Question

Calculate the following using THE DIRECT STIFFNESS METHOD :

a) The reaction forces, bending moments and deflections at nodes 1, 2 and 3.

E=200 Gpa

I= 4x10^-5 m^4

The force acting in the center of the beam is 115 kg x 9.81 = 1 128,15 N

(2 O2ISM 14 2154

Answer 1

SOLUTION:

-> Given, 1128.15 N 0.215m + 0-215m &=200 apa I= uxlosm step D Node Representation, 02 303 4 W2 aut w3 element 1 Node i-2 2-3 2 Element stiffness Matrix 12 GL k=E I L 3 422 GL -12 62 -6l -12 -6h 12 -ol 62 212 -6L 422 222 CS

2) for element 1 L=0.215m Ex I = 2008 109 x 4x105 Ex I = 8x106 Nm?. Therefore κι: εΙ 12 (0.21533 1.29 -12 1-29 1:29 -12 1-29 0-1849 -1.29 0-09245 -1-29 12 -1-29 0.09245 -1.29 0-1849 ki=8xlo 12 Co. 215) 3 1.29 -12 1.29 1:29 0-1849 -1.29 0-09245 a12 1.29 12 -1-29 1-29 0.09245 -1-29 0.1849 ki=8.05x108 12 1:29 -12 1.29 1-29 -12 0.1849 -1-29 -1029 12 009245 -129 1-29 O-09245 -1.29 0.1849 ki= 10 ol Col 96.6 10.3845 –96.6 10.3845 10.3845 1.488445 -10.3845 0.74422 Co2 -96.6. -10-3845 96.6 -10.3845 10.3845 0.74422 -10.3845 az 1.488445 02

for element 2 Since, L=0.215m and E I = 8x10 Nm. Therefore, stiffness Matrix. K2=ki 8 k2= 10 Coz 96.6 10.3845 10.3845 1-488445 -96.6 -10.3845 10.3845 0:74422 w3 -96-6 -10.3845 96.6 -10.3845 10.3845 0774422 -10.3845 1.488445! 3038 3 Global stiffness Matrix, Cul Co2 Cu 3 8 k=10 این 62 Co3 03

cu 3 Q3 96-6 10-3845 -96.6 10.3845 10.3845 1.488445 710.3845 074422 o2 10.3845 - 96.6 116.3845 laß.2 -a6-6 k=10 10.3845 0.74422 o 2.98 Flo.3845 6.74422 96.6 -10.3845 Co3 |- 96-6 -10.3845 110.3845 6.74422 -110.3845 1-488445/03 CS Scanned with CamScanner

Applying sc's, At node 1 fixed support, CUI= QI=0 At node 3w fixed support, cu 33 033 9) Global Displacement Matrixy da Cul Cu2 CZ 02 Cu3 10/0/0 S) ریاهان Forces, F1 = 0 = 63 And F2 = 1128-15 N. using, the equation [k][a] =F After reduction of global stiffness 0 where - 1128.13

10+ [1922 1921 F2 [1128.15) After Solving, los [ 1a3.2 cu 2 ] = •1128-15 qu23-5.84 x loom 10° 2.9862] = 0 Tazz o rad 6 Reactions And Moments D At. Node I R= [k] [d]-[FT Ri=108 [96.6 10.3845 -96.6 10.3845 X O -5.84x1ā8 {o} ol ololo R = 564-144 N

Mi=108 [10.3845 1.4884145 -10.3845 0.74422 07 O X -5.84x1 o8 {o} O O Mi= 60.6455 nm 2) At Node 3 R3=108 Loo -96-6-10.3845 96-6 -10-3fug] X -5.84x108 To ololo R3= 564.144 N

3= \о? Со о 10.3 RS -\о. з чS с.CC, 2 (- це? 445 X -S: R4x48 (१ ГАЗ- 60.6455 Nm. CS Scanned with CamScanner

ANSWERS:

1). AT NODE 1

REACTION (R1) = 564.144 N

MOMENT (M1) = 60.6455 Nm

SLOPE (w1) = DEFLECTION ($\Theta$1) = 0

2). AT NODE 2

FORCE (F2) = -1128.15 N -(GIVEN)

MOMENT (M2) = 0 Nm

SLOPE (w2) = -5.84 * $10^{-8}$ m

DEFLECTION ($\Theta$2) = 0 rad

3). AT NODE 3

REACTION (R3) = 564.144 N

MOMENT (M3) = -60.6455 Nm

SLOPE (w3) = DEFLECTION ($\Theta$3) = 0

VERIFICATION OF FORCES AND MOMENTS:

SUM OF ALL FORCES = 0 , SUM OF MOMENTS =0

1. R1 - F + R3 = 564.144 - 1128.15 + 564.144 = 0.1 N(CLOSE TO ZERO)

(ERROR DUE TO ROUNDING OFF THE VALUES)

2. M1 + M2 - M3 = 60.6455 + 0 - 60.6455 = 0 Nm

HENCE, VERIFIED.