.zoho.com/getimage.do?genurl=o0753e0442d2179529fe155db0f7e8c6c&image=o0753e0442d2179529fe155db0f7e8c6c_001.png}
    Fx=9KN Fy=-13KN
    Mx=5200N.M My=3600N.M
    A=(2)(150)(13)+(13)(75-26)=4537*10-6m2
    Ix=2112150133+1501337.5-6.52+1121375-263=3.9303*10-6m4
    Iy=7.3215*10-6m4
    For point A
    Qx=Qy=0
    For point B
    A*=1360=780mm2
    x=-45mm y=31mm
    ∴Qx=24.18*10-6m3
    Qy=-35.1*10-6m3
    At point A
    σA=MxYIx-MyxIy=86.5MPa
    τA=0
    At point B
    σB=MxYIx-MyxIy=57MPa
    τB=VxQyIyt+NyQxIxt=9.47MPa
    {https://blogger.googleusercontent.com/tracker/1972821090574092599-260382311407418906?l=aero103.blogspot.com}

    (view changes)
    11:49 am

Sunday, May 29

  1. page transformation of plane stress 2 edited The following was originally published May 30, 2011 at http://aero103.blogspot.com/2011/05/chapter…
    The following was originally published May 30, 2011 at http://aero103.blogspot.com/2011/05/chapter7transformation-of-plane-stress.html
    Calculator Introduction of the stresses
    Given the stresses at a space point in the body, sx, sy, and txy, this calculator computes the stresses of the same space point in a rotated coordinate system, sx', sy', and tx'y'. {http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/images/StressTransform.gif}
    Equations behind the Calculator
    The following coordinate transformation equations were used,
    {http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/images/TransStressAll.gif}
    principale stresses
    Given the stress components sx, sy, and txy, this calculator computes the principal stresses s1, s2, the principal angle qp, the maximum shear stress tmax and its angle qs. It also draws an approximate Mohr's cirlce for the given stress state.
    {http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/images/PrincipalStress.gif}
    The Mohr's circle associated with the above stress state is similar to the following figure. However, the exact loaction of the center sAvg, the radius of the Mohr's circle R, and the principal angle qp may be different from what are shown in the figure.
    {http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/images/MohrCircleCase1.gif}
    Equations behind the Calculator
    The formulas used in this calculator are,
    {http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/images/PrincipalAngle.gif}
    {http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/images/PrincipalStressEq.gif}
    {http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/images/MaxShearStressEq.gif}
    {http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/images/MaxShearAngle.gif}
    {http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/images/MaxShearStress.gif}
    {https://blogger.googleusercontent.com/tracker/1972821090574092599-1534002788081940855?l=aero103.blogspot.com}

    The following was originally published May 30, 2011 at http://aero103.blogspot.com/2011/05/chapter7transformation-of-plane-stress.html
    Calculator Introduction of the stresses
    (view changes)
    10:12 pm
  2. page transformation of plane stress 2 edited The following was originally published May 30, 2011 at http://aero103.blogspot.com/2011/05/chapter…
    The following was originally published May 30, 2011 at http://aero103.blogspot.com/2011/05/chapter7transformation-of-plane-stress.html
    Calculator Introduction of the stresses
    Given the stresses at a space point in the body, sx, sy, and txy, this calculator computes the stresses of the same space point in a rotated coordinate system, sx', sy', and tx'y'. {http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/images/StressTransform.gif}
    Equations behind the Calculator
    The following coordinate transformation equations were used,
    {http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/images/TransStressAll.gif}
    principale stresses
    Given the stress components sx, sy, and txy, this calculator computes the principal stresses s1, s2, the principal angle qp, the maximum shear stress tmax and its angle qs. It also draws an approximate Mohr's cirlce for the given stress state.
    {http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/images/PrincipalStress.gif}
    The Mohr's circle associated with the above stress state is similar to the following figure. However, the exact loaction of the center sAvg, the radius of the Mohr's circle R, and the principal angle qp may be different from what are shown in the figure.
    {http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/images/MohrCircleCase1.gif}
    Equations behind the Calculator
    The formulas used in this calculator are,
    {http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/images/PrincipalAngle.gif}
    {http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/images/PrincipalStressEq.gif}
    {http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/images/MaxShearStressEq.gif}
    {http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/images/MaxShearAngle.gif}
    {http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/images/MaxShearStress.gif}
    {https://blogger.googleusercontent.com/tracker/1972821090574092599-1534002788081940855?l=aero103.blogspot.com}

    (view changes)
    10:12 pm
  3. page transformation of plane stress edited The following was originally published May 30, 2011 at http://aero103.blogspot.com/2011/05/chapter…
    The following was originally published May 30, 2011 at http://aero103.blogspot.com/2011/05/chapter7-transformation-of-plane-stress.html
    Mohr's Circle
    Introduced by Otto Mohr in 1882, Mohr's Circle illustrates principal stresses and stress transformations via a graphical format,
    {http://1.bp.blogspot.com/-kC_9J8wcRT0/TeMaV3UgdRI/AAAAAAAAAEQ/xIpWD3jJBEo/s320/image001.png}
    The two principal stresses are shown in red, and the maximum shear stress is shown in orange. Recall that the normal stesses equal the principal stresses when the stress element is aligned with the principal directions, and the shear stress equals the maximum shear stress when the stress element is rotated 45° away from the principal directions.
    As the stress element is rotated away from the principal (or maximum shear) directions, the normal and shear stress components will always lie on Mohr's Circle.
    Mohr's Circle was the leading tool used to visualize relationships between normal and shear stresses, and to estimate the maximum stresses, before hand-held calculators became popular. Even today, Mohr's Circle is still widely used by engineers all over the world.
    Derivation of Mohr's Circle
    To establish Mohr's Circle, we first recall the stress transformation formulas for plane stress at a given location,
    {http://1.bp.blogspot.com/-RYYBbTFHu_0/TeMaV7XUehI/AAAAAAAAAEY/blWeYPp55Ls/s320/image002.png}
    Using a basic trigonometric relation (cos22q + sin22q = 1) to combine the two above equations we have,
    {http://4.bp.blogspot.com/-Lq6ca6wWxr4/TeMaWJAKobI/AAAAAAAAAEg/bJWDA5cjLF0/s320/image003.png}
    This is the equation of a circle, plotted on a graph where the abscissa is the normal stress and the ordinate is the shear stress. This is easier to see if we interpret sx and sy as being the two principal stresses, and txy as being the maximum shear stress. Then we can define the average stress, savg, and a "radius" R (which is just equal to the maximum shear stress), {http://3.bp.blogspot.com/-b7fTbkZzIec/TeMaWIIl69I/AAAAAAAAAEo/BCt2C-BhkAY/s320/image004.png}
    {http://2.bp.blogspot.com/-GNFbcFgmMr8/TeMaWbPTJNI/AAAAAAAAAEw/V4nzmNujK4k/s320/image005.png}
    The circle equation above now takes on a more familiar form,
    {http://2.bp.blogspot.com/-HmK6iSYLU_A/TeMa6IvUK-I/AAAAAAAAAE4/oeOTsCsjVYU/s320/image006.png}
    The circle is centered at the average stress value, and has a radius R equal to the maximum shear stress, as shown in the figure below,
    {http://1.bp.blogspot.com/-kC_9J8wcRT0/TeMaV3UgdRI/AAAAAAAAAEQ/xIpWD3jJBEo/s320/image001.png}
    Cylindrical Pressure Vessel
    Consider a cylindrical pressure vessel with radius r and wall thickness t subjected to an internal gage pressure p.
    {http://1.bp.blogspot.com/-F7J08Y3efpk/TeMa6Xpmg6I/AAAAAAAAAFA/Wf9lfGIcQks/s320/image007.png}
    The coordinates used to describe the cylindrical vessel can take advantage of its axial symmetry. It is natural to align one coordinate along the axis of the vessel (i.e. in the longitudinal direction). To analyze the stress state in the vessel wall, a second coordinate is then aligned along the hoop direction.
    With this choice of axisymmetric coordinates, there is no shear stress. The hoop stress sh and the longitudinal stress sl are the principal stresses.
    To determine the longitudinal stress sl, we make a cut across the cylinder similar to analyzing the spherical pressure vessel. The free body, illustrated on the left, is in static equilibrium. This implies that the stress around the wall must have a resultant to balance the internal pressure across the cross-section. {http://1.bp.blogspot.com/-vTkBedbUEu8/TeMfDKDflUI/AAAAAAAAAFo/hUwgfHKWmso/s320/image008.png}
    Applying Newton's first law of motion, we have, {http://2.bp.blogspot.com/-i5fISuN9ThU/TeMa6gsClcI/AAAAAAAAAFQ/Ca9yHVsSCds/s320/image009.png}
    and <this strees is a n axial stress&get;
    · To determine the hoop stress sh, we make a cut along the longitudinal axis and construct a small slice as illustrated on the right.
    The free body is in static equilibrium. According to Newton's first law of motion, the hoop stress yields,
    {http://4.bp.blogspot.com/-5kdNuHcwpM4/TeMa699LVyI/AAAAAAAAAFY/CbU6J9NCEd0/s320/image010.png}
    {http://1.bp.blogspot.com/-zEM9-aMLpAA/TeMbHYYGm0I/AAAAAAAAAFg/jTlNN2fYHS0/s320/image011.png}
    {http://2.bp.blogspot.com/-i5fISuN9ThU/TeMa6gsClcI/AAAAAAAAAFQ/Ca9yHVsSCds/s320/image009.png}
    {https://blogger.googleusercontent.com/tracker/1972821090574092599-8476115280072446130?l=aero103.blogspot.com}

    (view changes)
    9:45 pm

Friday, April 29

  1. 12:02 am

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