Question

PROBLEM #2 (35 points) Using Castigliano's method, determine the vertical deflection at the middle of the simply supported beam (point E) shown in the figure due to the vertical load P applied at point E. The second area moments for sections AC, CD and DB are 21, I and 21, respectively. Ar 21 - 21B I 1/4 1 1/4 1 1/4 1 1/4

Answer 1

given Problem: 21 2T le you choose STEP 1: We need to determine the deflection at E. Lets draw the Shear forre diagram and Berding Moment diagram for the beam. Hoa Let Reactions at Supports be RA and RB. Applying Equilibrium equations, { Fy zo Rat RB = P EMA=0 = P. L - RB.L=0 - RB = P/2 & RA=P-RB & RAFP/ The two reactions Ra and RB are / and P/2. P-P/2 =P-RB = and are :

Step 2: We know that dv = -w and dy = v Between A,B,, wzo =) dx = 0 =) Shear Forte diagram will be straight line Between BG, Wzo 2) d V O dx a) Shear Force diagram will be straight line. But due to load at By, the shear force direction. will be changed. I Now dM = V. Between A, B, V-positive and Constant; hence slope of Bending moment dia agram will be positive. dx Between BIG , value of Shear force is negative and constant, Hence the Bending moment diagram Between B, C will be negative.

Step 3: Drawing the Shear Force and Bending Moment diogram Shear Force Diagram hiba IPLA PLIA Bending Moment Diagram : from We can see from the Bending moment diagram that the value of Bending moment at ang distance x support A is Pa

Step 4: Draw the following Table- IM ар Region Moment (M) Limits flexural stiffness I AL PR/2 x/2 otolu 261 CE Pafz / 44 to 4 EI This table has been made upto point & because the beam is symmebricat. We shall multiply the final result with 2 to get the exac value.

Steps: By Castigliano's Theorem, уч (1) * 4 | ) () ZEI 44 q = P x² dx + P x² da o AEI) 4 EI 6)®-6] + Rs 50° 40'] 12 ET

2 ) P CAT 12ET 了。 [ ] [ 」 自 | P3 IP.3 152EET 168EI -) - PC + 14 PL? 153CEL S PL3 1536 ET. 152/CT Noco, St: 2 . 2x 19 PL | S3 EI ) 2x p2 424x64 EL SP3 ASKET