Question

4. A rigid square plate of side 2a is supported on four identical coil springs each of stiffness k at the corners A, B, C, D, as shown in Figure 4. A vertical force P is now applied at a point with the coordinates (x, y). Find the forces in the four springs as functions of x and y. РТУ, > ? Um Figure 4

Answer 1

Answer: Determination of forces in four Springs Load on each spring due to symmetry uniformity → RC = P2 - RA RA + Rc - Pl Rot RD = Pla cew Exporession for the moment is → (RB x 2a) + (Rcx 20) - [Px Cy+a] -0 Substitute Rc 20 RB + (PL-RA) aa = P[y+a] 20 R8 + PA – ZARA = PY + Pa Ca) 20 | Ca CRB - R) = PM +Pa – Pa .: RB – RA - Doy - → (RAX 2a) + (Rex 24).– [px Ca-x)] = 0 20RA + 2aRB = Pla-~) PA+R$ = (2-x) RB = L (2-x) – RA Substitute this equation in (a) Per le CA-x) =-Ra] - RA - en pCont) 2Ra - Home PRA = Plan de 20 2RA [pca-x) – py] RA - "a [rca-x)– py] - 0

Substitute RA in (m) RB - ta [P(a-x) – P(y)] = py R$ = polyes + Ha [p(a->) – Pcy)] RB = 204 + P(a-) –py 4a RB = PY+ Pla-2) 40 Substitute RB in RB + RD = P12 40 PY +P (-) + Rp = { RD = { [RY + -)] Rp = { bayan Polat 40 - consider, (R8x 20)+ (Rcx 20) - [Px Cy+a)] -0 20R8+ 20Rc-Ply+a)=0 2a (Rg+RC) = P(y+a) RB+RC = P(4+a) 20 Substitute RB Py + Pla-1) + Rc 4a Ply+a) +RC = Ply+a) 20 ча RC = plu+a)_ Py + P(c-1) Referent them be - Plant Rc = pu a + e - Planet - 112,3,4 equations are force in spring A, B, C, D in terms of X,Y.