Question

Use the stiffness method to analyse the elastic frame ABC shown below. Use a model made up of 2 the elements (AB and CB) and the axis indicated in the figure. All members have the following properties: E=2.108 kPa, A = 0.005 m², 1 = 1.5e - 4 m4. Also the lengths of the elements are the same: AB = BC = L = 4.7 m and w=17 kN/m. - B Structure, loading, member axes Degrees of freedom

Answer 1

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E = 2X108 kPa w = 2X108 kN/m? A=0.005 m2 - 15 x 10-4 mg AB = BC = L = 47 m 20 = 17 kN/m AE = 0'005 x 2x10 - 1x of kv ES = 2x108 x 15 x10 = 30109 kN m S y Steps Identity degree of freedom and numbering the Step II Moke all displacements equal to 2000 by dixing joint B, and calculate stiffnen condicients. Apply, unit displacement along degree of freedom t. E +265 Kno

Kw = A + DES - Axo + 12X3X10" = 216233140 Ka = - GER - 63X10 ' - - 8148048 472 Apply, mit rotation of degree of freedom 2. Kiz = - 60 Kaj = -8148148. 3X109 x2 = 51063'83. K32 = Set 8148048 Reply, mit displacement at degree of freedom 3. t a ner - ME

Kig =0 K23 = ce = 3148:48 K33 = P + 128 - 14.00 + 12 XFX/10? 23 = 216233.40 There fore Stiftnes matrix 21623340 -814848 K- -814848 5706383 814848 lo 3148 48 21623340 Here, k, = 21623340 kotte k z2 = 5106383 K33 = 216233140 Step III Colculate fixed and reactions corresponding to unknown joint displacements 20 PF er RF = = 17847 - 39.95 KN RP = -1942 - 17X,4172 = - 3129 kom 12

Therefore lood vector SpA 2 S 39:957 = -3129 Step Now, applying principal of superposition " {p} = [K] {n} + {pr} where {p} - rector of external load of unknown degree of freedom {u} - displacement rector. This, qu} = [1] [ip} - {p*}] 216 233'40 - 8148'48 717 (3995 1-214842 51063 83 8148:48 RB 31294 0 8 148:48 2162 72-3995-2 P= P = P = 0 since there is no external lood at joint B 1216 23340 -8148.480 7 1-8148-48 51063183 2148-48. 3129 10 9148.48 21623340 [4.65 x10 1'47 x 107 - 2:81 x 10-8 1-39195-7 7'47x107 1.98x10-5 – 747x10 7 3 31:29 -281x103 3744x107 465x156 | - -O'0001632 O'000 590 (-0'000022 1-163 x 10-4 2 35'90x/0-4 1 - 0'22 X10-4 rod

Displacements are U, = -1063x10-4 m Uz = 5.90 x10" sad Ug = 0'22 X 10^9 m . Step D. Calculate moment at joint i n Moment at A due to displacement at 1. MA = - x U, = - 3148048 x 61'63x104) = 133 km m Moment at A due to rotation at 2. ma - 2EF xv = _2x 3X10" X 5'-90x10-4 7'53 km 1 Moment at A due to displacement at 3 Fixed end moment at 'A of wh? PAA 12 - - 17x4'= 31'29 km T2 There fore moment at A Ma = Ma't ma + ma me = 33 + 753 +0 + 31'29 = 40115 kNm