Indeterminate Structures — Question 4

Section 12-5 · Physics pages 1858–1859

Question 4 A slack support engages only beyond a load threshold Two vertical elastic ties connect the same load hook to a fixed support. Tie 1 has stiffness k1 = 1000 N/m and no slack. Tie 2 has stiffness k2 = 2000 N/m and slack g = 2 mm. Their tensions are T1 = k1δ and T2 = k2 max(δ −g, 0). δ P g k1g k1 slope k1 + k2 Load–displacement law: the second tie engages after the initial slack is removed. (a) Find the downward load at which tie 2 begins carrying force. (b) For a load of 10 N, find the hook displacement and both tensions.
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Question 4 - Solution A slack support engages only beyond a load threshold δ P g k1g k1 slope k1 + k2 (a) Derivation and calculation The second tie engages when δ = g = .002 m. Before engagement, P = k1δ, T2 = 0. Thus the threshold is Pc = k1g = 1000(.002) = 2.00 N . (b) Result and interpretation Since 10 > Pc, both ties carry load: P = k1δ + k2(δ −g) ⇒δ = P + k2g k1 + k2 . δ = 10 + 2000(.002) 3000 = 4.67 mm , T1 = 14/3 = 4.67 N , T2 = 16/3 = 5.33 N . Physical check The tensions sum to 10 N, and the calculated displacement exceeds the 2 mm slack.

Question 4

A slack support engages only beyond a load threshold

Two vertical elastic ties connect the same load hook to a fixed support. Tie 1 has stiffness k1=1000N/mk_1=1000\,\mathrm{N/m} and no slack. Tie 2 has stiffness k2=2000N/mk_2=2000\,\mathrm{N/m} and slack g=2mmg=2\,\mathrm{mm}. Their tensions are T1=k1δT_1=k_1\delta and T2=k2max⁡(δ−g,0)T_2=k_2\max(\delta-g,0).

Diagram shown in the original SVG worksheet.

  1. Find the downward load at which tie 2 begins carrying force.

  2. For a load of 10N10\,\mathrm N, find the hook displacement and both tensions.

Question 4 - Solution

A slack support engages only beyond a load threshold

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

The second tie engages when δ=g=.002\delta=g=.002 m. Before engagement, P=k1δ,T2=0.P=k_1\delta,\qquad T_2=0. Thus the threshold is Pc=k1g=1000(.002)=2.00N.\boxed{P_c=k_1g=1000(.002)=2.00\ \mathrm N}.

(b) Result and interpretation

Since 10>Pc10>P_c, both ties carry load: P=k1δ+k2(δ−g)⇒δ=P+k2gk1+k2.P=k_1\delta+k_2(\delta-g)\Rightarrow \delta=\frac{P+k_2g}{k_1+k_2}. δ=10+2000(.002)3000=4.67mm,\boxed{\delta=\frac{10+2000(.002)}{3000}=4.67\ \mathrm{mm}}, T1=14/3=4.67N,T2=16/3=5.33N.\boxed{T_1=14/3=4.67\ \mathrm N},\qquad\boxed{T_2=16/3=5.33\ \mathrm N}.

Physical check

The tensions sum to 10 N, and the calculated displacement exceeds the 2 mm slack.