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2. Compound Bars

  • 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, \(F_{s} + F_{c} = F_{total}\).

  • These different materials are constrained to deform together such that they satisfy compatibility condition, \(\epsilon_{material1} = \epsilon_{material2}\).

  • Modular ratio, \(m\), is given by \(\frac{E_{material1}}{E_{material2}}\).

\[\begin{align}\tag{9} F_{s} + F_{c} = F_{total} \end{align}\]
\[\begin{align}\tag{10} \epsilon_{material1} = \epsilon_{material2} \end{align}\]

2.1. A concrete column with steel reinforcement

Idea

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 \(=\ 4\ m\).

Solution:

stresses in steel (\(\sigma_{s}\)) and concrete (\(\sigma_{c}\))
Equilibrium condition Equation 9: \(F_{s} + F_{c} = F\)
Compatibility condition Equation 10: \(\epsilon_{s} = \epsilon_{c}\)

\[\begin{align*} \frac{\sigma_s}{E_s} &= \frac{\sigma_c}{E_c} \\ \frac{F_s}{A_sE_s} &= \frac{F_c}{A_cE_c} \\ F_s &= F_c \times \frac{A_sE_s}{A_cE_c} \\ F_s &= (F - F_s) \times \frac{A_sE_s}{A_cE_c} \\ \frac{F_s}{F - F_s} &= \frac{A_sE_s}{A_cE_c} \\ \frac{1}{\frac{F}{F_s} - 1} &= \frac{A_sE_s}{A_cE_c} \\ F_s &= F \times \frac{A_sE_s}{A_cE_c + A_sE_s} \\ \sigma_s &= \frac{F_s}{A_s} = \frac{FE_s}{A_cE_c + A_sE_s} \\ A_s &= 8 \times \pi \big(\frac{D}{2} \big)^2 = 2\pi D^2 \\ A_c &= (\text{dimension} \times \text{dimension}) - 2\pi D^2 \\ \sigma_c &= \frac{F_c}{A_c} = \frac{F-Fs}{A_c} = \frac{F-(\sigma_sA_s)}{A_c} \end{align*}\]

elastic shortening (\(\Delta L\))

\[\begin{align*} \epsilon_s &= \frac{\Delta L_s}{L_s} \\ \epsilon_s &= \frac{\sigma_s}{E_s} \\ \Delta L_s &= L_s \times \frac{\sigma_s}{E_s} \\ \Delta L_c &= L_c \times \frac{\sigma_c}{E_c} \\ L_s &= L_c = L \text{ and } \epsilon_{s} = \epsilon_{c} \\ \Delta L &= L \times \frac{\sigma_s}{E_s} = L \times \frac{\sigma_c}{E_c} \text{ (compression) } \end{align*}\]

2.2. Encased composite bar

Idea

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 \(=\ 100\ GPa\).

Solution:

stresses in steel (\(\sigma_{s}\)) and cast iron (\(\sigma_{I}\))
Equilibrium condition Equation 9: \(F_{s} + F_{I} = F\)
Compatibility condition Equation 10: \(\epsilon_{s} = \epsilon_{I}\)

\[\begin{align*} \frac{\sigma_s}{E_s} &= \frac{\sigma_I}{E_I} \\ \frac{F_s}{A_sE_s} &= \frac{F_I}{A_IE_I} \\ F_s &= F_I \times \frac{A_sE_s}{A_IE_I} \\ F_s &= (F - F_s) \times \frac{A_sE_s}{A_IE_I} \\ \frac{F_s}{F - F_s} &= \frac{A_sE_s}{A_IE_I} \\ \frac{1}{\frac{F}{F_s} - 1} &= \frac{A_sE_s}{A_IE_I} \\ F_s &= F \times \frac{A_sE_s}{A_IE_I + A_sE_s} \\ \sigma_s &= \frac{F_s}{A_s} = \frac{FE_s}{A_IE_I + A_sE_s} \\ A_s &= \pi \big(\frac{d}{2} \big)^2 \\ A_I &= \frac{\pi D^2}{4} + \frac{\pi d^2}{4} = \frac{\pi}{4} (D^2 - d^2) \\ D &= d + (2 \times \text{thickness}) \\ \sigma_I &= \frac{F_I}{A_I} = \frac{F-Fs}{A_I} = \frac{F-(\sigma_sA_s)}{A_I} \end{align*}\]

elongation of the assembly (\(\Delta L\))

\[\begin{align*} \epsilon_s &= \frac{\Delta L_s}{L_s} \\ \epsilon_s &= \frac{\sigma_s}{E_s} \\ \Delta L_s &= L_s \times \frac{\sigma_s}{E_s} \\ \Delta L_I &= L_I \times \frac{\sigma_I}{E_I} \\ L_s = L_I &= L \text{ and } \epsilon_{s} = \epsilon_{I} \\ \Delta L &= L \times \frac{\sigma_s}{E_s} = L \times \frac{\sigma_I}{E_I} \text{ (extension) } \end{align*}\]

2.3. A steel rod and brass bush

Idea

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, \(E = 100\ GN/m^2\).

Solution:

Please attempt this assignment.

2.4. Buttwelded stanchion

Idea

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
(b) If \(E\) for steel is \(200\ GN/m^2\) and this value is twenty times that for the concrete, find when the stanchion carries a load of \(368.8\ kN\),

(i) The stress in the concrete
(ii) The stress in the steel
(iii) The amount the stanchion shortens over a length of \(2\ m\).

Solution:

Please attempt this assignment.



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Comments (5)

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Brenda

Ae we doing chapters 6, 7, 8 & 9?

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Dr. Sam

No. That's for Solid II.

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Elsie

Kindly confirm for us the formula for volumetric strain

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Brenda

Hello, are the notes here all we will have for SSM2 or is there a possibility for addition?

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Dr. Sam

Solid Mechanics II starts from chapter 6.