06-Mohr Example

06-Mohr Example - A ‘Real-Life’ Example of How Mohr’s...

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Unformatted text preview: A ‘Real-Life’ Example of How Mohr’s Circle Would Be Used Buildings are designed to resist gravity forces (load from people, self-weight of a building, etc...) and to resist lateral forces (earthquakes, wind etc..). One typical framing system that is used in reinforced concrete and masonry buildings is to incorporate very stiff wall that is capable of withstanding large amounts of lateral (and vertical) load. These walls are referred to as ‘shear walls”. They can commonly be found around stairwells, around elevator shafts, or on the end of buildings. A description of shear walls and lateral load resisting systems is found at the end of this example. This example will focus on determining the most critical state Compute” “EQUii’alem” . — . Weight of the Heat = 3,000 ibi’ft of stress for Point A on the shear wall shown below. Smce these shear walls are typically made of very brittle materials (such as concrete) they are typically reinforced with steel. Brittle materials are known however to be the most sensitive to- failure (cracking) under normal stress, therefore it is important to determine the direction of the maximum normal stress since 5% cracking may be expected to occur perpendicular to this ,. _____________ fl, ‘ Computed “Equivalent” ' " Wind Load : 250,000 lb direction. While a full analysis would be needed to find the PointA points of maximum stress, lets begin by supposing that an j engineer decides that they want to know the IIIOSI Cl'itiCEll Slam 5' """"""" ":i I 15ft of stress at point A because they want to the magnitude of the stress and the direction that they should place external reinforcement on the wall in the direction of the principal stresses for the loading state shown to the right. Note that the Wall is 0.07 ft (8 in) thick To begin this problem you can judge from the dimensions that it is relatively thin and a 2 dimensional approximation may be appropriate. Since there are no forces acting in the Z direction you can assume plane stress. Next, you would want to cut the structure apart and Evflfqfi‘tsfdtIZEqFéiEE'ingmo mm determine the internal. forces acting ‘on a plane that passes through point A. While in theory this could be done 1n any ‘— Computed “Equivalent” direction it is probably the easiest to ‘do this using a plane that is Wind Lead = 2501009") horizontal with the base of the wall Since this will be in the same direction as the applied loads. If this is done the internal forces 15“ and moments can be determined by solving the equations (1-3) 50 it of equilibrium as: Point A _ R. 2 F, = 0; — 250,000lb + Vim = 0 vi“ = 25000010 2 F, = 0; — 3.000% t 30 ft — Fint = 0 Equation 1—3 MIN, = 25000010 * 35 0 = 0,750,000 0-10 EM, =0; 250,000!b*35fi—Min, =0 0.0? it (a in) thick Figure anaermmatim oflnmmalmes Now that the internal forces are known the stresses acting on point A can be determined. First, (although not something we will cover in this class) it can be argued that the moment will not contribute to the stress at point A since this point is in the center of the wall along the Neutral Axis of the wall. Therefore, the moment will not be considered in any further discussions. Second, is to determine the effect of the internal normal force. Since this force is acting uniformly along the wall it could be shown that the stress is acting uniformly along the cross— section. Therefore the state of normal stress acting in the y—direction can be computed using equation 4: or = 5 fl = —31.25psi Equation 4 3" A : (gin a: 30}? a: 12mm) Now, to determine the shear stress you could make the same assumption that we have made in class that the shear stress acts uniformly across the entire section and the average shear stress can be computed as shown in Equation 5 . T = Y— ——:§O’—W—— = 86.8psi Equation 5 “W A _ (8in*30ft*12i”1fl} It turns out that although this is a perfectly valid assumption for CE 231 you will learn in other classes (CE 270 that the shear stress is higher in the center of a member than it is on the edges of the member and therefore this stress is more appropriately computed using equation 6. Do not get hung up on this difference for the sake of this class. For this class you are to assume that the average shear stress as determined using equation 5 is fine to use for any problem given.) However for this problem I will continue with the shear stress that was computed using equation 6. 25090005? : 1302103!- Equation 6 You now have two of the three stresses to define the state of stress assuming the 2D approximation. To find the state of stress acting in the X—Direction you could cut the structure apart one more time, this however is more complicated than what we should look at in CB 231 for this example. Although it can be shown. I will not go into detail to explain why the normal stress acting in the X—Direction can be assumed to be equal to zero. The state of stress acting at Point A can be illustrated as shown in Figure 3. This now looks like the states of stress that we have been working with. It can be seen that this state of stress is not in the principal state (i.e., shear stress is not equal to zero) or in the maximum shear state (i.e., the normal stresses are not equivalent). You can now begin to solve this problem Figure 3; Stress At point A just like you would solve the problems that we have been doing in class. — " —" —_' ' ' Before going any further this will be plotted using a .. - | Moms Circle approach as shown in Figure 4. It can ' be seen that the state of stress for each face is | represented by a single point. To locate these stresses look at the state of stress for the element in I Figure 3. It can be seen that the stresses acting on the element perpendicular to the y axis are actually on the face of the element that we will call the Y— | Face. Similarly the face that intersects the x—axis I will be called the x—Face. This is simply | convention. These points can be located on the . circle if the Normal Stress (Tension Positive) and | Shear Stress (CW positive) assumptions are made Shear Stress on the Y Face (psi) for the axis. These points can be connected with a “D'ma' Stress “thEYFacelps'l ' line that will act as the diameter for Mohrs Circle which can be drawn as shown in Figure 5. It can be seen that the center of the circle will pass through a ' point that is at an x value of —15.6 psi. Figure 4: State of Stress Acting on the Faces of Point A | Figure 5 shows the plot after Mohrs Circle has been drawn. The points of Maximum Shear Stress and the Points of Maximum Tensile and Compressive Stress (Principal Stresses) can be determined. If these values just need to be estimated you could just read these values off the shear and normal stress axis directly. However if you need to compute these values you could use the ._ _ __. __ _. equations derived in class (equations 7 and 8). It should Maximum «mm—-——n—.~2ee--—:~~-—~-—-—T~--m-l-m-e-e—m«WT—He be noted that if the circle were drawn to scale the Compressive Stress ' Maximum Shear -i -—i—- numbers would be exactly the same. | 0 + 0 _ 7 E t 5 JMAX.MIN = —X‘2—}/" i (0251’ )‘ + T3,}; Equation 7 a; _ i _ I E. ‘ : I a: i— {I emmm = 1 15.51935; — 146.3 psi 5 f _ g Maximum ! __ Tensile Stress g : I | s .a.._.—-- .—... -......... . __...l._. _,..,.1 _ _ 0' —D' 2 . TMAX — R — ( X; Y )2 + 73y Equation 8 I Normal Stress on the Y Face (pel) _ i 0—31.25 _ 2 s TMAX ( 2 )2 + ( 1302) Figure 5:Mehrs Circle of All Stress at Point A ; :MAX =l3l.3psi i ‘ Now this tells us the magnitude of the stresses in the wall under the applied load. If we want to know the orientation in which these stresses occur we would need to know how far the element needed to be rotated to get to the principal state. If we look at the plot shown in Figure 6 it can be seen that if the element was rotated some amount (<in in real life and 2 (pp in Mohr’s Circle CCW) it would be at the state of principal stress. To determine the magnitude of the angle the simple equation shown in class could be used as performed in equation 9. The angle is negative implying that the element should be rotated in the angle opposite to CCW (i.e., CW) 41.60. This would imply that the angle would be rotated opposite to what was derived in class. The X” and Y’ faces would be shown as seen in figure 7. As the diameter is rotated to be parallel with the X—axis the Y-Face becomes the Y’ Face and the X Face becomes the X’Face. This will allow you to determine what stresses should be placed on the rotated element as shown next to Figure 7. tan (29F ) = 21.” Equation 9 ox ~— 0'}, 2(--l 30.2) tan(2¢P )= m Zed-1.6 Now the same type of computations can be i performed to obtain the angle of rotation to obtain the maximum shear stress as shown in Figure 8. Sketches of the element in its final state have been shown to illustrate what this means in real life on the following page. Shear Stress on the Y Face (psi) I I ........ul-i...._,..L20E}. .__._.._.. - — 0 Normal Stress on the Y Face (psi) . tafl(2¢s )= _ 0 X Y = 3.4 Equation 10 ZTXY Figure 6:Mohrs Circle Illustration of Principal Angle -2:50 -2p0 -1 Shear Stress on the Y Face (psi) Normal Stress on the Y Face (psi) | Figure ?:ililohrs Circie for Use with Element Rotation - Principal Stresses 7 X‘- Face | Shear Stress on the Y Face (psi) Normal Stress on the Y Face (psi) ‘ Figure S:Mohrs Circle for Maximum Shear Stress Direction L _ __ _ _ _ _. _. Computed “Equivalent” Weight of the Fleet = 3,000 Ibr'ft _ ‘ Computed “Equivalent” Wind Load = 250,000 It: a? 50ft Note that the Wall is 0.67 ft {8 in) thick Figure 11: Final State of Principal Stress o’x=115.5 psi Figure 8: a} Stress At Point A in the Original Coordinate System And b) the Stress in the Principal Direction Figure 10: a) Stress At Point A in the Original Coordinate System And b) the Stress in the Principal Direction Computed “Equivalent” Weight of the Roof = 3,000 lbr’tt ‘ Computed “Equivalent” ' Wind Load = 250,000 lb 50ft Point A Note that the Wall is 0.67 ft [0 in) thick LATERAL STABILITY ' A burldsngfe etruetunef elemente mural; be cenfigur‘ed ta liar-rm a étfil’lfl etruotum under any gamble lead membrane. Thererere’, while 3 " k “‘ ebmofure! eyetem 16 dementia pmmomly be carry K Vertrezl qrawty (cede, Ll; muen‘; filea be 3H5 en «a wrthebend lateral Wffltl er earbhaluaka Farrow. There are three baa-3m meehsmeme Per“ 1': eheurmg lateral étfib’fll‘by. 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R’Jgtd Hemzenfifil Amphnagmej 36“}ng 36 flat? glee? interwan beam-e, Aspen between éhmr Mlle, Them: are “fib'l'z'WWWWt omeeéfiry {er branm‘er" lateral leade- {rem new lead—beefing wall-r: to We leadflmrrymg eheer mile, Lateral made “band {:0 be more errtresl m the ehert director: an" rectangular bmldmge and the more affmenb meefiememe (eheer walla er braced frames‘») are [lead m the duree‘bmn, In the tang Air-genera? other elmilnr' elemente er a r:ng frame can be used. dwtmd if The arrangement 31‘ [eternal webmzm elemente awn“; firm :6 Impertanb to the etelnitty 9F 2 etrueture .36 3 whple, Ar: aeymmetrmfil leyaulfi, where the cantwrd 0f the a pineal fierce» re nab eameudenb wrth the eentmmffi the reelétmg means} can cause mime] effwte. A eyrnmefmeal 3W?” mail? of later}! Jemimaer eiement re there are alwaye deflmlflle Tine Frmmfle :9 eepecrsly Impertan-I: her be” buildinge. ‘33er mofiraeal {aye " Aeymmfibmezl layer {'2 From Ching, F.D.K.,.“Building Construction Illustrated, Second Edition”, Van Nostrand Reinhold, New York © 1991 ...
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06-Mohr Example - A ‘Real-Life’ Example of How Mohr’s...

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