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Example 2-9: - A \( 30-\mathrm{m} \) tape weighs \( 12.5 \mathrm{~g} / \mathrm{m} \) and has a cross-section of \( 0.022 \mathrm{~cm}^{2} \). It measures correctly when supported throughout under a tension of 8.0 kg and at a temperature of \( 20^{\circ} \mathrm{C} \). When used in the field, the tape is only supported at its ends, under a pull of 9.0 kg , and at an average temperature of \( 28^{\circ} \mathrm{C} \). Determine the distance between the zero and \( 30-\mathrm{m} \) marks, if \( E=2.10 \times 10^{6} \mathrm{~kg} / \mathrm{cm}^{2} \) \& \[ \mathrm{C}=0.0000116 / 1{ }^{\circ} \mathrm{C} \] - Given: \( \Rightarrow N L=30 \mathrm{~m} \) ; \( \mathrm{w}=12.5 \mathrm{~g} / \mathrm{m} \) - \( \mathrm{Ps}=8.0 \mathrm{~kg} \) - \( \mathrm{Pm}=9.0 \mathrm{~kg} \) \( \mathrm{E}=2.10 \times 10^{\circ} \mathrm{kg} / \mathrm{cm}^{2} \) \[ \begin{array}{l} \text { Ts }=20{ }^{\circ} \mathrm{C} \\ T=28^{\circ} \mathrm{C} \\ \mathrm{~A}=0.022 \mathrm{~cm}^{2} \\ \mathrm{~L}=30 \mathrm{~m} \text { (dist. to be measured) } \\ \mathrm{C}=0.0000116 / 1{ }^{\circ} \mathrm{C} \end{array} \] - Req'd: L' - Solution: - \( \mathrm{C}_{\rho}=\frac{\left(\mathrm{P}_{n}-\mathrm{P}_{\mathrm{s}}\right) L}{\mathrm{AE}}=\frac{(\theta-8) 30}{0.022\left(2.10 \times 10^{6}\right)}=6.494 \times 10^{-4} \mathrm{~m} \) - \( \mathrm{C}_{\mathrm{F}}=\mathrm{CL}\left(\mathrm{T} \mathrm{T}_{\mathrm{S}}\right)=0.0000116(30)(28-20) \) - \( C_{1}=2.784 \times 10^{3} \mathrm{~m} \) - \( C_{1}=5.6034 \times 10^{-3} \mathrm{~m} \) - \( \mathrm{L}^{\prime}=\mathrm{L} \) : \( \mathrm{C}_{\mathrm{L}} \) \( \mathrm{L}^{\prime} \mathrm{L}^{\prime}=30+0.0056=30,0056 \mathrm{~m} \)

          Example 2-9:
- A \( 30-\mathrm{m} \) tape weighs \( 12.5 \mathrm{~g} / \mathrm{m} \) and has a cross-section of \( 0.022 \mathrm{~cm}^{2} \). It measures correctly when supported throughout under a tension of 8.0 kg and at a temperature of \( 20^{\circ} \mathrm{C} \). When used in the field, the tape is only supported at its ends, under a pull of 9.0 kg , and at an average temperature of \( 28^{\circ} \mathrm{C} \). Determine the distance between the zero and \( 30-\mathrm{m} \) marks, if \( E=2.10 \times 10^{6} \mathrm{~kg} / \mathrm{cm}^{2} \) \&
\[
\mathrm{C}=0.0000116 / 1{ }^{\circ} \mathrm{C}
\]
- Given:
\( \Rightarrow N L=30 \mathrm{~m} \)
; \( \mathrm{w}=12.5 \mathrm{~g} / \mathrm{m} \)
- \( \mathrm{Ps}=8.0 \mathrm{~kg} \)
- \( \mathrm{Pm}=9.0 \mathrm{~kg} \)
\( \mathrm{E}=2.10 \times 10^{\circ} \mathrm{kg} / \mathrm{cm}^{2} \)
\[
\begin{array}{l}
\text { Ts }=20{ }^{\circ} \mathrm{C} \\
T=28^{\circ} \mathrm{C} \\
\mathrm{~A}=0.022 \mathrm{~cm}^{2} \\
\mathrm{~L}=30 \mathrm{~m} \text { (dist. to be measured) } \\
\mathrm{C}=0.0000116 / 1{ }^{\circ} \mathrm{C}
\end{array}
\]
- Req'd: L'
- Solution:
- \( \mathrm{C}_{\rho}=\frac{\left(\mathrm{P}_{n}-\mathrm{P}_{\mathrm{s}}\right) L}{\mathrm{AE}}=\frac{(\theta-8) 30}{0.022\left(2.10 \times 10^{6}\right)}=6.494 \times 10^{-4} \mathrm{~m} \)
- \( \mathrm{C}_{\mathrm{F}}=\mathrm{CL}\left(\mathrm{T} \mathrm{T}_{\mathrm{S}}\right)=0.0000116(30)(28-20) \)
- \( C_{1}=2.784 \times 10^{3} \mathrm{~m} \)
- \( C_{1}=5.6034 \times 10^{-3} \mathrm{~m} \)
- \( \mathrm{L}^{\prime}=\mathrm{L} \) : \( \mathrm{C}_{\mathrm{L}} \)
\( \mathrm{L}^{\prime} \mathrm{L}^{\prime}=30+0.0056=30,0056 \mathrm{~m} \)

Example 2-9:
- A 30-m tape weighs 12.5 g / m and has a cross-section of 0.022 cm^2. It measures correctly when supported throughout under a tension of 8.0 kg and at a temperature of 20^∘C. When used in the field, the tape is only supported at its ends, under a pull of 9.0 kg , and at an average temperature of 28^∘C. Determine the distance between the zero and 30-m marks, if E=2.10 × 10^6 kg / cm^2 &

C=0.0000116 / 1^∘C

- Given:
⇒ N L=30 m
; w=12.5 g / m
- Ps=8.0 kg
- Pm=9.0 kg
E=2.10 × 10^∘kg / cm^2

Ts =20^∘C

T=28^∘C
A=0.022 cm^2
L=30 m (dist. to be measured)
C=0.0000116 / 1^∘C

- Req'd: L'
- Solution:
- Cρ=((Pn-Ps) L)/(AE)=((θ-8) 30)/(0.022(2.10 × 10^6))=6.494 × 10^-4 m
- CF=CL(TTS)=0.0000116(30)(28-20)
- C1=2.784 × 10^3 m
- C1=5.6034 × 10^-3 m
- L^'=L : CL
L^'L^'=30+0.0056=30,0056 m

Added by Ver-Nica N.

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