Question 2.5: solid bar 50 mm in diameter and 2000 mm long consists of a s...
solid bar 50 mm in diameter and 2000 mm long consists of a steel and an aluminum section, as shown in Figure 2.34. When axial force P is applied to the system, a strain gauge attached to the aluminum indicates an axial strain of 0.000873 m/m. (a) Determine the magnitude of applied force P, and (b) if the system behaves elastically, find the total elongation of the bar.
Given: Dimensions of composite bar and measured normal strain.
Find: Applied force, P, and elongation of bar, δ.
Assume: Hooke’s law applies.
The diameter of the bar is 50 mm, so the cross-sectional areas of both parts are equal:
A_{St}=A_{Al}=\frac{\pi }{4}(0.05 \textrm{m})^2 = 0.00196 \textrm{m}^2 .
The elastic moduli for aluminum and steel may be looked up in Table 2.1 or in another reference.
Table 2.1
Approximate Design Values (Reflecting Proportional Limits) of Elasticity and Shear Moduli, in Linear Regimes (SI)
| Material | Modulus of Elasticity E (MPa) | Modulus of Rigidity G (MPa) |
| California redwood | 7600 | |
| Steel (carbon) ASTM A36 | 207,000 | 83,000 |
| Stainless steel | 200,000 | 80,000 |
| Aluminum 6061-T7 | 70,000 | 28,000 |
| Glass | 48,000–83,000 | 19,000–35,001 |
| Polycarbonate | 2400 | 800 |
| Concrete | 21,500 | 8970 |
| Bone | 1–16,000 | 4–8000 |
E_{Al} ≈ 70 GPa, and E_{St} ≈ 200 GPa.
If Hooke’s law applies, we can relate the strain measured in the aluminum portion to the stress induced by P in that portion:
\varepsilon _{_{Al}}= 0.000873=\frac{\sigma _{_{Al}}}{E_{Al}}=\frac{(P/A_{Al})}{E_{Al}} ,
so
P = (0.000873)(70 × 10^9 \textrm{Pa})(0.00196 \textrm{m}^2 ) = 120 kN.
We can exploit Hooke’s law and superpose the displacements of both portions of the bar:
\delta = \sum{\frac{PL}{AE}}=\left\lgroup\frac{PL}{AE}\right\rgroup _{St}+\left\lgroup\frac{PL}{AE}\right\rgroup _{Al}=\frac{120,000 \textrm{N}}{0.00196 \textrm{m}^2}\left[\frac{1.5 \textrm{m}}{200\times 10^9 \textrm{N/m}^2}+\frac{0.5 \textrm{m}}{70\times 10^9 \textrm{N/m}^2}\right]
= 459 × 10^{-6} \textrm{m} + 437 × 10^{-6} \textrm{m} = 896 × 10^{-6} \textrm{m}
δ = 896 μm, or 0.896 mm.
Note: The aluminum section is only a third as long as the steel, but it
deforms nearly as much!