2. Compound Bars
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A compound or composite bar consists of one or more materials bonded together rigidly such that the straining action of the external load is shared by the materials such that they satisfy equilibrium condition, \(F_{s} + F_{c} = F_{total}\).
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These different materials are constrained to deform together such that they satisfy compatibility condition, \(\epsilon_{material1} = \epsilon_{material2}\).
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Modular ratio, \(m\), is given by \(\frac{E_{material1}}{E_{material2}}\).
2.1. A concrete column with steel reinforcement
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Example 1 |
A reinforced cement concrete column of dimension \(625\ mm \times 625\ mm\) has eight steel rods of diameter \(30\ mm\) as reinforcement. Find the stresses in steel and concrete, and the elastic shortening of the column if \(E =\ 200,000\ N/mm^2\) for steel and \(10,000\ N/mm^2\) for concrete. Load on column \(=\ 3000\ kN\) and length \(=\ 4\ m\). |
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Solution:
stresses in steel (\(\sigma_{s}\)) and concrete (\(\sigma_{c}\))
\[\begin{align*}
\frac{\sigma_s}{E_s} &= \frac{\sigma_c}{E_c} \\
\frac{F_s}{A_sE_s} &= \frac{F_c}{A_cE_c} \\
F_s &= F_c \times \frac{A_sE_s}{A_cE_c} \\
F_s &= (F - F_s) \times \frac{A_sE_s}{A_cE_c} \\
\frac{F_s}{F - F_s} &= \frac{A_sE_s}{A_cE_c} \\
\frac{1}{\frac{F}{F_s} - 1} &= \frac{A_sE_s}{A_cE_c} \\
F_s &= F \times \frac{A_sE_s}{A_cE_c + A_sE_s} \\
\sigma_s &= \frac{F_s}{A_s} = \frac{FE_s}{A_cE_c + A_sE_s} \\
A_s &= 8 \times \pi \big(\frac{D}{2} \big)^2 = 2\pi D^2 \\
A_c &= (\text{dimension} \times \text{dimension}) - 2\pi D^2 \\
\sigma_c &= \frac{F_c}{A_c} = \frac{F-Fs}{A_c} = \frac{F-(\sigma_sA_s)}{A_c}
\end{align*}\]
elastic shortening (\(\Delta L\))
\[\begin{align*}
\epsilon_s &= \frac{\Delta L_s}{L_s} \\
\epsilon_s &= \frac{\sigma_s}{E_s} \\
\Delta L_s &= L_s \times \frac{\sigma_s}{E_s} \\
\Delta L_c &= L_c \times \frac{\sigma_c}{E_c} \\
L_s &= L_c = L \text{ and } \epsilon_{s} = \epsilon_{c} \\
\Delta L &= L \times \frac{\sigma_s}{E_s} = L \times \frac{\sigma_c}{E_c} \text{ (compression) }
\end{align*}\]
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2.2. Encased composite bar
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Example 2 |
A steel rod of diameter \(65\ mm\) and length \(1\ m\) is encased by a cast iron (CI) sleeve \(9\ mm\) thick and of internal diameter \(65\ mm\). The assembly is subjected to a load of \(45\ kN\). Find the stresses in the two materials and the elongation of the assembly. \(E\) for steel \(=\ 200\ GPa\) and \(E\) for cast iron \(=\ 100\ GPa\). |
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Solution:
stresses in steel (\(\sigma_{s}\)) and cast iron (\(\sigma_{I}\))
\[\begin{align*}
\frac{\sigma_s}{E_s} &= \frac{\sigma_I}{E_I} \\
\frac{F_s}{A_sE_s} &= \frac{F_I}{A_IE_I} \\
F_s &= F_I \times \frac{A_sE_s}{A_IE_I} \\
F_s &= (F - F_s) \times \frac{A_sE_s}{A_IE_I} \\
\frac{F_s}{F - F_s} &= \frac{A_sE_s}{A_IE_I} \\
\frac{1}{\frac{F}{F_s} - 1} &= \frac{A_sE_s}{A_IE_I} \\
F_s &= F \times \frac{A_sE_s}{A_IE_I + A_sE_s} \\
\sigma_s &= \frac{F_s}{A_s} = \frac{FE_s}{A_IE_I + A_sE_s} \\
A_s &= \pi \big(\frac{d}{2} \big)^2 \\
A_I &= \frac{\pi D^2}{4} + \frac{\pi d^2}{4} = \frac{\pi}{4} (D^2 - d^2) \\
D &= d + (2 \times \text{thickness}) \\
\sigma_I &= \frac{F_I}{A_I} = \frac{F-Fs}{A_I} = \frac{F-(\sigma_sA_s)}{A_I}
\end{align*}\]
elongation of the assembly (\(\Delta L\))
\[\begin{align*}
\epsilon_s &= \frac{\Delta L_s}{L_s} \\
\epsilon_s &= \frac{\sigma_s}{E_s} \\
\Delta L_s &= L_s \times \frac{\sigma_s}{E_s} \\
\Delta L_I &= L_I \times \frac{\sigma_I}{E_I} \\
L_s = L_I &= L \text{ and } \epsilon_{s} = \epsilon_{I} \\
\Delta L &= L \times \frac{\sigma_s}{E_s} = L \times \frac{\sigma_I}{E_I} \text{ (extension) }
\end{align*}\]
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2.3. A steel rod and brass bush
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Assignment 1 |
A \(75\ mm\) diameter compound bar is constructed by shrinking a circular brass bush onto the outside of a \(50\ mm\) diameter solid steel rod. If the compound bar is then subjected to an axial compressive load of \(160\ kN\) determine the load carried by the steel rod and the brass bush and the compressive stress set up in each material. For steel, \(E = 210\ GN/m^2\); for brass, \(E = 100\ GN/m^2\). |
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Solution:
Please attempt this assignment. |
2.4. Buttwelded stanchion
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Assignment 2 |
A stanchion is formed by buttwelding together four plates of steel to form a square tube of outside cross-section \(200\ mm \times 200\ mm\). The constant metal thickness is \(10\ mm\). The inside is then filled with concrete. (a) Determine the cross-sectional area of the steel and concrete (i) The stress in the concrete |
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Solution:
Please attempt this assignment. |
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Comments (7)
Ae we doing chapters 6, 7, 8 & 9?
No. That's for Solid II.
Kindly confirm for us the formula for volumetric strain
Hello, are the notes here all we will have for SSM2 or is there a possibility for addition?
Solid Mechanics II starts from chapter 6.
Could you confirm if in example 1 the shear force diagram is correct since from my shear force calculations the forces are positive meaning the diagram ought to be upwards
Yes, in example 1, the shear force diagram is correct. There are two point loads acting downwards, which tend to shear the bar in the negative direction.