91Ó°ÊÓ

A post having a hollow, circular cross section supports a load \(P=3.2 \mathrm{kN}\) acting at the end of an arm that is \(b=1.5 \mathrm{m}\) long (see figure). The height of the post is \(L=9 \mathrm{m},\) and its section modulus is \(S=2.65 \times 10^{5} \mathrm{mm}^{3}\) Assume that the outer radius of the post is \(r_{2}=123 \mathrm{mm}\) and the inner radius is \(r_{1}=117 \mathrm{mm}\) (a) Calculate the maximum tensile stress \(\sigma_{\max }\) and maximum in-plane shear stress \(\tau_{\max }\) at point \(A\) on the outer surface of the post along the \(x\) axis due to the load \(P .\) Load \(P\) acts at \(B\) along line \(B C\) (b) If the maximum tensile stress and maximum inplane shear stress at point \(A\) are limited to \(90 \mathrm{MPa}\) and \(38 \mathrm{MPa}\), respectively, what is the largest permissible value of the load \(P ?\)

Step by step solution

01

Calculate Maximum Bending Moment

The maximum bending moment at point A due to load P is calculated using the formula: \[ M = P \times b \]Substitute the given values:\[ M = 3.2 \text{kN} \times 1.5 \text{m} \] Convert kN to N: \(3.2 \times 1000 = 3200 \text{N}\)So, \(M = 3200 \times 1.5 = 4800 \text{Nm}\) or \(4800 \times 1000 = 4800000 \text{Nmm}\).

02

Compute Maximum Tensile Stress

The maximum tensile stress at point A is calculated with the bending stress formula:\[ \sigma_{\max} = \frac{M}{S} \]Substitute \(M = 4800000 \text{ Nmm}\) and \(S = 2.65 \times 10^5 \text{ mm}^3 \):\[ \sigma_{\max} = \frac{4800000}{2.65 \times 10^5} \approx 18.11 \text{ MPa} \]

03

Calculate Torsional Moment and Shear Stress

To find the in-plane shear stress, calculate the torsional moment. Since the post is only acted by bending moment in this problem, the torsional moment is zero. Thus:\[ \tau_{\max} = 0 \text{ MPa} \]

04

Solve for Maximum Load

The problem asks for the maximum permissible load given stress limits. Let's first solve for \(P\) under tensile stress limit.\[ \sigma_{\max,limit} = 90 \text{ MPa} \]Using \(\sigma_{\max} = \frac{M}{S}\):\[ 90 = \frac{Pb}{S} \]Reform the equation for \(P:\)\[ P = \frac{90 \times S}{b} \]Substitute the values \( S = 2.65 \times 10^5 \text{ mm}^3\) and \(b = 1.5 \times 1000 \text{ mm}\)\[ P = \frac{90 \times 2.65 \times 10^5}{1500} \approx 15.9 \text{ kN} \]

05

Final Step: Evaluate Under Shear Stress Limit (Optional)

The second condition of the problem restricts by shear stress, but shear due to bending is zero (as bending load doesn't produce shear). Since the maximum shear stress was found to be zero, the shear condition is automatically satisfied.Thus, the limiting factor in this problem is the tensile stress condition. Therefore, the final allowable maximum load \( P \) is determined to be approximately \(15.9 \text{ kN} \).

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Key Concepts

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

Tensile Stress

Tensile stress, often symbolized by \( \sigma \), is a measure of how much force is spread out over a given area. It's the force that tries to stretch or elongate a material. In simple terms, when you pull a rubber band, the stress you apply is tensile stress. This concept is crucial for understanding material limits and ensuring structures do not fail under loading.

Mathematically, tensile stress is determined by the formula:

• \( \sigma = \frac{F}{A} \)

Where:

• \( F \) is the force applied (in Newtons)
• \( A \) is the cross-sectional area (in square meters or square millimeters)

In the context of bending, the tensile stress in a beam due to bending moment is calculated using the bend stress formula, as shown:

• \( \sigma_{\text{max}} = \frac{M}{S} \)

Here, \( M \) is the bending moment applied, and \( S \) is the section modulus. Understanding tensile stress helps in determining whether a material can withstand specific loads without stretching or breaking.

Shear Stress

Shear stress, usually denoted by \( \tau \), acts parallel to the surface, trying to slide layers of the material past one another. Imagine cutting a block of cheese; the force you apply causes shear stress as the knife pushes against the cheese layers. In many structures, shear stress can arise due to torsion or similar forces acting transversely.

The mathematical expression for shear stress is:

• \( \tau = \frac{F_{\text{shear}}}{A} \)

Where:

• \( F_{\text{shear}} \) is the transverse force (in Newtons)
• \( A \) is the area over which the force is applied

For the original problem, since we only deal with bending loads, the torsional effects (which contribute to shear) are considered zero. Therefore, the maximum in-plane shear stress \( \tau_{\text{max}} \) is zero MPa in this scenario. In other applications, understanding shear stress is vital for checking how forces might cause parts to 'slide' or 'break' internally.

Section Modulus

The section modulus, represented by \( S \), is a geometric property of a cross-section that reflects its ability to resist bending. It is a critical factor in designing beams and structures. The larger the section modulus, the more resistant the beam is to bending and less likely it is to deform under stress.

It's calculated by the formula:

• \( S = \frac{I}{c} \)

Where:

• \( I \) is the moment of inertia (a measure of an object's resistance to changes to its rotation)
• \( c \) is the distance from the neutral axis to the outermost fiber (in meters or millimeters)

In practical terms, for a hollow circular section like in the exercise, the presence of both an inner and outer radius ensures optimal use of material while maintaining structural integrity. Knowing the section modulus helps engineers determine how much a beam will bend when subjected to certain forces, and select materials accordingly to prevent any failure.