ARCE 306: MATRIX STRUCTURAL ANALYSIS. Homework 1 (Due )

Winter Quarter ARCE 6: MATRIX STRUCTURA ANAYSIS January 4, Problem Homework (Due -6-) k k/ft A B C D E ft 5 ft 5 ft 5 ft () Use the slope deflection method to find the bending moment diagram for the continuous beam above (you may or may not use the pin modification). The beam has uniform flexural stiffness. () Find the vertical displacement at mid-span BC using the moment diagram of () and the principal of virtual forces (remember to work with a determinate base structure for the virtual loading). Solution: Δ= 576 k-ft () Use statics to obtain the shear force diagram from the moment diagram. (4) Find the maximum moment in span BC. Problem B w C.85 rad = 6 k-ft ft deflected shape not to scale ϕ A 5 ft D Under a uniformly distributed load with intensityw applied to the girder the two-hinge frame above deflects as shown. Find (a) the moment diagram of the frame, (b) the intensityw of the distributed load, and (c) the rotationϕ at the base. Solution: w = k/ft, ϕ A =.48 rad /4/ :58 PM C:\calpoly\arce6\homework\winter\hw.doc

Winter Quarter Problem (hand calculations required, but check results using MATAB) (a) Calculate C = ABwith 4 A = 4 5 6 B = 5 6 7 8 9 7 8 9 4 (b) Find the determinate of matrix 4 9 A = 6 5 6 49 64 8 (c) Use Gauss elimination to solve the system of equations Ax = B with 4 5 6 7 9 A = B = 6 8 9 5 4 7 9 58 (d) Calculate the inverse of the following matrix: A = (e) For which values ofa andb does the system of equations 8 4 x 6 7 x = a 5 x b have one solution no solution infinitely many solutions /4/ :58 PM C:\calpoly\arce6\homework\winter\hw.doc

Winter Quarter ARCE 6: MATRIX STRUCTURA ANAYSIS January 6, Problem k Homework -Solution k/ft A B C D ft 5 ft 5 ft 5 ft () Use the slope deflection method to find the bending moment diagram for the continuous beam above (work with the minimum number of unknowns) () Find the vertical displacement at mid-span BC using the moment diagram of () and the principal of virtual forces (remember to work with a determinate base structure for the virtual loading) Solution: Δ= 576 k-ft () Use statics to obtain the shear force diagram from the moment diagram. (4) Find the maximum moment in span BC. () Moment diagram Fixed-end moments 5 MBA = = 5. MBC = MCB = = 4.67 End moments, say = 5 5 MBA = MBA + ϕb = 5. + ϕb 5 5 MBC = MBC + (4 ϕb + ϕc ) = 4.67 + 4 ϕb + ϕc 5 5 M = M + ( ϕ + 4 ϕ ) = 4.67 + ϕ + 4 ϕ 5 5 MCD = 4 ϕc = 4ϕC 5 CB CB B C B C Equilibrium M = = 7ϕ + ϕ + 54.67 M = = ϕ + 8ϕ 4.67 B B C C B C 7 ϕb 54.67 ϕb 7.7, ϕc 9.95 8 ϕ = = = C 4.67 M M M BA BC CB M CD = 5. 7.7 =.6 k-ft = 4.67 4 7.7 + 9.95 =.6 k-ft = 4.67 7.7 + 4 9.95 = 79.8 k-ft = 4 9.85 = 79.8 k-ft C:\calpoly\arce6\homework\winter\hw_sol.doc /5/ :47 AM - -

Winter Quarter. + max M = k-ft 79.8 + 9.9 M [k-ft] B P = C 5 ft () Deflection at midspan (a) 5 Δ= MM dx = 6.5 (.6 + 79.8) + 56.5 5 = 576 k-ft ( ans) 4 + 6.5 M () Shear force diagram..67.67 + 4.79 4.79 6.9 V [k] += (4) Find the maximum moment in span BC. fromm andv diagrams:. k max M =. k-ft + =. +.4 =. k-ft(ans) k/ft or 6.9 k max M = 79.8 k-ft + = 79.8 + 8.9 =.k-ft(ans) discrepancy due to rounding k/ft C:\calpoly\arce6\homework\winter\hw_sol.doc /5/ :47 AM - -

Winter Quarter Problem w B C.85 rad = 6 k-ft ft deflected shape not to scale ϕ A 5 ft D Under a uniformly distributed load with intensityw applied to the girder the two-hinge frame above deflects as shown. Find (a) the moment diagram of the frame, (b) the intensityw of the distributed load, and (c) the rotationϕ at the base. (a) M BA 6 = ϕb =.85 =.8 k-ft( ans) (b) w MBC = MBC + ( 4ϕB + ϕc ) = + ( 4ϕB ϕb ) M = M + M = M = M =.8 k-ft B BC BA BC BA w 5 6.85 =.8 w = k/ft( ans ) 5 (c) MAB = ( 4ϕA + ϕb ) = ϕa = ϕb =.48 rad( ans)... M [k-ft] 8.. Note: Frame is non-sway due to symmetry. C:\calpoly\arce6\homework\winter\hw_sol.doc /5/ :47 AM - -

Winter Quarter Problem (a) Calculate C = ABwith 4 A = 4 5 6 B = 5 6 7 8 9 7 8 9 4 (b) Find the determinate of matrix 4 9 A = 6 5 6 49 64 8 (c) Use Gauss elimination to solve the system of equations Ax = B with 4 5 6 7 9 A = B = 6 8 9 5 4 7 9 58 (d) Calculate the inverse of the following matrix: A = (e) For which values ofa andb does the system of equations 8 4 x 6 7 x = a 5 x b have one solution no solution infinitely many solutions C:\calpoly\arce6\homework\winter\hw_sol.doc /5/ :47 AM - 4 -

Winter Quarter (a) 6 8 4 C = 5 5 65 8 95 7 94 48 44 466 (b) det(a)= 6 (c) ` x = [ 5] T (d).58..8 7 4 - A =.467.6667.8 or 5 8.8..467 4 5 (e) a 8 one solution (for all b) a = 8, b = 4 infinitely many solutions a = 8, b 4 no solution C:\calpoly\arce6\homework\winter\hw_sol.doc /5/ :47 AM - 5 -

Winter Quarter ARCE 6: MATRIX STRUCTURA ANAYSIS January 6, Problem Homework (Due --) k/ft k/ft ft 5 ft 5 ft 4 ft 5 () Determine the x element stiffness matrices k corresponding to the four spans of the continuous beam above. () Determine the x element load vectors D for each span (fixed-end moments). () Assemble the 5x5 structure stiffness matrix K. (4) Assemble the 5x structure load vector F. (5) Reduce the 5x5 structure stiffness matrix K to a 4x4 matrix K free by accounting for the boundary condition. (6) Solve (electronically) the system of equilibrium equations (expressing equilibrium at the nodes) to obtain the 4x vector qfree of node rotations of the free nodes. Solution: qfree = [ 564.8 46.9 54.8 4.] T (7) Sketch the deflected shape. (8) Expand the 4x vector qfree to a 5x vector q by adding zero for the rotation of the fixed node. (9) Extract the x vectors d of element rotations out of the 5x vector q of structure rotations. () Calculate the vector D of element end moments. () Draw the bending moment diagram. Problem k/ft k/ft 4 5 ft 6 ft 5 ft 5 ft ft () Find the 4x4 structure stiffness matrix Kfree of the above structure by modifying the 4x4 structure stiffness matrix of Problem. Ignore axial deformation, i.e. the structure is a non-sway frame. () Solve (electronically) the equilibrium equations and draw the deflected shape. Solution: qfree = [ 47. 7.5 49.6 99.] T Problem 4 = k-ft 5 9 r = ft 6 8 () In a 9x9 grid, mark by a x the non-zero elements in the structure stiffness matrix K of the above wheel (modeled as a non-sway frame). All connections are moment connections. () Without calculating any element stiffness matrices, find the values K99, K89, K, K87, K of the structure stiffness matrix () Calculate the complete 9x9 structure stiffness matrix. K =, K =, K =, K = 6., K = 445. (all k-ft) Solution: 99 89 87 7 C:\calpoly\arce6\homework\winter\hw.doc /4/ :7 PM

Winter Quarter ARCE 6: MATRIX STRUCTURA ANAYSIS January, Problem Homework -Solution k/ft k/ft ft 5 ft 5 ft 4 ft 5 () Determine the x element stiffness matrices k corresponding to the four spans of the continuous beam above. () Determine the x element load vectors D for each span (fixed-end moments). () Assemble the 5x5 structure stiffness matrix K. (4) Assemble the 5x structure load vector F. (5) Reduce the 5x5 structure stiffness matrix K to a 4x4 matrix K free by accounting for the boundary condition. (6) Solve (electronically) the system of equilibrium equations (expressing equilibrium at the nodes) to obtain the 4x vector qfree of node rotations of the free nodes. (7) Sketch the deflected shape. (8) Expand the 4x vector qfree to a 5x vector q by adding zero for the rotation of the fixed node. (9) Extract the x vectors d of element rotations out of the 5x vector q of structure rotations. () Calculate the vector D of element end moments. () Draw the bending moment diagram. () k.. 4 = = = = k.6.8 k.6.8 k.....8.6.8.6.. () w / 66.67 w / 4.7 () () () (4) D =,,, = = = = = 66.67 w 4.7 / D D w / D ()....6.8 K =.8..8.8.6... (4) -66.67 66.67 F = -4.7 4.7 (5).. -66.67..6.8 66.67 K free = [ k-ft].8..8 F free = [ -4.7 k-ft].8.6 4.7 (6) C:\calpoly\arce6\homework\winter\hw_sol.doc // :46 PM - -

Winter Quarter Solve: Kfreeqfree = F free q free [ 564.8 46.9 54.8 4.] T = [rad] (7) (8) [ 564.8 46.9 54.8 4. ] T q = [rad] (9) d 4 = -564.8 d = 46.9 d = -54.8 d = 4. 46.9-54.8 4. ().56 49.97 8. 4 D = D = D = D = -.56-49.97-8. 4. or in matrix form.56 49.97 8. D = -.56-49.97-8. 4. ().65 49.97 8. 56.5 M 4. [k-ft] C:\calpoly\arce6\homework\winter\hw_sol.doc // :46 PM - -

Winter Quarter Problem k/ft k/ft 4 5 ft 6 ft 5 ft 5 ft ft () Find the 4x4 structure stiffness matrix Kfree of the above structure by modifying the 4x4 structure stiffness matrix of Problem. Ignore axial deformation, i.e. the structure is a non-sway frame. () Solve (electronically) the equilibrium equations and draw the deflected shape. () Column 6 adds rotational stiffness k = 4 / 6 = 4 / =. to node. The structure stiffness matrix is thus K free......6 +..8..56.8 = =.8..8.8..8.8.6.8.6 () Solve: free free = free K q F q free [ 47. 7.5 49.6 99.] T = C:\calpoly\arce6\homework\winter\hw_sol.doc // :46 PM - -

Winter Quarter Problem 4 = k-ft 5 9 r = ft 6 8 7 () In a 9x9 grid, mark by a x the non-zero elements in the structure stiffness matrix K of the above wheel (modeled as a non-sway frame). All connections are moment connections. () Without calculating any element stiffness matrices, find the values K99, K89, K, K87, K of the structure stiffness matrix () Calculate the complete 9x9 structure stiffness matrix. Solution: () see () () 45 a = r sin = sin.5 = 7.654 ft 4 4 K99 = 8 = 8 = k-ft( ans) K89 = = = k-ft( ans) r r K = K87 = = = 6. k-ft( ans) K = 4 4 a 7.654 + = + r a 7.654 = 445 k-ft( ans) () 445 6 6 6 445 6 6 445 6 6 445 6 6 445 6 6 445 6 99 6 445 6 6 6 445 units: (k-ft) C:\calpoly\arce6\homework\winter\hw_sol.doc // :46 PM - 4 -

Winter Quarter ARCE 6: MATRIX STRUCTURA ANAYSIS January, Homework (Due --) Problem Develop a MATAB algorithm that assembles the structure stiffness matrix of an arbitrary non-sway frame structure (one degree-of-freedom per node). Test your work using the "wheel"-structure of HW. Below are a few suggestions for your algorithm that I hope you find helpful. %vector of element lengths (as many components as there are elements) = [..]; %vector of i-nodes (as many components as there are elements) ni =[..]; %vector of j-nodes (as many components as there are elements) nj =[..]; %flexural stiffness = %initialize structure stiffness matrix K = zeros(..) numel = length(ni); %assembly loop for el = :numel %calculate element stiffness matrix kele of element el kele = end %place element stiffness matrix at proper location into structure %stiffness matrix (this is the key step) Do not submit anything for this HW. Use your work as a starting point for ARCE 5, HW. // C:\calpoly\arce6\homework\winter\hw.doc - -

Winter Quarter 9 ARCE 6: MATRIX STRUCTURA ANAYSIS January, 9 Problem Homework -Solution The figure shows a uniform slab supported on four columns rigidly attached to the slab and clamped at the base. The slab is rigid in plane and out of plane. Each column is of circular cross section, and its second moment of cross-sectional area about any diametrical axis is as noted. With the three degrees-of-freedom selected as ux, uy, u θ at the center of the slab, formulate the stiffness matrix K in terms of the lateral stiffness k = / h of the smaller column. ux =, uy =, uθ = kxx = 6k kyx = k = θx ux =, uy =, uθ = kxy = kyy = 6k b b kθy = k k = kb ux =, uy =, uθ = kxθ = b b kyθ = k k = kb b b b b b b b b kθθ k k k 4k = + + + = kb 6 K = k 6 b,with k = ( ans) h b b C:\calpoly\arce6\homework\Winter9\hw_sol.doc //9 :6 PM

Winter Quarter 9 Problem u k u u k k k ft ft ft ft 4 4 k = k/ft rigid k For the structure above, find the stiffness matrix K with respect to the three degrees-of-freedom shown. k = k = k = k.6.7 7. k/ft = = k =.6.8 k = 9.k = 9 k k = + k = k = k (.8 ) 4.8 4.8 k/ft = = k = + k = k = (.8 ) 7 7 k-ft u = k k.6k.6k ft ft ft ft 4 4 k 7. 9 K = 4.8 9 7.8k u = k k k k k ft ft ft ft 4 4 k.8k u = k k k.8k = 6k.8k = 6k k k ft ft ft ft 4 4 Make sure you understand why there are the three units k/ft, k, and k-ft in the stiffness matrix. C:\calpoly\arce6\homework\Winter9\hw_sol.doc //9 :6 PM

Winter Quarter 9 Problems P, Δ EA, rigid α A tension rod stabilizes a rigid post as shown. (a) For given values EAα,,,, derive a parameterk such that P = k Δ. (b) Find the angle α that maximizesk. F rod P = cos α F P EA EA cos α sin α Δ P EA cos α EA cos α sin α rod Δ rod = rod = rod Δ = = P = cos αsin αδ( ans) k Δ hor = Δ α Δ AB f α = α α ( ) cos sin ( ) f α = α α+ α = α α+ α = ( ) cos sin cos cos sin cos α α trial and error ( ) ( ) α= : sin + cos =.5 +.75 =.5 α=5 : sin 5 + cos 5 =.658 +.67 =. (good enough, ans) C:\calpoly\arce6\homework\Winter9\hw_sol.doc //9 :6 PM

Winter Quarter ARCE 6: MATRIX STRUCTURA ANAYSIS January, Problem Homework 4 (Due -8-) The figure shows a uniform slab supported on four columns rigidly attached to the slab and clamped at the base. The slab is rigid in plane and out of plane. Each column is of circular cross section with moment of inertiai. With the three degreesof-freedom selected as ux, uy, u θ at the center of the slab, formulate the stiffness matrix K in terms of the lateral stiffness k = / h of the smaller column. 6 Solution: K = k 6 b b b u Problem For the structure above, find the stiffness matrix K with respect to the three degrees-of-freedom shown. 7. k/ft 9 k Solution: K = 4.8k/ft 9k 7k-ft Problems P, Δ k u k k k ft ft ft ft 4 4 k = k/ft u rigid k EA, rigid α A tension rod stabilizes a rigid post as shown. (a) For given values EAα,,,, derive a parameterk such that P = k Δ. (b) Find the angle α that maximizesk. EA Solution: k = cos αsin α C:\calpoly\arce6\homework\winter\hw4.doc // 9:57 AM

Winter Quarter ARCE 6: MATRIX STRUCTURA ANAYSIS January, Problem Homework 4-Solution The figure shows a uniform slab supported on four columns rigidly attached to the slab and clamped at the base. The slab is rigid in plane and out of plane. Each column is of circular cross section, and its second moment of cross-sectional area about any diametrical axis is as noted. With the three degrees-of-freedom selected as ux, uy, u θ at the center of the slab, formulate the stiffness matrix K in terms of the lateral stiffness k = / h of the smaller column. ux =, uy =, uθ = kxx = 6k kyx = k = θx ux =, uy =, uθ = kxy = kyy = 6k b b kθy = k k = kb ux =, uy =, uθ = kxθ = b b kyθ = k k = kb b b b b b b b b kθθ k k k 4k = + + + = kb 6 K = k 6 b,with k = ( ans) h b b C:\calpoly\arce6\homework\winter\hw4_sol.doc // 9: AM

Winter Quarter Problem u k u u k k k ft ft ft ft 4 4 k = k/ft rigid k For the structure above, find the stiffness matrix K with respect to the three degrees-of-freedom shown. k = k = k = k.6.7 7. k/ft = = k =.6.8 k = 9.k = 9 k k = + k = k = k (.8 ) 4.8 4.8 k/ft = = k = + k = k = (.8 ) 7 7 k-ft u = k k.6k.6k ft ft ft ft 4 4 k 7. 9 K = 4.8 9 7.8k u = k k k k k ft ft ft ft 4 4 k.8k u = k k k.8k = 6k.8k = 6k k k ft ft ft ft 4 4 Make sure you understand why there are the three units k/ft, k, and k-ft in the stiffness matrix. C:\calpoly\arce6\homework\winter\hw4_sol.doc // 9: AM

Winter Quarter Problems P, Δ EA, rigid α A tension rod stabilizes a rigid post as shown. (a) For given values EAα,,,, derive a parameterk such that P = k Δ. (b) Find the angle α that maximizesk. F rod P = cos α F P EA EA cos α sin α Δ P EA cos α EA cos α sin α rod Δ rod = rod = rod Δ = = P = cos αsin αδ( ans) k Δ hor = Δ α Δ AB f α = α α ( ) cos sin ( ) f α = α α+ α = α α+ α = ( ) cos sin cos cos sin cos α α trial and error ( ) ( ) α= : sin + cos =.5 +.75 =.5 α=5 : sin 5 + cos 5 =.658 +.67 =. (good enough, ans) C:\calpoly\arce6\homework\winter\hw4_sol.doc // 9: AM

Winter Quarter ARCE 6: MATRIX STRUCTURA ANAYSIS January, Homework 5 (Due -7-) Problem 4k/ft k 5 k k-ft 4 4 ft 5 ft.5 ft.5 ft 5 ft 5 () Find the moment and shear force diagrams of the beam using the slope deflection method. Use the pin modification for joint if you wish. () Use the moment diagram and PVF to find the rotation and deflection of joint. See (5) for the solution. Model the continuous beam using the beam element with four degrees-of-freedom ( dofs per node). Using hand calculations: () Find the x load vector of the structure. T Solution: F = [. 5. 8. 6..5 5. 56.5 ] k, k-ft (4) Without calculating any stiffness matrices find the elements K77, K88, K67, K78 of the x structure stiffness matrix K Solution: K77 =.56, K88 =., K67 =.96 K78 = [k/ft, k, k-ft] (5) If the 5x vector of free structure displacements is free [ 5.466.77.7.6.76] T q = ft, rad find the element end forces for elements 4 and 45 by matrix (hand) calculations (4x4 element stiffness matrix times 4x vector of element displacements) and show that they match the results of (). C:\calpoly\arce6\homework\winter\hw5.doc // :5 PM

Winter Quarter ARCE 6: MATRIX STRUCTURA ANAYSIS January, Problem Homework 5-Solution 4k/ft k 5 k k-ft 4 4 ft 5 ft.5 ft.5 ft 5 ft 5 () Find the moment and shear force diagrams of the beam using the slope deflection method. Use the pin modification for joint if you wish. () Use the moment diagram and PVF to find the rotation and deflection of joint. See (5) for the solution. Model the continuous beam using the beam element with four degrees-of-freedom ( dofs per node). Using hand calculations: () Find the x load vector of the structure. T Solution: F = [. 5. 8. 6..5 5. 56.5 ] k, k-ft (4) Without calculating any stiffness matrices find the elements K77, K88, K67, K78 of the x structure stiffness matrix K Solution: K77 =.56, K88 =., K67 =.96 K78 = [k/ft, k, k-ft] (5) If the 5x vector of free structure displacements is free [ 5.466.77.7.6.76] T q = ft, rad find the element end forces for elements 4 and 45 by matrix (hand) calculations and show that they match the results of (). C:\calpoly\arce6\homework\winter\hw5_sol.doc /8/ 5:5 PM - -

Winter Quarter () let = 5(to get pleasing numbers for / ) M M M 4 4 = k ft 8 4 k/ft 5 ft = 5 66.67 = 6.67 k-ft 5 5 = = 56.5 k-ft 8 = 56.5 k-ft M M M M M = M + ϕ = 6.67 + ϕ = 6.67 + 6ϕ 4 M4 ϕ 5 5 = 5 + ( 4 + ϕ4) = 56.5 + ( 4ϕ + ϕ4) = 56.5 + 8ϕ + 4ϕ4 5 5 = M + ( ϕ + 4ϕ ) = 56.5 + ( ϕ 5 + 4ϕ ) = 56.5 + 4ϕ + 8ϕ 5 = ( 4ϕ + ) = ( 4ϕ 5 + ) = 8ϕ 5 = ( ϕ + ) = ( ϕ 5 + ) = 4ϕ 4 4 4 4 4 45 4 4 54 4 4 M = = M + M4 = 6.67 + 6ϕ + 56.5 + 8ϕ + 4ϕ4 = 9.58 + 4ϕ + 4ϕ4 M4 = = M4 + M45 = 56.5 + 4ϕ + 8ϕ4 + 8ϕ4 = 56.5 + 4ϕ + 6ϕ4 + ϕ ϕ = 4.7 4.7 5 6. = = ( ans) = 4.547 4.547 5 7.6 = = ( ans) M = M = 4.4 k-ft, M = 6.8 k-ft, M = 6.8 k-ft, M = 8.9 k-ft 4 4 45 454 Moment and shear force diagrams 4.4 6.4 C:\calpoly\arce6\homework\winter\hw5_sol.doc /8/ 5:5 PM - -

Winter Quarter () from moment diagram Mx ( ) 4 x x 4 = + 5x x = + 5x x 5 5 MΔ () x = + x 5 Mϕ ( x) = + x = MΔ ( x) 5 5 4 4 4 = + + = 5 75 M () x M()d x x 9x 6x x x dx 8. Δ = + 8. = 547 checks ok ϕ = + 8. = 78. ccw checks ok (), D = D D,, 4, D = 5 8. 4 5 5. 6 5 8. 8. = = 5 8. 5. 4 5+ 5 5. 5. 5 5. 5 5 56. 8 = = 5 5. 56. 5 5 8 C:\calpoly\arce6\homework\winter\hw5_sol.doc /8/ 5:5 PM - -

Winter Quarter. 5 5. 8. 8. 5 5 6. F = from joint loading F = from element loading F = total ( ans) 5 56.5.5 5 5. 56.5 56.5 (4) 4 K77 = + = =.56 ( ans) 5 K K K 88 4 45 4 4 8 = + = =. ( ans) 5 4 45 6 6 = = =.96 ( ans) 67 4 5 6 6 6 6 = + = + = ( ans) 78 4 45 5 5 (5) Member end forces -. 5. 5..8. 4.4 6.8 D =. 4.99 4.8 -.8 -. -4.4-6.4 8.9 C:\calpoly\arce6\homework\winter\hw5_sol.doc /8/ 5:5 PM - 4 -

Winter Quarter ARCE 6: MATRIX STRUCTURA ANAYSIS February, Homework 6 (Due -4-, 6 pm) Problem () Read and understand the handout on the D truss element. () Use the principal of virtual work and the member forces given in the handout to verify the three displacements qfree of Eq. (5.45) of the example. () Modify your MATAB program for the beam element with four degrees-of-freedom (two dofs per node) by implementing the truss element with four degrees-of-freedom (two dofs per node) and test your algorithm by reproducing the results of the example. From now on, we define the structure as we do in RISA, that is by inputting the x - andy coordinates of the nodes (as vectors) in addition to the connectivity information. We do no longer input the length of the members directly. For the simple truss on the handout, the vectors of coordinates are x = [ 8 4], y = [ ]; Whenever we need the length of an element we should calculate it by using Pythagoras' theorem ( ) ( ) j i j i = Δ x +Δ y = x x + y y (a) Save you function elestiff.m as elestiff.m and turn the 4x4 element stiffness matrix for the beam element into the local 4x4 element stiffness matrix k for the truss element according to Eq. (5.) of the handout (in the weeks to come we will refer to the D truss element as element type ). The result should be the function function k = elestiff(e,a,) where parameters EAare,, the modulus of elasticity of the material, the cross-sectional area and length of the element. (b) Write a MATAB function function = elelength(xs,xe,ys,ye) that calculates the length of a member where input parameters xs, xe, ys, ye are the x - andy- coordinates of the start and end nodes of the element, respectively. (c) Write a MATAB function function T = transele(xs,xe,ys,ye) that calculates the 4x4 transformation matrix T of a truss element according to Eq. (5.) of the handout. Caution: Do not calculate the angleϕ before calculating sin ϕ and cos ϕ. You should calculate sin ϕ and cos ϕ directly according to Eq. (5.8). (d) Save any of your previous MATAB work that uses the beam element to hw6.m and make the necessary modifications. Note that the transformation matrix T comes into play twice: () in the first loop to calculate the T global element stiffness matrix k = T kt according to Eq. (5.7) and () in the second loop when calculating the local element displacements d = Tdwhich are then multiplied by the local element stiffness matrix k to obtain the member forces D (see Eq. (5.46)). C:\calpoly\arce6\homework\winter\hw6.doc // : PM

Winter Quarter Problem Without calculating any stiffness matrices, calculate the elements K5,5, K6,6, K6,5, K7,7, K8,8, K8,9, K8,8, K8,7, K8,6 of the 44x44 structure stiffness matrix K. The nodes of the bottom chord lie on a quadratic parabola with zero slope at node. The truss structure has constant EA. K5,5 K6,6 K6,5 K7,7 K8,8 K8,9 K8,8 K8,7 K8,6 Solution:.5797.76.699.754.76 x EA Submit: Hand calculations, no MATAB code C:\calpoly\arce6\homework\winter\hw6.doc // : PM

Winter Quarter ARCE 6: MATRIX STRUCTURA ANAYSIS February 4, Homework 6-Solution Problem () Read and understand the handout on the D truss element. () Use the principal of virtual work and the member forces given in the handout to verify the three displacements qfree of Eq. (5.45) of the example. () Modify your MATAB program for the beam element with four degrees-of-freedom (two dofs per node) by implementing the truss element with four degrees-of-freedom (two dofs per node) and test your algorithm by reproducing the results of the example. From now on, we define the structure as we do in RISA, that is by inputting the x - andy coordinates of the nodes (as vectors) in addition to the connectivity information. We do no longer input the length of the members directly. For the simple truss on the handout, the vectors of coordinates are x = [ 8 4], y = [ ]; Whenever we need the length of an element we should calculate it by using Pythagoras' theorem ( ) ( ) j i j i = Δ x +Δ y = x x + y y (a) Save you function elestiff.m as elestiff.m and turn the 4x4 element stiffness matrix for the beam element into the local 4x4 element stiffness matrix k for the truss element according to Eq. (5.) of the handout (in the weeks to come we will refer to the D truss element as element type ). The result should be the function function k = elestiff(e,a,) where parameters EAare,, the modulus of elasticity of the material, the cross-sectional area and length of the element. (b) Write a MATAB function function = elelength(xs,xe,ys,ye) that calculates the length of a member where input parameters xs, xe, ys, ye are the x - andy- coordinates of the start and end nodes of the element, respectively. (c) Write a MATAB function function T = transele(xs,xe,ys,ye) that calculates the 4x4 transformation matrix T of a truss element according to Eq. (5.) of the handout. Caution: Do not calculate the angleϕ before calculating sin ϕ and cos ϕ. You should calculate sin ϕ and cos ϕ directly according to Eq. (5.8). (d) Save any of your previous MATAB work that uses the beam element to hw6.m and make the necessary modifications. Note that the transformation matrix T comes into play twice: () in the first loop to calculate the T global element stiffness matrix k = T kt according to Eq. (5.7) and () in the second loop when calculating the local element displacements d = Tdwhich are then multiplied by the local element stiffness matrix k to obtain the member forces D (see Eq. (5.46)). C:\calpoly\arce6\homework\winter\hw6_sol.doc /4/ :47 PM - -

Winter Quarter Solution () Virtual Work P = 5 kn kn. m.65 T.65C.5 T member forces from handout 4. m 4. m.8c P =.8C EA = MN.667 T 64 T [kn] 9 T 8 C virtual forces by simple statics (see right) Horizontal displacement of node EAq q = 8 8 = 464 knm 464 knm = kn =.46 mm q =.46 mm (left) checks ok 6 Horizontal displacement of node EAq 5 = 8.5 8 + 64.65 5 9.65 5 = 44 knm q 44 knm = =.44 mm q5 =.44 mm (right) checks ok kn 5 6 Vertical displacement of node EAq q 6 = 8.667 8 64.8 5 9.8 5 = 4447 knm 4447 knm = kn = 4.45 mm q = 4.45 mm(up) checks ok 6 6 6 C:\calpoly\arce6\homework\winter\hw6_sol.doc /4/ :47 PM - -

Winter Quarter Problem y Without calculating any stiffness matrices, calculate the elements K5,5, K6,6, K6,5, K7,7, K8,8, K8,9, K8,8, K8,7, K8,6 of the 44x44 structure stiffness matrix K. The nodes of the bottom chord lie on a quadratic parabola with zero slope at node. The truss structure has constant EA. K5,5 K6,6 K6,5 K7,7 K8,8 K8,9 K8,8 K8,7 K8,6 Solution:.5797.76.699.754.76 x EA Solution Some geometrical calculations.5 y(9) =.5 + 7. =.75 m 8,9 =.75 m..5 y() =.5 + 8. =. m 9, =. m..5 y() =.5 + 9. =.55 m, =.55 m. ϕ ϕ 9,9 9, 9,9 9,.75 = tan ( ) = 59.9.55 = tan ( ) = 68.9 = +.75 =.994 m = +.55 =.76 m K 5,5 x -force at node 8 due to x - displacement at node 8 EA EA EA EA K5,5 = EA( ans) + = + = K 6,6 78 89 y -force at node 8 due to y - displacement at node 8 EA EA K6,6 = = =.5797 EA( ans).75 8,9 K 6,5 y -force at node 8 due to x - displacement at node 8 C:\calpoly\arce6\homework\winter\hw6_sol.doc /4/ :47 PM - -

Winter Quarter K6,5 = ( ans) K 7,7 x -force at node 9 due to x - displacement at node 9 EA EA EA EA K7,7 = + + cos ϕ9,9 + cos ϕ9, 8,9 9, 9,9 9, = EA + + cos 59.9 + cos 68.9.994.76 =.76 EA( ans) K 8,8 y -force at node 9 due to y - displacement at node 9 EA EA EA K = + sin + sin = EA + sin 59.9 + sin 68.9 =.699 EA( ans) 8,8 ϕ9,9 ϕ9, 9, 9,9 9,..994.76 K 8,9 y -force at node 9 due to x - displacement at node K 8,9 = K 8,8 y -force at node 9 due to y - displacement at node 9 EA K8,8 = sin ϕ9,9 = EA sin 59.9.754 EA( ans) =.994 9,9 K 8,7 y -force at node 9 due to x - displacement at node 9 EA K8,7 = sinϕ9,9 cosϕ9,9 9,9 = EA sin 59.9 cos 59.9.994 =.76 EA( ans) 8,6 K y -force at node 9 due to y - displacement at node 8 (no connection between nodes 9 and 8 C:\calpoly\arce6\homework\winter\hw6_sol.doc /4/ :47 PM - 4 -

Winter Quarter ARCE 6: MATRIX STRUCTURA ANAYSIS February, Problem Homework 7 (Due -6-) Joints 9 and 9 experience horizontal and vertical displacements as given below. The nodes of the bottom chord lie on a quadratic parabola with zero slope at node. The truss structure has constant EA. Find Δ -56.4 9, x 9, y Δ -47. = Δ 9, x 4.6 Solution: F9 9= 8.6 kn (T) /EA [ m] Δ9, y -958.4 Find the axial force in member 9-9 (hand calculations for (a) and (b)). (a) using scalar trigonometry, i.e. the way you would solve the problem by hand without using any matrices and vectors. (b) using the 4x4 local element stiffness matrix k, the 4x4 transformation matrix T and the 4x vector d of global element displacements. Problem Implement into MATAB the procedure to account for homogeneous boundary conditions (prescribed zero displacements at the fixed degrees-of-freedom) approximately (large value method, see notes) and the procedure to calculate reaction forces. Test your algorithm by analyzing the truss structure below. Check the member forces against the given results. Calculate the reaction forces and check them by statics (submit nothing). 4 kn 6 EA = kn 4 kn 666 C 55 T 96 C m m 78 T 78 T 8 m 8 m // 9:5 AM C:\calpoly\arce6\homework\winter\hw7.doc

Winter Quarter Problem EA α T = 6 kn =. x 5 K 4 m m 8 m 8 m (a) Using hand calculations, assemble the structure load vector for a temperature change of Δ T = K in all members. (b) Use MATAB to find the nodal displacements and member forces of the truss due to a () temperature change of Δ T = K in all members, () displacement Δ, x = 5mmof node in the horizontal direction (node, 5mm, to the right). () displacement Δ, y = 5mmof node in the vertical direction (node, 5mm, down). For () and () you have to implement into MATAB the procedure to account for non-homogeneous boundary conditions (prescribed non-zero displacement, see notes). Submit: hand calculations for (a), deflected shape and member forces for each of the three load cases, comment on the member forces you obtain. // 9:5 AM C:\calpoly\arce6\homework\winter\hw7.doc

Winter Quarter ARCE 6: MATRIX STRUCTURA ANAYSIS February, Problem Homework 7 (Due -6-)-Solution Joints 9 and 9 experience horizontal and vertical displacements as given below. The nodes of the bottom chord lie on a quadratic parabola with zero slope at node. The truss structure has constant EA. Find Δ -56.4 9, x 9, y Δ -47. = Δ 9, x 4.6 Solution: F9 9= 8.6 kn (T) /EA [ m] Δ9, y -958.4 Find the axial force in member 9-9 (hand calculations for (a) and (b)). (a) using scalar trigonometry, i.e. the way you would solve the problem by hand without using any matrices and vectors. (b) using the 4x4 local element stiffness matrix k, the 4x4 transformation matrix T and the 4x vector d of global element displacements. Solution: (a) Δ = 56.4 cos(59.9 ) 47. sin(59.9 ) 4.6 cos(59.9 ) + 958.4 sin(59.9 ) = 6.6/ EA (elongation) Δ 6.6 ε = = = 8.6/ EA EA.994 P = EAε = EA 8.8/ EA= 8.6 kn( ans)tension (b) cos 59.9 sin 59.9-56.4 sin 59.9 cos 59.9 EA -47. D= kd = ktd = /EA.994 cos 59.9 sin 59.9 4.6 sin 59.9 cos 59.9-958.4 = 8.6 kn( ans) /8/ :6 PM C:\calpoly\arce6\homework\winter\hw7_sol.doc

Winter Quarter Problem Implement into MATAB the procedure to account for homogeneous boundary conditions (prescribed zero displacements at the fixed degrees-of-freedom) approximately (large value method, see notes) and the procedure to calculate reaction forces. Test your algorithm by analyzing the truss structure below. Check the member forces against the given results. Calculate the reaction forces and check them by statics (submit nothing). 4 kn 6 EA = kn 4 kn 666 C 55 T 96 C m m 78 T 78 T 8 m 8 m /8/ :6 PM C:\calpoly\arce6\homework\winter\hw7_sol.doc

Winter Quarter (b) () zero member forces, structure is determinate, only displacements.88.8 q = mm 5.76.88.6 Deflected shape due to temperature change of Δ = K (no forces generated) T ().5-4.4 q = mm 5...5-4. member forces due to Δ = 5 mm F F F = F = = = - -4 F 4-4, x 9.kN T 5.9 kn C = 64.7 kn T Deflected shape due to horizontal displacement of Δ, x = 5mm () zero member forces, structure is determinate, only displacements.94 -.5 q = mm -5..87 -.5 Deflected shape due to vertical support settlement Δ, y = 5mm (no forces generated) /8/ :6 PM C:\calpoly\arce6\homework\winter\hw7_sol.doc

Winter Quarter ARCE 6: MATRIX STRUCTURA ANAYSIS February 7, Problem Write a MATAB function Homework 8 (Due -9-) function kbar = elestiff(e,a,i,,flag) that calculates the 6x6 local element stiffness matrix k of a -D frame element (also called beam-column element) for different moment release conditions at the element end (use if-statements), where flag = : fixed-fixed; flag = : fixed-pinned, flag = : pinned-fixed, flag = 4: pinned-pinned (truss element) fixed-fixed fixed-pinned 6 6 6 4 6 EA k = + k = EA... + 6 6 6 6 4 pinned-fixed pinned-pinned (truss) EA EA k = + k = Test your function by typing kbar=elestiff(9,,4,8,). The result should be 9-9 48-48 48 57777-48 8888 k = -9 9 - -48-48 48 8888-48 57777 /7/ : PM C:\calpoly\arce6\homework\winter\hw8.doc

Winter Quarter Test your function by typing kbar=elestiff(9,,4,8,). The result should be 9-9 5 74-5 74 9-74 k = -9 9-5 -74 5 Write a MATAB function function T = transele(xs,xe,ys,ye) that calculates the 6x6 transformation matrix of a -D frame element. T = cosϕ sinϕ sinϕ cosϕ cosϕ sinϕ sinϕ cosϕ Caution: As for the truss element, do not calculate the angleϕ before calculating sin ϕ and cos ϕ. You should calculate sin ϕ and cos ϕ directly according to Eq. (5.8). /7/ : PM C:\calpoly\arce6\homework\winter\hw8.doc

Winter Quarter Problem F 4 8 F F 6 = k-ft 7 6 6 ft ft ft 5 ft (a) Analyze the three-story frame in MATAB using the -D frame element. Draw the bending moment, shear force and axial force diagrams. You may use RISA but you need to show that your MATAB results match those from RISA. Calculate the story drift ratios in %. Ignore axial deformation by assigningea a large value (use EA = ). The total base shear 8 is F = 6 k. Use what you learned in ARCE 7 to distribute the base shear to the floor levels. Floors and have identical weights, the roof weighs 5% of the floor. Partial Solution F = 87.5 k, M = M = 64 k-ft(cw), DR =.9% 6 6 (b) Analyze the frame for pinned columns and determine the story drift ratios. You don't have to draw the internal force diagrams. (c) What is the required flexural stiffness of a grade beam -5 to limit the first story drift ratio of the frame to.5%. Joints and 5 are pinned as in (b). Use the MATAB function deflected_shape_frame(x,y,ns,ne,q,scale)(emailed to you) to plot the deflected shape of the three frame structure (use the same scale factor). Submit: Handcalcs, diagrams, deflected shape (in short: decent documentation), no MATAB. /7/ : PM C:\calpoly\arce6\homework\winter\hw8.doc

Winter Quarter 9 ARCE 6: MATRIX STRUCTURA ANAYSIS February, 9 Problem Homework 8-Solution = knm kn/m 4 EA = 5 kn 5 6 [m] 8. 4. 6. Element i Node j Node 4 4 5 5 6 4 6 5 6 5 7 6.5 m 4 5 6 7 8 9 4 5 6 7 8 @. m 9. m all forces F = kn k/ft II III EAI,, 4 I 5 k IV 5 V k/ft 6 VI 7 ft 6 ft Solution : R = 9.79 k R x y M R R R R 5x 5y 7x 7y =.4 k =5. k-ft ccw =.74 k = 8.6 k =4.47 k = 9.5 k ft ft Implement the "large value approach" to account for the boundary conditions into your MATAB program and calculate the support reactions of the above structures. Ignore axial deformation for structure. /5/9 9: AM C:\calpoly\arce6\homework\Winter9\hw8Sol.doc

Winter Quarter 9 Problem w I I B I x x I When using continuous girders in bridge design it is common to make the section at the interior supports significantly deeper than the sections in the span (see pictures). This practice results in architecturally more pleasing and more efficient structures than using girders with uniform section depth. Since the region near the supports is stiffer than that near midspan, the support "attracts" bending moment resulting in a larger support moment than for uniform moment of inertia along the beam. The objective of this assignment is to quantify the relative increase in bending moment at the support compared to the corresponding value for uniform stiffness ( M = w ). B /8 () Use the beam element implemented in MATAB to analyze a two-span beam with varying second moment of inertia I. For simplicity we work with piecewise constant moment of inertia as shown in the figure. Submit a single figure showing the ratio MB / M B (on they- axis) as a function of the normalized length x/ (on the x - axis) where M B and M B are the bending moments at B for uniform and non-uniform moments of inertia, respectively. Plot four lines (for I / I =,4,6,8 ) and use < x / <.5. Interpret the results. () For I / I= 6and = ft, x = 6 ft analyze the two-span beam by hand and draw the bending moment diagram. Make sure your result matches that of (). Solution M M B B I, 4, 6, 8 I = x /5/9 9: AM C:\calpoly\arce6\homework\Winter9\hw8Sol.doc

Winter Quarter 9 Hand calculations for = ft, x = 6 ft Base structure w = x Moments for base structure (letw = for simplicity).7. M 4.5 4.5 + M 4. 4 ' Displacements for base structure (consider only half of structure) δ = MM dx =.7 4 + (.7 + +.7.) 6 6 =.7 Δ = MM dx = 4.7 4 4 (.7 + ) 6 6 6 = 4.5.7 4 4.5(.7 + ) 6 6 = 6.58 MB 6.58 [ ] x = MA = = 78.4 k-ft( ans) M B.7.8 uniform I.7 M A 78.4 5. w = = = 5. k-ft( ans) 8 8 =.57( ans).6.5.4 I, 4, 6, 8 I = hand calcs.......4.5 x [ ] /5/9 9: AM C:\calpoly\arce6\homework\Winter9\hw8Sol.doc

Winter Quarter ARCE 6: MATRIX STRUCTURA ANAYSIS February 9, Problem Homework 9 (Due -4-) = knm kn/m 4 EA = 5 kn 5 6 [m] 8. 4. 6. Element i Node j Node 4 4 5 5 6 4 6 5 6 5 7 6 Use the plane frame element to analyze the above structure in MATAB. Draw the bending moment, shear force, and axial force diagrams. Pay special attention to the member end conditions and use the given element definition: Solution Problem kn/m [m] 5 8 kn kn/m 6 = knm 4. 4. 4 4. 4. 5. y x () D 5 () D 5 = [. 4.7. 4.7] kn, knm = [ 5 4.7 5 4.7] kn, knm (4) F(7) = 5 kn F(8) = 6kN F(9) = 87.5 knm F() = 4kN F() = 5. knm F() = 5 kn F(4) = kn F(5) = 4.7 knm all other components of F are zero (5) K K K 8,8 8,, = 88 knm = 5.5 knm = T T For the plane frame structure above, calculate (hand calcs): () The size of the structure stiffness matrix. K8,6 = 5.5 knm () The 6x element force vector D (fixed-end forces) of element 5 in local (element) coordinates. () The 6x element force vector D (fixed-end forces) of element 5 in global (structure) coordinates. (4) The non-zero components of the structure force vector (value and location). (5) Elements K, K and K 8,8 8,, of the structure stiffness matrix. (6) The element of the structure stiffness matrix that relates the rotation of node to the moment at node 6. (6) /9/ 4:6 PM C:\calpoly\arce6\homework\winter\hw9.doc

Winter Quarter Problem w I I B I x x I When using continuous girders in bridge design it is common to make the section at the interior supports significantly deeper than the sections in the span (see pictures). This practice results in architecturally more pleasing and more efficient structures than using girders with uniform section depth. Since the region near the supports is stiffer than that near midspan, the support "attracts" bending moment resulting in a larger support moment than for uniform moment of inertia along the beam. The objective of this assignment is to quantify the relative increase in bending moment at the support compared to the corresponding value for uniform stiffness ( M = w ). B /8 () Use the beam element implemented in MATAB to analyze a two-span beam with varying second moment of inertia I. For simplicity we work with piecewise constant moment of inertia as shown in the figure. Submit a single figure showing the ratio MB / M B (on they- axis) as a function of the normalized length x/ (on the x - axis) where M B and M B are the bending moments at B for uniform and non-uniform moments of inertia, respectively. Plot four lines (for I / I =,4,6,8 ) and use < x / <.5. Interpret the results. () For I / I= 6and = ft, x = 6 ft analyze the two-span beam by hand and draw the bending moment diagram. Make sure your result matches that of (). Solution.8.7 M M B B.6.5.4. I, 4, 6, 8 I =......4.5 x /9/ 4:6 PM C:\calpoly\arce6\homework\winter\hw9.doc

Winter Quarter ARCE 6: MATRIX STRUCTURA ANAYSIS February 9, Problem Homework 9-Solution = knm kn/m 4 EA = 5 kn 5 6 [m] 8. 4. 6. Element i Node j Node 4 4 5 5 6 4 6 5 6 5 7 6 Use the plane frame element to analyze the above structure in MATAB. Draw the bending moment, shear force, and axial force diagrams. Pay special attention to the member end conditions and use the given element definition: 74.4 Solution Dbar = -. 7.6. 5. -7.6.69-8.58 4.4 -..46-4.77 7.9 74.4.8. -74.4 4.76 kn, knm. -7.6 -. -5. 7.6 -.69 8.58 7.6. -.46 4.77-7.9. -.8. -4.76 5. C. C 74.4 6 = 45. 8 M [knm] 4.8 7.6 7.6. T.7 C 7.6 7.6 T T C 8.6 5. 8.6 N [kn] 7.6 V..5 7. [kn] 7. 4.4 4.8 4.8 /6/ :47 PM C:\calpoly\arce6\homework\winter\hw9sol.doc

Winter Quarter Problem 8 kn kn/m 5 [m] kn/m 6 = knm 4. 4. For the plane frame structure above, calculate (hand calcs): () The size of the structure stiffness matrix. () The 6x element force vector D (fixed-end forces) of element 5 in local (element) coordinates. () The 6x element force vector D (fixed-end forces) of element 5 in global (structure) coordinates. (4) The non-zero components of the structure force vector (value and location). (5) Elements K, K and K 8,8 8,, of the structure stiffness matrix. (6) The element of the structure stiffness matrix that relates the rotation of node to the moment at node 6. 4 4. 4. 5. y x Some calculations = 4 = 6.4 m 5 sin α = 5 4 =.789 cos α = 4 4 =.647 4m 5 w = kn/m = 4.7 knm 6.4 m 5 w = kn/m =. kn 8 m 4 w = kn/m = 5. knm /6/ :47 PM C:\calpoly\arce6\homework\winter\hw9sol.doc

Winter Quarter () K = 8 x 8 () D 5 = [. 4.7. 4.7] kn, knm T (). sin α =..789 = 5 kn. cos α =..647 = kn D 5 = [ 5 4.7 5 4.7] kn, knm T (4) F(7) = 5 kn F(8) = 4 = 6 kn F(9) = 4.7 5. = 87.5 knm F() = 4kN F() = 5. knm F() = 5 kn F(4) = 8 = kn F(5) = 4.7 knm all other components of F are zero (5) K K K 8,8 8,, 4 4 = 4 = 4 4 = 88 knm = = 4 = 5.5 knm = (6) Moment at joint 6 (dof 8) due to unit rotation at joint (dof 6) K 8,6 = = = 5.5 knm 4 /6/ :47 PM C:\calpoly\arce6\homework\winter\hw9sol.doc

Winter Quarter Problem w I I B I x x I When using continuous girders in bridge design it is common to make the section at the interior supports significantly deeper than the sections in the span (see pictures). This practice results in architecturally more pleasing and more efficient structures than using girders with uniform section depth. Since the region near the supports is stiffer than that near midspan, the support "attracts" bending moment resulting in a larger support moment than for uniform moment of inertia along the beam. The objective of this assignment is to quantify the relative increase in bending moment at the support compared to the corresponding value for uniform stiffness ( M = w ). B /8 () Use the beam element implemented in MATAB to analyze a two-span beam with varying second moment of inertia I. For simplicity we work with piecewise constant moment of inertia as shown in the figure. Submit a single figure showing the ratio MB / M B (on they- axis) as a function of the normalized length x/ (on the x - axis) where M B and M B are the bending moments at B for uniform and non-uniform moments of inertia, respectively. Plot four lines (for I / I =,4,6,8) and use < x / <.5. Interpret the results. () For I / I= 6and = ft, x = 6 ft analyze the two-span beam by hand and draw the bending moment diagram. Make sure your result matches that of (). Solution.8.7 M M B B.6.5.4. I, 4, 6, 8 I =......4.5 x /6/ :47 PM C:\calpoly\arce6\homework\winter\hw9sol.doc

Winter Quarter Solution () MATAB (code not shown here) () Hand calculations for = ft, x = 6 ft Use force method n==>one unknown Slope deflection method appears to be more work here since structure is a sway structure due to the change in I => two unknowns Base structure w = x Moments for base structure (letw = for simplicity).7. M 4.5 4.5 + M 4. 4 ' Displacements for base structure (consider only half of structure) δ = MM dx =.7 4 + (.7 + +.7.) 6 =.7 6 Δ = MM dx = 4.7 4 4 (.7 + ) 6 4.5.7 4 4.5(.7 + ) 6 6 6 6 = 6.58 6.58 x = MB = = 78.4 k-ft( ans).7 MB [ ] M I B, 4, 6, 8 uniform I I = M M M B B B w = = 8 8 = 5. k-ft( ans) 78.4 = 5. =.57( ans).8.7.6.5.4.. hand calcs.....4.5 x [ ] /6/ :47 PM C:\calpoly\arce6\homework\winter\hw9sol.doc

Winter Quarter ARCE 6: MATRIX STRUCTURA ANAYSIS February 4, Problem Homework (Due --) System E ν t wall =, 8 ksi =.5 = ft rigid rigid rigid System E E ν wall steel =, 8 ksi = 9, ksi =.5 t = ft Find the three forces F, F, F and the lateral deflection of the two structural systems. (a) by hand calculation. Use the force method and select the forces in the rigid links as the unknowns. Then show that the forces are given as a function of the relative stiffness (rigidity) of the lateral force resisting elements. Fi = k ki / ki i= Solution: F = 67.7 k F = 5. k F = 7. k Δ =. in System F = 89. k F =.7 k F =. k Δ =.6 in System (b) using your MATAB implementation of the -D frame element after adding shear deformation (flag 5). To avoid type errors copy and paste the electronic version (has been emailed to you) of the statements into your function elestiff. Neglect shear and axial deformations in the moment frame. Consider the braced frame a truss. /4/ :9 PM C:\calpoly\arce6\homework\winter\hw.doc

Winter Quarter ARCE 6: MATRIX STRUCTURA ANAYSIS February 5, Problem Homework -Solution System E ν t wall =, 8 ksi =.5 = ft rigid rigid rigid System E E ν wall steel =, 8 ksi = 9, ksi =.5 t = ft Find the three forces F, F, F and the lateral deflection of the two structural systems. (a) by hand calculation. Use the force method and select the forces in the rigid links as the unknowns. Then show that the forces are given as a function of the relative stiffness (rigidity) of the lateral force resisting elements. Fi = k ki / ki i= Solution: F = 67.7 k F = 5. k F = 7. k Δ =. in (System ) (b) using your MATAB implementation of the -D frame element after adding shear deformation (flag 5). To avoid type errors copy and paste the electronic version (has been emailed to you) of the statements into your function elestiff. Neglect shear and axial deformations in the moment frame. Consider the braced frame a truss. /4/ : PM C:\calpoly\arce6\homework\winter\hwSol.doc

Winter Quarter SYSTEM Deflections for each wall due to unit force Δ = MM dx k VV dx + GA P 6 P = + 5 GA 5 = (.65+.868) 5 =.59 ft wall 5 5 = (.6667+.889) = 4.56 ft wall 5 5 = (.457+.48) = 4.66 ft wall Formal force method analysis δ δ δ W W W W W W =Δ +Δ = 5.575 =Δ +Δ =.87 = Δ = 4.56 Δ = Δ =.59 Δ = 5 4 5.5575.455 x.59 x. k, x 7. k.455.87 x = = = F =. = 67.7 k( ans) F F =. 7. = 5.7 k ( ans) = = 7. k( ans) Deflection of wall 5 Δ = 67.7.59 ft =. ft =. in ( ans) /4/ : PM C:\calpoly\arce6\homework\winter\hwSol.doc

Winter Quarter Direct approach using relative stiffness K = Δ = 6588 k/ft wall = 4657 k/ft wall = 68 k/ft wall K = 6588 + 4657 + 68 = 977 k/ft F F F 6588 = k 977 = 67.7 k ( ans) 4657 = k 977 = 5. k ( ans) 68 = k 977 = 7. k ( ans) 67.7 k 5. k.k Δ = = = =. ft =. in ( ans) 6588k/ft 4657 k/ft 68 k/ft Results match those of the force method /4/ : PM C:\calpoly\arce6\homework\winter\hwSol.doc

Winter Quarter SYSTEM Deflections for each wall due to unit force Shear wall GA = 7 44 =, 6, 8 k = 8 44 =, 6, k-ft Δ = MM dx k VV dx + GA P 6 P = + 5 GA 4 6 4 = +, 6, 5, 6, 8 =.5855 ft/k k = = = 7, 8 k/ft Δ.5855 ft/k Moment frame = 9, 45 /44 = 85, 59 k-ft Δ = MM dx = 7 4 7 5 + 6.667 ft = 85, 59 k-ft =.75 ft/k k = = = 7.9 k/ft Δ.75 ft/k Braced frame for the braced frame, we can calculate the stiffness directly EA 9,.8 k = cos α = = 95 k/ft + 4 + 4 /4/ : PM C:\calpoly\arce6\homework\winter\hwSol.doc

Winter Quarter Formal force method analysis δ δ δ =.5855 +.75 =.7 =.75 + =.75 +.595 =.777 95 = =.75 Δ =.5855 =.5855 Δ =.7.75 x.585.75.777 x = x =.8 k, x =. k F =.8 = 89. k ( ans) F F =.8. =.7 k ( ans) =. k( ans) k.8 k.8 k.k.k 89. k.7 k.k Deflection of wall Δ = 89..5855 ft =.5 ft =.6 in ( ans) /4/ : PM C:\calpoly\arce6\homework\winter\hwSol.doc

Winter Quarter Direct approach using relative stiffness F F F 7, 8 = k = 89. k ( ans) 7, 8 + 7.9 + 95 7.9 = k =.7 k ( ans) 7, 8 + 7.9 + 95 95 = k =. k ( ans) 7, 8 + 7.9 + 95 89. k.7 k.k Δ = = = =.5 ft =.6 in ( ans) 78 k/ft 7.9 k/ft 95 k/ft Results match those of the force method /4/ : PM C:\calpoly\arce6\homework\winter\hwSol.doc

Winter Quarter ARCE 6: MATRIX STRUCTURA ANAYSIS March, Problem Homework (Due --) Wall Wall 4 Rigid Diaphragm Wall Wall E ν t wall =, 8 ksi =.5 = 6in You are working with a somewhat eccentric architect on a single-story (cantilever) shear wall building and asked to evaluate its seismic behavior. Find the x stiffness matrix in units of kips and inches with respect to the three degrees-of-freedom uvθ,, shown. () Use unit displacements along the three global degrees-of-freedom (as learned early this quarter) to find the global stiffness matrix directly (hand analysis). () Use local degrees-of-freedom u, v, u, v, u, v, u4, v4 defined somewhere along each wall along the global directions, corresponding local wall stiffnesses and transformation matrices (hand analysis, except matrix products for which you can use MATAB). You may want to explain to yourself why it doesn't matter where along the wall you define the local dofs. The story height is 4 ft. Ignore any out-of-plane stiffness of the shear walls. Solution 8,794 -,55 6,7 -,4,679 -,55 -,4,679 5,76,88 units: kips and feet or 9,66 -,55,5 -,4,679 9 -,55 -,4,679.469 units: kips and inches // 6:5 AM C:\calpoly\arce6\homework\winter\hw.doc

Winter Quarter ARCE 6: MATRIX STRUCTURA ANAYSIS March, Problem Homework -Solution Wall Wall 4 Rigid Diaphragm Wall Wall E ν t wall =, 8 ksi =.5 = 6in You are working with a somewhat eccentric architect on a single-story (cantilever) shear wall building and asked to evaluate its seismic behavior. Find the x stiffness matrix in units of kips and inches with respect to the three degrees-of-freedom uvθ,, shown. () Use unit displacements along the three global degrees-of-freedom (as learned early this quarter) to find the global stiffness matrix directly (hand analysis). () Use local degrees-of-freedom u, v, u, v, u, v, u4, v4 defined somewhere along each wall along the global directions, corresponding local wall stiffnesses and transformation matrices (hand analysis, except matrix products for which you can use MATAB). You may want to explain to yourself why it doesn't matter where along the wall you define the local dofs. The story height is 4 ft. Ignore any out-of-plane stiffness of the shear walls. Solution 8,794 -,55 6,7 -,4,679 -,55 -,4,679 5,76,88 units: kips and feet or 9,66 -,55,5 -,4,679 9 -,55 -,4,679.469 units: kips and inches /4/ :4 PM C:\calpoly\arce6\homework\winter\hwsol.doc

Winter Quarter () Using direct method F = K = k +.8.8k +.6.6k x = k = 5497 = 8, 794 k/ft F = K =.8.6k.8.6k y = M = K = k +.8.8k.6.6k z + 4.8.6k =.4k =.4 54, 97 =, 55 k-ft/ft F = K = K = x Fy = K = k +.6.6k +.8.8k = k + k = 54, 97 + 7, 74 = 6,7 k/ft M = K = 4k.8.6k + 4.8.8k z.6.8k = 4k.k = 4 7, 74. 54, 97 =, 4, 67 k-ft/ft F = K = K x F = K = K y M z + +.8 4 + 4.8 4 = K = 4 k k k k.6 4 k = 6k + 67k = 6 7, 74 + 94 54, 97 = 5, 76, k-ft/rad 8,8 -,6 6, -,44, units: kips and feet -,6 -,44, 5,7, or 9,66 -,6,5 -,44, units: kips and inches 9 -,6 -,44,.469 /4/ :4 PM C:\calpoly\arce6\homework\winter\hwsol.doc

Winter Quarter () Using transformation of stiffness Use units of kips and feet Displacement due to unit force Δ = MM dx k VV dx + GA P 6 P = + 5 GA 5 =.88 ft for walls,,4 5 =.94 ft for wall k = Δ = 54, 97 k/ft for walls,,4 = 7, 74 k/ft for wall ocal element stiffness matrices k = 54, 97 k k k = 7, 74.8.8.6 484-6 = 54, 97 =.8.6.6-6 958.6.8.6 958 6 = 54, 97 =.8.6.8 6 484 4 Transformation matrices - T = T = -4 5 T = - - T 4 = 4 /4/ :4 PM C:\calpoly\arce6\homework\winter\hwsol.doc

Winter Quarter Global element stiffness matrices k 5497-69 = -69 4895777 k k k 4 = 774-86968 - 86968 4784 484-6 444 = - 6 958-78 444-78 99 958 6 45698 = 6 484 695 45698 69 5 669 Structure stiffness matrices 8,794k/ft -,55 k/rad K k k k k -,55k - ft/ft -,4,679 k - ft/ft 5,76,88 k - ft/rad (ans) = + + + 4 = 6,7 k/ft -,4,679 k/rad or 9,66 -,6 K =,5 -,44, 9 -,6 -,44,.469 units: kips and inches The two methods give identical results. /4/ :4 PM C:\calpoly\arce6\homework\winter\hwsol.doc

Winter Quarter ARCE 6: MATRIX STRUCTURA ANAYSIS March 5, Homework (Due --) q q q q q q = const. = const. = const. (a) (b) (c) () For the above frame structures, use MATAB to find the stiffness matrix with respect to the two degrees-of-freedom shown using static condensation. () If the girders of structure (b) are considered rigid find the stiffness matrix with respect to the degrees-of-freedom shown by hand (think). () For structures (a) and (c), use hand calculation to find the flexibility matrix F with respect to the degrees-of-freedom shown. Show that the stiffness and flexibility matrices are inverse to each other. Neglect axial deformation for all structures. Solution () 7.47.74.77.4646 (Str. a) K = (Str. b) =.74.74 K.4646.66..97 (Str. c) K =.97. ()..888 K =.888.888 () 7. 7..556. (Str. a) F = (Str. c) 7. 54. F =..556 Use the following commands for (). Kfree = K(free,free); Kee = Kfree(e,e); Krr = Kfree(r,r); Kre = Kfree(r,e); Ker = Kfree(e,r); Kstar = Krr - Kre*inv(Kee)*Ker (derived in Friday, 5-mar class) C:\calpoly\arce6\homework\winter\hw.doc /5/ :9 PM

Winter Quarter ARCE 6: MATRIX STRUCTURA ANAYSIS March 9, q Homework (Due --) q q q q q = const. = const. = const. (a) (b) (c) () For the above frame structures, use MATAB to find the stiffness matrix with respect to the degrees-of-freedom shown. () If the girders of structure (b) are considered rigid find the stiffness matrix with respect to the degrees-of-freedom by hand (think). () For structures (a) and (c), use hand calculation to find the flexibility matrix F with respect to the degrees-of-freedom shown. Neglect axial deformation for all structures. Solution () 7.47.74.77.4646 (Str. a) K = (Str. b) =.74.74 K.4646.66..97 (Str. c) K =.97. ()..888 K =.888.888 () 7. 7..556. (Str. a) F = (Str. c) 7. 54. F =..556 Use the following commands for (). Kfree = K(free,free); Kee = Kfree(e,e); Krr = Kfree(r,r); Kre = Kfree(r,e); Ker = Kfree(e,r); Kstar = Krr - Kre*inv(Kee)*Ker C:\calpoly\arce6\homework\winter\hwSol.doc /8/ :6 PM

Winter Quarter Solution () / H + / H q =, q = q =, q = / H / H / H + / H K K / H / H K H / H / K / H / H K K K = + = + = =. 7 H H 4 = = =.888 7 H 4 = = =.888 7 H..888 K = ( ans).888.888 C:\calpoly\arce6\homework\winter\hwSol.doc /8/ :6 PM