Question

$HM_2_Q.jpg$

The following system is composed of several beams with two nodes fixed. The cross-section area is A = 0.006m², moment of inertia I=6.87x10-5m4 elastic modulus is E = 2.1x108kN/m², initial length L=5m. Calculate evaluate the nodal force, displacement and moment based on Euler and Timoshenko beam

Answer 1

*****if this is helpful to you please like it.*****

5.00 m 2 3 #1 5.00 m #2

Current Member member-1 #2 5.00 m # 5.00 m Local Stiffness: Member-1 = 0.0000 from node-1 = (0.00, 0.00) m to node-2 = (0.00, 5.00) m Member length, L = 5.00 m cos = x2 - x1 = (0.00, 0.00) 5.00 sine = y2 - y1 = (5.00, 0.00) L 5.00 [K] = [T)'[k'][T] , where [k'] = Local Member Stiffness = 1.0000 1 1 2 3 AE/L 0 0 0 12 EI/LY 6EI/L2 06E1/L2 4EI/L -AEL 0 0 0 - 12E1/L-6E1/L? 06E1/L2 2AE/L 4 5 6 -AE/L 0 0 0 -12 EI/L 6EI/L? 0 -6E1/L2 2 EI/L AE/L 0 0 0 12 EI/L? -6EI/L2 0-6EI/L 4EI/L 3 4 5 6 where, A = 0.0060 m2 E = 210.0000 kN/m2 | = 6.870e-05 m L = 5.00 m AEL = (0.0060) 210.0000)/5.0000 = 0.2520 12E1/L? = 12(210.0000)(6.870e-05)/5.0000} = 0.0014 6EI/L? = 6(210.0000)(6.870e-05)/5.00002 = 0.0035 4E1/L = 4(210.0000)(6.870e-05)/5.0000 = 0.0115 2E1/L = 2(210.0000)(6.870e-05)/5.0000 = 0.0058

Current Member member-1 #2 5.00 m # 5.00 m ZEI/L = 21ZTU.UUUUJ6.870e-0575.0000 = 0.0058 3 6 0.25 (0.25) 1 2 0.00 0.00 0.00 0.01 (0.00) 0.00 (0.00) 0.01 [k'] = (0.25) 0.25 0.00 (0.00) 3 4 5 6 (0.00) (0.00) 0.00 0.01 - (0.00) 0.01 [T] = Transformation Matrix 0 cos sine -sino cos 0 0 0.00 1.00 1.00 0.00 0 0 0 0 0 1 0 0 0 cos sino 0 -sin cose 0 0 0 0 0 0 0 0 0.00 1.00 - -1.00 0.00 0 0 0 0 0 1 [T] = Transposed [T] Matrix 0 0 0 0 0.00 -1.00 1.00 0.00 0 0 coso -sino sino cos 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 cose -sino O sino cose 0 0 0 0 0.00 -1.00 1.00 0.00 0 1

Current Member member-1 6 #2 5.00 m 5.00 m 0 - 1.00 0.00 -sine cose 0 0 0 1 Oooo 0 0 0 1 0 0 0 cos sine 0 -sin cose 0 0 0 0 0 0.00 1.00 - - 1.00 0.00 0 0 1 [T] = Transposed [T] Matrix 0 0 0 0 0.00 - 1.00 1.00 0.00 0 0 0 0 0 = coso -sine sin cos 0 0 0 0 0 0 0 0 1 0 0 0 cos -sino 0 sino cose 0 0 0 0 0 -0.00 -1.00 - 1.00 0.00 1 1 2 [K] = Global Member Stiffness 2 4 5 6 0.00 -0.00 -0.00 -0.00 0.25 -0.25 -0.00 0.01 0.00 0.01 -0.00 0.00 0.00 0.00 - -0.25 0.25 -0.00 0.01 0.00 0.01 - 5 6

Current Member member-2 6 #2 5.00 m 5.00 m Local Stiffness: Member-2 from node-2 = (0.00, 5.00) m = 1.0000 to node-3 = (5.00, 5.00) m Member length, L = 5.00 m coso = x2 - x1 = (5.00, 0.00) 5.00 sin = y2 - y1 = (5.00, 5.00) L 5.00 [K] = [T'[k'][T] , where [k'] = Local Member Stiffness = 0.0000 4 4 5 6 AE/L 0 0 0 12 EI/L 6EI/L? 0 6EI/L 4EI/L -AE/L 0 0 0 -12 EI/LP -6EI/L? 0 6E1/L2 2AE/L 7 8 9 -AEL 0 0 0 -12 EI/L3 6E1/L2 0-6EI/L 2 EI/L AE/L 0 0 0 12 EI/L -6EI/L? 0-6EI/L? 4EI/L 5 6 7 8 9 where, A = 0.0060 m² E = 210.0000 kN/m² | = 6.870e-05 m L = 5.00 m AE/L = (0.0060)(210.0000)/5.0000 = 0.2520 12E1/L? = 12(210.0000)(6.870e-05)/5.0000} = 0.0014 6E1/L² = 6(210.0000)(6.870e-05)/5.00002 = 0.0035 4E1/L = 4(210.0000)(6.870e-05)/5.0000 = 0.0115 2 EI/L = 2(210.0000) (6.870e-05)/5.0000 = 0.0058

Current Member member-2 6 2 3 1 #2 5.00 m # 1 5.00 m ZEIL סטטט.מכטטטטטטטטט.טו == U.UUPO 4 5 4 5 6 7 8 9 0.25 (0.25) - 0.00 0.00 (0.00) 0.00 0.00 0.01 - (0.00) 0.01 (0.25) 0.25 - (0.00) (0.00) 0.00 (0.00) 0.00 0.01 (0.00) 0.01 [k'] = 6 7 8 9 [T] = Transformation Matrix cos sine -sino cos 0 0 0 0 0 0 0 1.00 0.00 -0.00 1.00 0 0 0 0 0 0 1 0 0 0 cos sino 0 -sin cos 0 0 0 0 1 1.00 0.00 - -0.00 1.00 0 0 0 0 0 1 1 [T] = Transposed [T] Matrix 0 0 0 0 cos -sine sino cose 0 0 1.00 -0.00 0.00 1.00 0 0 0 0 0 1 0 1 0 0 O cos -sine 0 sino cose 0 0 0 0 0 0 0 0 0 0 1 1.00 -0.00 0.00 1.00 0 1

Current Member member-2 6 #2 5.00 m 5.00 m 0 0 0 0 -0.00 1.00 0 1 -sine cose 0 0 0 0 0 0 0 0 0 1 0 0 cos sino 0 -sin cos 0 0 0 1.00 0.00 -0.00 1.00 0 1 [T] = Transposed [T] Matrix 0 0 0 0 coso -sine sine cose 0 0 1.00 -0.00 0.00 1.00 0 0 0 0 0 0 0 0 1 0 0 0 cose -sino 0 sino cose 0 0 0 1.00 -0.00 0.00 1.00 0 0 0 0 0 1 4 [K] - Global Member Stiffness 6 7 8 9 0.25 -0.25 0.00 0.00 --0.00 0.00 0.00 0.01 -0.00 0.01 -0.25 0.25 -0.00 -0.00 0.00 -0.00 0.00 0.01 - -0.00 0.01 5 6 7 8 9

-2.00 KN 2.50 kN/m 1.00 KN 9 4.00 KN 5.00 m 5.00 m Frame Stiffness: Matrix [Q] = [K][D] + [QF] where, [Q] = Joint Load Matrix where, [K] = Frame Stiffness Matrix where, [D] = Joint Displacements Matrix where, (QF] = Equivalent Load Fixed-end Forces Matrix 4 1.00 2.00 0.0 5 2.5e+05 0.0 3,462.48 0.0 2.5e+05 3,462.48 3,462.48 3,462.48 23,083.20 D4 D5 D6 -2.00 6.25 2.71 6 solve the matrix linear equation for [D] m D4 D5 0.00001 -0.00002 -0.00012 m D6 rad

Current Member member-1 5.00 m 5.00 m [q] = [k][d] + [qF] where, [9] = Member Force in Global Coordinate where, [k] = Member Stiffness in Global Coordinate where, [d] = Member End Displacement in Global Coordinate where, [F] = Member End Force in Global Coordinate 1 1,384.99 1 -2.00 2 8832 1 -3,462.48 - 1,384.99 2 3 4 -3,462.48 -1,384.99 2.5e+05 11,541.60 3,462.48 3,462.48 1,384.99 -2.5e+05 5,770.80 3,462.48 5 -3,462.48 -2.5e+05 - 5,770.80 3,462.48 2.5e+05 11,541.60 0.0000 0.0000 0.0000 1.3439e-05 51 -1.5173e-05 -0.00012 4 0.00 2.50 -2.00 0.00 -2.50 5 6 -3,462.48 -1.61 3.82 1.87 -2.39 3.82 KN KN KN-m KN KN

Current Member member-1 #2 | ա 00°C 5.00 m -1.61 3.82 1.87 -2.39 -3.82 -3.80 KN KN KN-m KN KN kN-m Transform, (q'] = [T)'[q] Member Local Force 3.82 1.61 1.87 -3.82 2.39 -3.80 KN KN KN-m KN KN KN-m

Current Member member-2 5 8 16 9 7 2 #2 5.00 m Et 1 5.00 m where, (q) = Member Force in Global Coordinate where, [k] = Member Stiffness in Global Coordinate where, [d] = Member End Displacement in Global Coordinate where, [F] = Member End Force in Global Coordinate 4 2.5e+05 4 5 6 5 6 7 8 9 -2.5e+05 1,384.99 3,462.48 - -1,384.99 3,462.48 3,462.48 11,541.60 -3,462.48 5,770.80 - 2.5e+05 -1,384.99 -3,462.48 1,384.99 -3,462.48 3,462.48 5,770.80 - -3,462.48 11,541.60 94 95 96 97 q8 99 = 4 1.3439e-05 -1.5173e-05 6 -0.00012 7 0.0000 8 0.0000 0.0000 0.00 6.25 5.21 0.00 6.25 -5.21 -2.5e+05 7 8 9 KN KN 3.39 5.82 3.80 -3.39 6.68 -5.94 KN-m KN KN KN-m

-2.00 KN 2.50 kN/m 1.00 KN -3.39 #2 -5.94 kN-m 6.68 kN 4.00 KN -1.61 kN. 1.87 kN-ml 5.00 m 3.82 kN Base Units: m, kN, KN-m, kN/m², m2, m4 +- Fx = 0 - 1.61 - 3.39 +4.00 + 1.00 = 0.0000 +1 {Fy = 0 + 3.82 + 6.68 + 2.00 - 12.50 = 0.0000 +- EM = 0 + 1.87 - 5.94 + (3.39)(5.00) + (6.68)(5.00) - (4.00)(2.50) - (1.00) (5.00) - (12.50) (2.50) = 0.0000