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Chapter 2: Problem 50

The concrete post \(\left(E_{c}=3.6 \times 10^{6} \mathrm{psi} \text { and } \alpha_{c}=5.5 \times 10^{-6} /^{\circ} \mathrm{F}\right)\) is reinforced with six steel bars, each of \(\frac{z}{8}\) -in. diameter \(\left(E_{x}=29 \times 10^{6} \mathrm{psi} \text { and } \alpha_{x}=6.5 \times 10^{-6} /^{\circ} \mathrm{F}\right) .\) Determine the normal stresses induced in the steel and in the concrete by a temperature rise of \(65^{\circ} \mathrm{F}\)

Short Answer

Expert verified
Steel stress: 1.885 ksi, Concrete stress: -0.234 ksi.

Step by step solution

01

Calculate the Change in Length for Concrete

For a temperature increase of \(65^{\circ} \mathrm{F}\), calculate the change in length \( \Delta L_c \) for the concrete using the formula:\[ \Delta L_c = \alpha_c \cdot \Delta T \cdot L \]Here, \( \Delta T = 65^{\circ} \mathrm{F} \) and \( \alpha_c = 5.5 \times 10^{-6} /^{\circ} \mathrm{F} \). Since \( L \) cancels out later, assume it is 1 for simplicity:\[ \Delta L_c = 5.5 \times 10^{-6} \times 65 \allowbreak= 3.575 \times 10^{-4} \]
02

Calculate the Change in Length for Steel

Similarly, calculate the change in length \( \Delta L_x \) for the steel:\[ \Delta L_x = \alpha_x \cdot \Delta T \cdot L \]Given \( \alpha_x = 6.5 \times 10^{-6} /^{\circ} \mathrm{F} \):\[ \Delta L_x = 6.5 \times 10^{-6} \times 65 \allowbreak= 4.225 \times 10^{-4} \]
03

Find the Difference in Strain Between Materials

Determine the difference in strain due to the different expansion rates:\[ \varepsilon_c = \Delta L_c / L = 3.575 \times 10^{-4} \]\[ \varepsilon_x = \Delta L_x / L = 4.225 \times 10^{-4} \]The difference in strain is:\[ \Delta \varepsilon = \varepsilon_x - \varepsilon_c = 4.225 \times 10^{-4} - 3.575 \times 10^{-4} = 6.5 \times 10^{-5} \]
04

Calculate the Force Needed to Equalize the Strain

Calculate the force \( F \) needed to counter the difference in strains using Hooke's Law. For the steel:\[ \sigma_x = E_x \cdot (\varepsilon_x - \varepsilon_{combined}) \]For the concrete:\[ \sigma_c = E_c \cdot (\varepsilon_c + \varepsilon_{combined}) \]Where:\[ \varepsilon_{combined} = -\Delta \varepsilon \allowbreak L_c/L_x \]Assume that because of balance:\[ \varepsilon_{combined} = \Delta \varepsilon \cdot \frac{E_c}{E_x} \]
05

Calculate Stresses in Steel and Concrete

Using the value of \( \varepsilon_{combined} \) from Step 4, substitute back into Hooke's Law to find stresses:**For steel:**\[ \sigma_x = E_x \cdot \Delta \varepsilon = 29 \times 10^6 \cdot 6.5 \times 10^{-5} = 1.885 \text{ ksi} \]**For concrete:**\[ \sigma_c = E_c \cdot (-\Delta \varepsilon) = 3.6 \times 10^6 \cdot (-6.5 \times 10^{-5}) = -0.234 \text{ ksi} \]
06

Conclusion: Final Results

The normal stress in the steel is approximately \(1.885 \text{ ksi}\), and the normal stress in the concrete is approximately \(-0.234 \text{ ksi}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Concrete Reinforcement
Concrete reinforcement refers to the process of strengthening concrete by embedding materials that are stronger than the concrete itself. In most cases, this is achieved using steel bars, also known as rebar. The reason for reinforcing concrete is that it can withstand compression well, but not tension. Thus, adding materials that prosper under tensile loads can significantly augment the durability and strength of concrete structures.

When incorporating steel bars into concrete, it is essential to consider the bonding between the steel and the concrete. The two materials must adequately adhere to each other to equally distribute loads and stresses across the composite structure. This is particularly important in the context of temperature changes, which can cause one material to expand or contract more than the other.
  • The compatibility of thermal expansion rates between concrete and reinforcing steel is crucial.
  • Proper alignment and spacing of reinforcing bars ensure balanced stress distribution.
  • Concrete covers over steel bars protect them from corrosion and external damage.
Thermal Expansion
Thermal expansion is a material's tendency to change its shape, area, and volume in response to a change in temperature. This phenomenon is critical to consider in construction as temperature variations can cause materials to expand or contract, thereby affecting the structure鈥檚 stability over time.

Each material has a specific coefficient of thermal expansion, determined by its atomic structure and binding forces. In this context:
  • Steel has a higher coefficient of thermal expansion ( \( \alpha_x = 6.5 \times 10^{-6} \/^{\circ} \mathrm{F} \) ) compared to concrete ( \( \alpha_c = 5.5 \times 10^{-6} \/^{\circ} \mathrm{F} \) ).
  • This difference results in varying length changes when the temperature fluctuates, as seen in the exercise.
  • The challenges arise when different parts of a structure expand at different rates, leading to potential stress and structural integrity issues.
Understanding and calculating the effects of thermal expansion helps engineers design structures that can withstand temperature changes without material failure.
Mechanical Properties
Mechanical properties refer to the characteristics of a material that describe its behavior under various forms of force and load. They are crucial in determining how a material will perform in real-world applications, particularly when subjected to forces beyond its static load.

In the realm of concrete and steel reinforcements, important mechanical properties include:
  • Elastic Modulus (E): This represents a material's ability to deform elastically (return to its original shape) after stress. Steel in this exercise has a much higher elastic modulus ( \( E_x = 29 \times 10^6 \text{ psi} \) ) compared to concrete ( \( E_c = 3.6 \times 10^6 \text{ psi} \) ), showing that steel is more resistant to deformation.
  • Stress and Strain: Stress is the force applied per unit area, while strain is the deformation or displacement. Monitoring these properties helps in understanding the responses of each material to stress, helping predict issues like cracking or buckling.
  • Tensile Strength: This is the resistance of a material to breaking under tension. Steel's high tensile strength makes it an ideal reinforcement for concrete.
By combining materials with complementary mechanical properties, such as steel and concrete, structures can achieve greater strength and durability.

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Most popular questions from this chapter

Link \(B D\) is made of brass \((E=105 \mathrm{GPa})\) and has a cross-sectional area of \(240 \mathrm{mm}^{2}\). Link \(C E\) is made of aluminum \((E=72 \mathrm{GPa})\) and has a cross-sectional area of \(300 \mathrm{mm}^{2}\). Knowing that they support rigid member \(A B C\), determine the maximum force \(\mathbf{P}\) that can be applied vertically at point \(A\) if the deflection of \(A\) is not to exceed \(0.35 \mathrm{mm}\)

A vibration isolation unit consists of two blocks of hard rubber with a modulus of rigidity \(G=19\) MPa bonded to a plate \(A B\) and to rigid supports as shown. Denoting by \(P\) the magnitude of the force applied to the plate and by \(\delta\) the corresponding deflection, determine the effective spring constant, \(k=P / \delta,\) of the system.

The steel rails of a railroad track \(\left(E_{3}=200 \mathrm{GPa}, \alpha_{5}=11.7 \times 10^{-6} /^{\circ} \mathrm{C}\right)\) were laid at a temperature of \(6^{\circ} \mathrm{C}\). Determine the normal stress in the rails when the temperature reaches \(48^{\circ} \mathrm{C}\), assuming that the rails \((a)\) are welded to form a continuous track, \((b)\) are \(10 \mathrm{m}\) long with \(3-\) mm gaps between them.

A block of 10 -in. length and \(1.8 \times 1.6\) -in. cross section is to support a centric compressive load \(\mathbf{P}\). The material to be used is a bronze for which \(E=14 \times 10^{6}\) psi. Determine the largest load that can be applied, knowing that the normal stress must not exceed 18 lssi and that the decrease in length of the block should be at most \(0.12 \%\) of its original length.

Bar \(A B\) has a cross-sectional area of \(1200 \mathrm{mm}^{2}\) and is made of a steel that is assumed to be elastoplastic with \(E=200 \mathrm{GPa}\) and \(\sigma_{Y}\) \(=250\) MPa. Knowing that the force \(\mathbf{F}\) increases from 0 to \(520 \mathrm{kN}\) and then decreases to zero, determine ( \(a\) ) the permanent deflection of point \(C,(b)\) the residual stress in the bar.
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