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# HONORS ENGR STATICS E M 274H

ISU

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This 77 page Class Notes was uploaded by Mariela Collier on Sunday September 27, 2015. The Class Notes belongs to E M 274H at Iowa State University taught by Staff in Fall. Since its upload, it has received 21 views. For similar materials see /class/214484/e-m-274h-iowa-state-university in Engineering Mechanics at Iowa State University.

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Date Created: 09/27/15

2w Nzcos60 a Wsin30 FM 1W2 MW N 2 200s60 a W ZFyZO F Nl Wcos30quot stin60 a0 W N17 Mtan60 a F W 2F 2 O F M 1Wsin30quot 0 FM 1W2 N1 x Mtan60 a N1 0 when 6 2 tan 60 a function M Cansa arg M floorsqrt3tanarg EDUgtgt a inspace0 55 1000 EDUgtgt mcansa EDUgtgt plota m 60 Principle of Virtual Work for A Rigid Body rigid body displacements in 2D 6 y a general displacement of any point in a rigid body consists of a translation plus a rotation If the translation and rotation are both small then the displacement at any point in a rigid body can be written as drdDdi9kgtltr l l small small translation rotation translation dr dD constant for all points in the body dD rotation dr has a magnitude dB 1 of rde direction of dr is perpendicular to r dr d6k x r where k is a unit vector along the zaxis if we have a series of forces Fi and couples Ci acting on a rigid body then the virtual work 6W done by these forces and couples if we give a small virtual rigid body displacement to the body is 2 F1 where 51 5D 66k X 1391 is the virtual displacement of each point where the forces act Thus 5W 5D Fij Fi 56eriicj 66k 391 j1 i1 1 but E 56kgtltri56kri gtltFi since abxcbcgtltacagtltb so the virtual work done is N N M 5W 5DZE 6kZri XE ch 1 1 171 7 11 7 i1 21 11 N N M 5W DZFJ 6kZri XE ZCJ If equilibrium of the rigid body is satisfied then we have and hence 6W 0 Thus if a rigid body is in equilibrium the virtual work done by all the forces and moments acting on the body is zero or in other words equilibrium gt 5wo 21 397 F1 N N M 6W5DZFJ 6kZrixl ZCJ 1 1 However we can also turn this relationship around say that If the virtual work done by all the forces and couples on a body is zero for all possible virtual riqid bodv displacements then the body must be in equilibrium ie 5W0 lgt equilibrium Principle of Virtual vvom for a Rigid Body To prove the principle of virtual work If we let 8DX 6D and all other components of the translation 0 and set the rotation 66 0 then for zero virtual work we must have N 502 0 l l i so that we also find 2 If instead we give the body a pure virtual rotation we find in xFi Jr cjz 0 which is just the moment equilibrium equation 11 z 11 for 2D problems 2M2 o Example find the equilibrium angle for a platform of weight 200b supported by two identical weightless links All connections are smooth pins 3 ft l 3 ft 200 lb T1 T2 equilibrium approach 2MP o 50 lt 9 9 TI 511193T2 sin63 0 T1 T2 T 200 ZFx 0 2Tcost9 50 0 sz0 1 tane 4 2Tsin6 20020 9760 virtual work approach T1 T2 200 l 200 6W50 Susin6 2006ucos6 20 gt tan 6 4 note tensions do no work in this assumed virtual rotation we can find the tensions if we want to by looking at additional virtual translations or rotations T1 T2 Su 50 9 3 59Tamp 6u366Wl 3 200 5W2 11 sin65uT2 sin65u0 same reSUIt as our preVIous moment T1 T2 T equation T1 T2 50 SU 5U 6U 1 lt lt V 200 l same result as first revious force e uation 6W 506u 1139cos66u 1392005620 39 39 Tlcos6T2cos650 5W T1 sin65u sint95u 2005u O Tlsin9Tzsin9200 same result as second previous force equa on 2H ABC P0 ZMx0o AR Pey BRZ CR20 ZMZo0 PexCR 2 BR 2O solving for ABC P A R26y P B R ey ex P C R ey exJ P NotezifeXeyOthen 14szng If the table is homogeneous and has a weight Wthen W P A R2 y W P B R ey 6x CgR ey ex P A RZey If we have WO and set AO for tipping impending then R 7 since this is A independent of ex A X because of symmetry have similar regions between AB and AC region of tipping between 80 If the weight is nonzero then we need a larger distance to make A O Rt W e 1 y 2 P MATLAB is a software package that makes it easy to manipulate matrices Vectors can also be easiI handled as a s ecial case A matrix in MATLAB is just a 2 D array of numbers Example gtgtA1 2 35 24 67 A LO IM 1 3 2 6 7 The dimensions of a matrix is NxM where N is the number of rows and N is the number of columns The matrixA above has dimensions 4x2 We can add or subtract matrices by simply adding or subtracting each element in the arrays located at the same position Example gtgtA1234 A 1 3 gtgtB3355 B C To add matrices they must be of the same size we can also multiply matrices together in the following fashion Let Amn be the element in an array A of dimensions MxN at the mth row and nth column position Similarly for a matrix B of dimensions NxK we have Bnk is the element at the nth row and kth column position Then the matrix product ofA and B CAB is defined for each element of C as N ka Z Amank n21 MxK MXN NxK Note for this to be meaningful we must have the number of columns ofA be equal to the number of rows of B but the other dimensions can be different Example gtgt A 2 3 1 5 gtgt B 2 2 6 3 gtgt CAB C 22 13 32 17 which comes from our multiplication rule C11 2 AHB11 1412321 2 2X2 3X6 2 22 C12 2 1411312 141sz2 2 2X2 3X3 13 C21 2 1421311 1422321 2 1X2 5X6 2 32 C22 2 1421312 1422322 2 D0 5X3 2 17 We can consider vectors to be just examples of matrices where one of the dimensions 1 We can either have row or column vectors gtgt v1 1 2 3 5 1x4 row vector gtgt v2 1235 v2 4x1 column vector 0100M we can multiply vectors A and B in the same fashion as other matrices as long as we adhere to the previous rule we must have the number of columns ofA be equal to the number of rows of B in the product AB Consider the two vectors gtgtv11234 gtgt v2 1234 gtgt dv1v2 1x4 d 4x1 30 1x1 scalar This is just the dot product or the square of the magnitude in this case Now consider reversing the order gtgtv11234 gtgt v2 1234 va 4x1 1x4 p 1 4x4 2 4 8 12 16 Here pm vzm V1 2x1 We can also multiply matrices with vectors if we follow the same rule gtgt M1 2 3 4 M 3 4 gtgt x 56 X 5 6 2X2 gtgt b Mx 4 2X1 b 17 39 This is equivalent to the linear system of equations Mllxl M12x2 b1 M21x1M22x2 b2 another way to write these equations is M11 M12 J 2 M21 M22 x2 2 We also could set gtgtX56 X 5 6 gtgt M1 2 3 4 M 4 3 4 1X2 Note that this b vector is different from before even though the x and M have b the same elements as 23 34 gt gtgt bxM 4 2X If we want to get the same result for b as before but now in the form of a row vector we need to multi I x b the trans ose of M MT where if M 2M11 M12 M21 M22 M M MT 11 21 Interchange rows and columns M12 M22 This works since for any two matrices or vectors A B where the product is defined we have ABT BTAT so if we take the transpose of M11 M12xlbl M21 M22 x2 b2 we get M M M leiM Mir1 M In MATLAB the transpose is MT M39 gtgt bXM39 17 39 same b as obtained originally but now in terms of a row vector For vector analysis MATLAB has builtin dot and cross product functions gtgt v1 1 2 6 gtgtv2352 gtgt d dotv1 v2 d 1 gtgt C crOSSV1 V2 here the vectors must be 3D vectors c ie 1x5 or 5x1 in dimensions 34 20 1 we saw before we could also get the dot product by calculating v1v2T gtgt v1v239 ans 1 In solving Statics equilibrium problems we will typically have to solve a set of linear equations for a set of unknown forces or moments We can do this easily in MATLAB Example Consider the following system of two equations for the two unknowns x1 and x2 91 2 C11 C12 J C21 C22 x2 If we define the C matrix of coefficients and the vector b in MATLAB then the solution is given by using a backslash operator gtgtC1237 gtgt b11 gtgt x Cb X 50000 20000 Principle of Virtual Work for a Deformable Body For a rigid body that body can only translate and rotate For a deformable body however the shape and dimensions of the body can change If we choose virtual displacements that are not just rigid body displacements then the virtual work done will not be zero Example Consider a spring of unstretched length L dx H F Hm If we apply a small elongation dx to the spring the force F will do work dW Fdx This work is not zero since by stretching the spring we increase its internal energy E also called the strain energy or the potential energy of the spring For a linear spring we have Fkx where k is the spring constant and x is the total stretch of the spring Thus the total work W in going from the unstretched length to a final length xf is W xff kxdx x0 1 2 2 f The quantity on the right side of this equation is just the total strain energy or potential U of the spring so we have WE total work done by the force on the spring is equal to the strain energy of the spring By differentiating this relationship we also have dW dE Fdx kxdx If we choose small virtual displacements that allow deformations of the body to occur then the virtual work done by the forces and couples acting on the body will not be zero but will be equal to the virtual change in the strain energy of the body ie 5vv 5E As in the case of rigid bodies for deformable bodies we can show that if we satisfy the above relationship for quotallquot possible small virtual displacements then equilibrium of the deformable body will be satisfied or in other words SW 6E Igt equilibrium Principal of Virtual Work for Deformable Bodies The reason this is important is that we can use this principle to solve both statically determinant and indeterminant problems In applying the virtual work principle for deformable bodies we generally express the strain energy of the body in terms of displacements and then use the virtual work principle to determine those displacements The manner in which this is done is as follows We first break the system we are considering into deformable quotelementsquot and use the virtual work principle to relate the forces acting on an element to the displacements at quotnodesquot of those elements in terms of a element quotstiffnessquot matrix We then combine assemble all the element stiffness matrices together into a single stiffness matrix for the entire system that is a set of equations that relate all the displacements to all the forces acting on the system We then apply the quotboundary conditionsquot ie conditions we know about dis lacements or forces to this set of e luations to obtain a set of linear equations for all the unknown displacements We solve this system of equations for the displacements We can then use those displacements to find all the forces in the individual elements The steps we will follow are 1 Describe the nodes and their connectivity for the system 2 determine the stiffness matrix of each element 3 assemble the global stiffness matrix 4 apply the boundary conditions 5 solve the equations 6 postI rocess the solution to Vet all the forces Example We are going to examine a problem where we have a series of interconnected linear springs and we wanted to solve for the displacements at the ends of each spring and the forces that produce these displacements Consider first a single spring displacements gt U U m m forces 39 EH on mth element 39 km nodei Odel mth element FM F10quot l gt km nodei Odel mth element The strain energy of this spring is 1 2 Ezakm U1 UZ if we give the displacements at each node a small virtual displacement then the work done by the forces will be 5W Elml Ui Fjlmle and by the virtual work principle 6W 26E za Ewi a Eau an 6U J 2 km U1 U16Uj km U1 U5U i Thus Elml Ui Fjlml Uj km U1 U1 5U km U1 U1 5Ul If this is true for all virtual displacements we must have Elm km Ul U1 Fl km U1 U1 J or in matrix form Fjlml km km U element stiffness matrix Km Ff quot km km U1 Ermgt km km HUJ note that equilibrium of the spring is satisfied since Eon Fltmgt 0 J also note that we can write the total strain energy of the spring in terms of this stiffness matrix E 2Ui Ujk kka5i m m J UTiKmiU Kmquot fr US m m IO show how we use this element stiffness matrix it best to take a specific example 100 lb A H B 20 Ibin 3O Ibin Determine the forces acting on the supports at A and B This is a statically indeterminant problem since A 100 B N H W AB100O is the only equilibrium equation so we cannot find A and B Thus we need to consider the deformations of the springs This is easily done with the virtual work principle 1 define the nodes and connectivity element number node i nodej 1 1 2 z z 5 2 define the element stiffness matrices 20 20 K1 20 20 30 30 Kg 30 3o 3 Assemble a global stiffness matrix from each element Here the variables we want to solve for are the displacements U1 U2 U3 U1 u2 U3 tWV NW so we want to form u a ilobal stiffness matrix for the whole s stem that has the form U1 Fl K U2 F2 U3 F3 Consider the first element contribution to the global equation U1 U2 1511 Fm gt 2 20 element number nodei nodej I 1 1 2 20 20 0 U1 E 2 2 3 20 20 0 U2 2 1721 0 0 0 U3 0 EM 2 km km Uz39 FJW km km U Now add in the second element contribution 2k11 k12Ui U2 U3 k21 k22 Uj Ff Flt2gt 3 30 20 20 0 U1 F1 20 2030 30 U2 2 F 1gtF2lt2gt 0 30 30 U3 F3lt2gt But note that we have U1 U2 U U 2 3 F10 F Flu 1122 F F39 gt gt 20 100 30 so 1721 1722 100 and we have 20 20 0 U1 F1 20 2030 30 U2 100 0 30 30 U3 Ff 4 apply the boundary conditions u1o l u2 U3o W 20 20 0 U1 F1 20 50 30 U2 100 0 30 30 U3 F39 so we can ignore the first and third equations and from the second equation we have only one nontrIVIaI equation 5 solve the remaining system of equations 20U1 50U2 30 100 and so the solution is U2 2 in We would like to automate the application of the boundary conditions so that we can solve the original system of equations One way to do this is as follows 1 initialize the right hand side to zero 2 If a force is specified at a node place it in the appropriate nodal force location 0 100 O 3 If a displacement is specified as zero at the nth node leave the right hand side value as zero and replace the Knn term in the stiffness matrix by a large value N For our previous example we would have 109 20 0 U1 0 20 50 30 U2 100 0 30 109 U3 0 If a displacement is specified as a nonzero value C at the nth node then replace the right hand side value by CxN where N is a large value and replace the Knn term in the stiffness matrix by the same large value N Example in our previous example if we had U3 3 instead of U3 0 we would have 109 20 0 U1 0 20 50 30 U2 100 0 30 109 U3 3x109 4 We can then solve the system of equations directly For example 109 20 0 U1 0 20 50 30 U2 100 0 30 109 U3 0 gt 000000004000000 U1 2 uU2 2 U3 U 200000005200000 t 000000006000000 exac y 109 20 0 U1 0 20 50 30 U2 2 100 0 30 109 U3 3x109 U1OU2 38U3 3 gt 0000000076300000 exactly 380000009880000 300000011400000 6 Postprocessing If we I lace the dis lacements back into our s stem of e Iuations we find 20 20 0 so we have 40 lb 20 0 0 40 50 30 2 100 30 30 0 60 EOE 40 1732 60 100b 60b which shows that that we have also satisfied equilibrium of the entire system Example 39 free body diagram 3000 lb OA4i 3i6k IOAIZM pOSItlon vectors OB 4i 6i 4k lOBI 0C0i5j0k OC5 e1eoA e2 eoB 4i e3 eoc Oi1j0k unit vectors leTlel T2T ZeZ T3 T3e3 so equilibrium gives 2F 2 0 T1T2T3 3000k0 in components ZFX 0 2Fyo 2w 4 4 3T1 KTz 0T3 0 solving 3 T 6 TT0 112234311 61 1 68 2 3 T222474lb 6 T1 4 T20T33000 T322700 this system of equations can be written in matrixvector form as 62x 62 62 63x I 0 e3y 7 0 e32 T3 3000 e MATLAB solution example39 position vectors r1 OA r2OB and r3OC respectively r1436 r2464 r3050 calculate unit vectors e1 r1normr1 e2 r2normr2 e3 r3normr3 form up matrix of coefficients of equilibrium eqs Ee139 e239 e339 known force on right side of equilibrium eqs column vector W O O 300039 solve for tensions IW gtgt example39 e1 05121 03841 07682 e2 04851 07276 04851 e3 0 1 0 E 05121 04851 0 03841 07276 1 0000 07682 04851 0 T 10e003 23431 24739 27000 Statically Indeterminant Beams If we have too many supports or support conditions a beam will be statically indeterminant and we will not be able to solve for the external reactions or the internal forces Example A w lblength l L two equations of equilibrium three unknown reactions Just as in the case of springs or trusses we can solve statically indeterminant problems we elate t e oaus to t e ueormation In a beam the vertical deflection of the beam vx is related to the internal bending moment Mx in the beam through the relation d2v a x2 vx l x E Young39s modulus of the beam a material property y El Mx I an area moment of the beam cross sectional Area A a geometry ro ert 14ysz A We can use this relationship and appropriate boundary conditions on the beam slope or vertical deflection to find all the reactions and make the problem solvable pin or roller support fixed clamped support v 0 no deflection V0 n0 deflection ClVClX sIope IS not constrained dvdx 0 no slope Consider our previous example y I w IbIength Mx Ax wx22 V00 gtC20 3 4 VL0 AL WL C1L0 6 24 2 3 gt WL C120 dx 6 Solving for A1 C1 3 A wL 1 C1 WL3 knowing A we can solve for B and MB W l B5wL8 A3WL8 3 4 3 We can also now EVx3WLx K x look at the beam 48 24 48 deflection E 3 x 3 1 x 4 1 x l l l l WL 48 L 24 L 48 L deflected shape exaggerated If we use singularity functions to write the internal bending moment we can solve statically indeterminant problems with various loading conditions relatively easy JEL 8 0 V 39 t X Ai a gtllt sl s39 gtIC from force and moment equilibrium l B 2502A 0250 A y 500 lb l 500 MxAltx Ogt1Bltx 6gt1 500ltx 91Cltx 121 2 Mx EI2AX o1 19x 61 500x 91 Cx 121 dv A 2 B 2 2 C 2 1915 39 0 3ltx 6gt 250x 9 3ltx 12gt C1 C A B 250 EIVEltx 03 Eltx 6gt3 Tltx 9gt3 Eltx 12gt3 Clx C2 The boundary conditions are note could drop this term throughout since it does not contribute to deflection VO 0 39gt 02 0 or slope for Oltxlt12 It is only really needed in the shear force expression v6 0 lgt 6A 501 0 W12 0 gt 288A363 225012C1 0 using the relationship between B and Awe can solve forA and C1 Solving for B and C then H 3900 39 390 B 3162 lb gives 0 2169 lb Principle of Virtual Work for Trusses Atruss is like a series of interconnected springs in 2D since the force in each bar is along the axis of the bar causing each bar to extend or compress just like a spring The spring constant k for a bar is given by k L E is the Young39s modulus for the bar a material property given in GPa GigaPascaIs in the SI system where 1 Pa 1 Newtonm2 For example for steel E 210 GPa21Ox1O9 Pa A is the crosssectional area of the bar in m2 L is the length of the bar in m2 In the English system E is measured in psi lbin2 A is measured in in2 and L is measured in inches For steel E 30x106 psi One difference between a truss and our previous 1D spring examples is that the displacements at the ends of each bar are all in different directions V We need to place all of these displacements in the same quotglobalquot coordinate system along the x and yaxes t Utj S V jth node ith node UXi 6 is the angle that the bar makes with the positive xaxis FV In the st coordinates we have the stiffness matrix for each element bar 111 k 0 k 0 Us FSZkUsi USJ E 0 0 0 0 Un which isthe FHO same as FS k 0 k 0 Us FSJ kUSjUsi Ft 0 0 0 0 Ur F 111 k 0 k 0 Us Fn o o o 0 U FS 2 k 0 k 0 Us Ft 0 o o o Ur again this relationship simply comes from the virtual work principle 5W 2 Fsi USl Fn5Un 17515US 36 1 2 5E 5 kug 415 J F kUSl 413 gt F3 kUSJ US 0 F C We need to transform the stiffness matrix below which is given in the st coordinates to one given in the global xy coordinates k 0 k 0 0 o o 0 Kg k 0 k 0 o o o o The easiest way to do this is to note that the strain energy of each element bar LIGII GIOU JG VVI ILLUII GO k 0 k 0 Usz39 1 1 O O O 0 U1 E UTKeU Um Un st39 Ur k 0 k 0 U o o o 0 Ur 1 klUsrUsigt2 UXi ith node To transform the displacements at the ith node we have U31c s U 020056 U s c in ssin6 and at the jth node 3H1 15 We can write these relations together as Usi c s O O U U s c 0 0 Uyl Us O O c s UXJ U0 0 O S c ij oras U QU Placing this relationship into the strain energy expression we find 1 T 1 EU KelU5UTiQiTiKeiiQiU UTKeU which shows that Ke KeQlTiI ltellQ Expanding this out we find C2 CS C CS 2 2 CS S CS S Kg k 2 2 C CS C CS CS S CS S element bar stiffness matrix in the global xy coordinates we now need to assemble these stiffness matrices into a global stiffness matrix Kg for the truss apply the boundary conditions and solve for the displacements U Then we can find the external forces on all the nodes pins in the xy coordinates including the reactions Kgmam the contribution to the global Kg matrix for an element containing the ith and jth nodes 1 Kg2i l2i 1 Kg2i 12i Kg2i 12j 1 Kg2i l2j U211 UM Kg2i2i 1 Kg2i2i Kg2i2j 1 Kg2i2j U U 2139 y Kg2j 12i 1 Kg2j 12i gm 12141 Kg2j 12j Um ij Kg2j2i 1 Kg2j2i Kg2j2j 1 Kg2j2j UM yj UN Kg2i12i1k11 components of the K6 stiffness matrix for an element Kg 2139 12z39 k12 etc Finally we can find the force in each bar from U since st kUsj Us1 if tensile if compressive and we have simply used the transformation of displacements to express this force in terms of the x and y components of the displacements

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