Question

Q.2: Taking A, B, C and D as the nodes for the grillage frame (AB and BC are parallel to Y axis and BD is parallel to X axis) shown below, evaluate the nodal load vector {P) and the stiffness matrix [K] of the structure taking same elastic modulus (E = 100GPa) and Poisson's ratio (v = 0.25) for all members. The members are having same rectangular cross-section (a = 200mm, b = 250 mm). Solve the governing equation [K]{X}={P} to calculate the nodal displacement vector X of the structure. Using {X} (along with the figures or equation given in slide 16 of the 2nd set of lecture slides), calculate the torsional moments at the two ends of each member and use these values to draw the torsion diagram of the structure. Use the attached table for calculating the torsional constant J (Note: J is not equal to the polar moment of inertia for a non-circular section) of the members. 50 KN 2.5m 20kN AB = 5m BC = 4m BD = 8 m ba Horizontal plane X-Y G= 2(1+V) The torsional constant (I) of a member having a rectangular section (axb): J = kab Table of Constants for Torsion of a Rectangular Bar 1.0 1.2 1.5 0.675 0.759 0.848 0.930 0.968 0.1406 0.166 0.196 0.229 0.249 0.208 0.219 0.231 0.246 0.258 0.985 0.997 0.999 1.000 1.000 0.263 0.281 0.291 0.312 0.333 0.267 0.282 0.291 0.312 0.333 2.5 The table is taken from “Theory of Elasticity, 3rd Edition, S.P. Timoshenko and J.N. Goodier".

Answer 1

solution, sign convertfensi Considering anti-clock wise member end moments as positive. Governΐνης αμαtion : (K] {x}={p} H where, {1} = { Po} - {f} 363 - joint load. matoix , {pl} a member end force maduix due to external loads skje stiffness matrix {x} - displacement matrix s p ray Toint load mataix : P150 ('. No moment acting, at joint B) P2=-20 1: 20 kN force is acting at joint B in 2-disetén) Member end force matrix; {R} - C:} v om 9770N um

fined end moments: MAB = Wt 0x MB 231.25 kNam (anti-clockwise) MBA. W L = -50 X5 MBA = -31125 kN-m [clock wise] MBC – Mcp = Map = MDBo lino external load in member L BC and BD) P4 = MBA + BC & M BD 331125 toto P1 = -31.25 kN-m 1 Ple a reaction et joint B in 7-direction : 325 KN (inte direction hence positive) se {P. 2 5-31,252 Le 1.25 J | {P3 = {{o} - Ele} -m ieszy! C 2 SP? 5.31.252 stiffness matoixi b le a) Allowing init rotation corresponding to degree of freedom 1 (lie along 05)

261 7 6 E2" B a A GEI kn = YEL +44 + 4E2 Kiny = 2.3 ET K121 = 01228 45. EL ) b) Allowing unit xotation corresponding to degree of freedom e Cię, along. Jes) REBALL un uz 43 43

ke :-GEL e + CEL :: 52 De 18 142 4 6E ko - 12 Eli .712ET Kipa wa 34 1262 12 62 1692 2 0.5069 EL Hence stiffness matoix o la 167= T kail Kiz 7 nowd aftene 01228 HSE 22815 (2.32 0122848ED) Lor 21815EL_013069.EE Given E=100 G pa. = 100X109N/m2 = 100 Xigé kn/m? 2- body { fon rectangular section] Where; b= width of rectangle .dz depth of rectangle. for this case, I abs py) a= 200 nim zovem a b 2250 mm 0.25m I= CHR 1 = 0,00026041:666 m4 BI - (100X106) 60.000 260 Û 1666) EI = 20041.667 KN-M2 stiffness mateix termys , ... Kil=213 ED 2213x26041,667. GAD Tkel 2.39895-5335

K12 - Kai = 0.82875 EL 20182875 726041.661 ), K2= K21 = 5954.031 k22 - 013069 EZ 2013069. X 2004).667 ke2 = 4992,188) 1K7 = 598851833 5957,03)/ 5957,031 79921188 Governing equilibolum equation ; (k) (x) = {p} [59885.833 9957,031 7 SOB? - $1,25? 159571031 79921 OB 0.00117 s o Ser=-0.00650m. 11. . the 1.08.20000117 Canti clockwise) S2B20100650m 20150 mm (down wards).