Indeterminate Structures — Question 2

Section 12-5 · Physics pages 1854–1855

Question 2 An eccentric load changes which elastic supports remain active Three identical compression-only springs support a rigid weightless beam at −a, 0, a, with level unloaded tops. A downward load W is at x = e ≥0. In the small-displacement model each contacting spring obeys R = kδ. −a 0 a W e Rigid beam on three initially level elastic supports; load position is specified in the text. (a) Find the reactions assuming all three contacts remain active, and the allowed range of e. (b) For e = .8a, find the actual reactions and check separation of the left spring.
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Question 2 - Solution An eccentric load changes which elastic supports remain active −a 0 a W e (a) Derivation and calculation For all contacts active, set δ(x) = d + ϕx. Balance gives 3kd = W, 2ka2ϕ = We. Consequently, R0 = W/3 , R± = W/3 ± We/(2a) . Compression-only contact requires R−≥0, hence 0 ≤e ≤2a/3 (strictly positive forces below the endpoint). (b) Result and interpretation At e = .8a, the left contact is inactive. Rebalance the two remaining supports: R0 + R+ = W, aR+ = .8aW. Thus R−= 0, R0 = .2W, R+ = .8W . Rigid-beam extrapolation gives δ−= 2δ0 −δ+ = −.4W/k < 0, confirming a gap at the left spring. Physical check A negative all-contact reaction must be replaced by a new contact regime, not merely set to zero without rebalancing.

Question 2

An eccentric load changes which elastic supports remain active

Three identical compression-only springs support a rigid weightless beam at −a,0,a-a,0,a, with level unloaded tops. A downward load WW is at x=e≥0x=e\ge0. In the small-displacement model each contacting spring obeys R=kδR=k\delta.

Diagram shown in the original SVG worksheet.

  1. Find the reactions assuming all three contacts remain active, and the allowed range of ee.

  2. For e=.8ae=.8a, find the actual reactions and check separation of the left spring.

Question 2 - Solution

An eccentric load changes which elastic supports remain active

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

For all contacts active, set δ(x)=d+ϕx\delta(x)=d+\phi x. Balance gives 3kd=W,2ka2ϕ=We.3kd=W,\qquad 2ka^2\phi=We. Consequently, R0=W/3,R±=W/3±We/(2a).\boxed{R_0=W/3},\qquad\boxed{R_\pm=W/3\pm We/(2a)}. Compression-only contact requires R−≥0R_-\ge0, hence 0≤e≤2a/3\boxed{0\le e\le2a/3} (strictly positive forces below the endpoint).

(b) Result and interpretation

At e=.8ae=.8a, the left contact is inactive. Rebalance the two remaining supports: R0+R+=W,aR+=.8aW.R_0+R_+=W,\qquad aR_+=.8aW. Thus R−=0,R0=.2W,R+=.8W.\boxed{R_-=0,\quad R_0=.2W,\quad R_+=.8W}. Rigid-beam extrapolation gives δ−=2δ0−δ+=−.4W/k<0\delta_-=2\delta_0-\delta_+=-.4W/k<0, confirming a gap at the left spring.

Physical check

A negative all-contact reaction must be replaced by a new contact regime, not merely set to zero without rebalancing.