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A steel bar of square cross section \((50 \mathrm{mm} \times 50 \mathrm{mm})\) carries a tensile load \(P\) (see figure) The allowable stresses in tension and shear are \(125 \mathrm{MPa}\) and \(76 \mathrm{MPa}\), respectively. Determine the maximum permissible load \(P_{\max }\).

Short Answer

Expert verified
The maximum permissible load \( P_{\max} \) is 190,000 N.

Step by step solution

01

Calculate Cross-sectional Area

The steel bar has a square cross-section with each side measuring 50 mm. To find the cross-sectional area \( A \), use the formula for the area of a square: \( A = ext{side}^2 \). So, the area is \((50 ext{ mm})^2 = 2500 ext{ mm}^2\).
02

Convert Area to Square Meters

Since stress is given in MPa (which is N/m²), convert the cross-sectional area from mm² to m². Use the conversion: \( 1 ext{ mm}^2 = 1 imes 10^{-6} ext{ m}^2 \). Thus, \( 2500 ext{ mm}^2 = 2500 imes 10^{-6} = 0.0025 ext{ m}^2 \).
03

Determine Load Based on Tensile Stress

The allowable tensile stress is given as \(125 ext{ MPa}\). The maximum permissible load based on tension is calculated using \( P = ext{tensile stress} imes A \). Therefore, \( P_{ ext{tension}} = 125 imes 10^6 imes 0.0025 = 312,500 ext{ N}\).
04

Determine Load Based on Shear Stress

The allowable shear stress is given as \(76 ext{ MPa}\). The maximum permissible load based on shear is \( P = ext{shear stress} imes A \), therefore \( P_{ ext{shear}} = 76 imes 10^6 imes 0.0025 = 190,000 ext{ N}\).
05

Identify Maximum Permissible Load

Compare the loads permissible in tension and shear. Since the maximum permissible load is governed by the lower of these two values, \( P_{ ext{max}} = ext{min}(312,500 ext{ N}, 190,000 ext{ N}) = 190,000 ext{ N}\).

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Key Concepts

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

Tensile Stress
Tensile stress occurs when a material is subjected to a force that attempts to stretch it. Imagine a steel bar being pulled at both ends; that pull is what we call tensile stress. The material's ability to withstand such stretching depends on its strength.To calculate tensile stress, use the formula:\[\text{Tensile Stress} = \frac{\text{Force}}{\text{Cross-sectional Area}} \]This equation tells us that as the force increases or the area decreases, the tensile stress increases. That's why a thick rope holds more weight than a thin one—it has a greater cross-sectional area.For example, in our steel bar scenario, if the tensile force is within the permissible limit of the material (in this case, 125 MPa), it will not deform or break. When designing or analyzing structures, engineers must ensure tensile stress does not exceed the material's allowable limit to prevent failure.
Shear Stress
Shear stress arises when a force is applied parallel or tangential to a surface. Think of it like using a pair of scissors; the blades apply a shear force to the paper. The formula for shear stress is similar to that of tensile stress:\[\text{Shear Stress} = \frac{\text{Force}}{\text{Cross-sectional Area}} \]When a material experiences shear stress, one part of it slides over another. It's crucial to evaluate shear stress carefully, as it can lead to different types of failure compared to tensile stress. In our exercise, the allowable shear stress for the steel bar is 76 MPa, which is less than the tensile stress capacity. This indicates that the material can withstand more pulling apart than sliding motion across its surface. Thus, designers must pay attention to shear stress, especially since it often dictates the maximum allowable load under certain conditions.
Cross-sectional Area
Cross-sectional area is a fundamental concept in understanding how materials react to loads. It is the area of the specific section of the material where the force is applied. Imagine cutting a loaf of bread into slices; each slice has its own cross-sectional area—it’s where you'd apply butter. The calculation of cross-sectional area depends on the shape of the object:
  • For a square or rectangle, multiply the base by the height.
  • For a circle, use the formula \(\pi r^2\).
In our problem, the steel bar is square-shaped with sides of 50 mm. Thus, the area is calculated as:\[50 \text{ mm} \times 50 \text{ mm} = 2500 \text{ mm}^2\]This area was then converted to square meters, resulting in 0.0025 m², which suits our stress calculations. A larger cross-sectional area means that the load applied to the material is more distributed, resulting in lower stress for the same amount of force. Conversely, a smaller area increases stress, making it more likely for the material to fail under the same force. That's why calculating the cross-sectional area is a critical step in engineering and design.

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Most popular questions from this chapter

A bar \(A B\) of length \(L\) is held between rigid supports and heated nonuniformly in such a manner that the temperature increase \(\Delta T\) at distance \(x\) from end \(A\) is given by the expression \(\Delta T=\Delta T_{B} \mathrm{x}^{3} / L^{3},\) where \(\Delta T_{B}\) is the increase in temperature at end \(B\) of the bar (see figure part a). (a) Derive a formula for the compressive stress \(\sigma_{c}\) in the bar. (Assume that the material has modulus of elasticity \(E\) and coefficient of thermal expansion \(\alpha\) ) (b) Now modify the formula in part (a) if the rigid support at \(A\) is replaced by an elastic support at \(A\) having a spring constant \(k\) (see figure part b). Assume that only bar \(A B\) is subject to the temperature increase.

A steel cable with nominal diameter \(25 \mathrm{mm}\) (see Table \(2-1\) ) is used in a construction yard to lift a bridge section weighing \(38 \mathrm{kN},\) as shown in the figure. The cable has an effective modulus of elasticity \(E=140 \mathrm{GPa}\) (a) If the cable is \(14 \mathrm{m}\) long, how much will it stretch when the load is picked up? (b) If the cable is rated for a maximum load of \(70 \mathrm{kN}\) what is the factor of safety with respect to failure of the cable?

A flat bar of width \(b\) and thickness \(t\) has a hole of diameter \(d\) drilled through it (see figure). The hole may have any diameter that will fit within the bar. What is the maximum permissible tensile load \(P_{\max }\) if the allowable tensile stress in the material is \(\sigma_{t} ?\)

A circular steel rod \(A B\) (diameter \(d_{1}=15 \mathrm{mm}\), length \(L_{1}=1100 \mathrm{mm}\) ) has a bronze sleeve (outer diameter \(d_{2}=\) \(21 \mathrm{mm},\) length \(L_{2}=400 \mathrm{mm}\) ) shrunk onto it so that the two parts are securely bonded (see figure) Calculate the total elongation \(\delta\) of the steel bar due to a temperature rise \(\Delta T=350^{\circ} \mathrm{C}\). (Material properties are as follows: for steel, \(E_{s}=210 \mathrm{GPa}\) and \(\alpha_{s}=12 \times 10^{-6 /^{\circ} \mathrm{C}}\) for bronze, \(E_{b}=110 \mathrm{GPa}\) and \(\alpha_{b}=20 \times 10^{-6 /^{\circ} \mathrm{C}}\)

A uniformly tapered tube \(A B\) of circular cross section and length \(L\) is shown in the figure. The average diameters at the ends are \(d_{A}\) and \(d_{B}=2 d_{A}\). Assume \(E\) is constant. Find the elongation \(\delta\) of the tube when it is subjected to loads \(P\) acting at the ends. Use the following numerical data: \(d_{A}=35 \mathrm{mm}, L=300 \mathrm{mm}, E=2.1 \mathrm{GPa}\) \(P=25 \mathrm{kN}\). Consider the following cases: (a) A hole of constant diameter \(d_{A}\) is drilled from \(B\) toward \(A\) to form a hollow section of length \(x=L / 2\) (b) A hole of variable diameter \(d(x)\) is drilled from \(B\) toward \(A\) to form a hollow section of length \(x=L / 2\) and constant thickness \(t=d_{A} / 20\)

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