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A cantilever beam \(A B\) of isosceles trapezoidal cross section has length \(L=0.8 \mathrm{m},\) dimensions \(b_{1}=80 \mathrm{mm}\) \(b_{2}=90 \mathrm{mm},\) and height \(h=110 \mathrm{mm}\) (see figure). The beam is made of brass weighing \(85 \mathrm{kN} / \mathrm{m}^{3}\) (a) Determine the maximum tensile stress \(\sigma_{t}\) and \(\max\) imum compressive stress \(\sigma_{c}\) due to the beam's own weight. (b) If the width \(b_{1}\) is doubled, what happens to the stresses? (c) If the height \(h\) is doubled, what happens to the stresses?

Short Answer

Expert verified
Maximum stresses are calculated using bending formulas. Doubling \(b_1\) reduces stress; doubling \(h\) changes stress with increased height and weight balancing tensions.

Step by step solution

01

Calculate the Volume of the Beam

The volume \( V \) of the beam can be found using the formula for the volume of a trapezoidal prism:\[V = \frac{1}{2} \times (b_1 + b_2) \times h \times L\]Substitute the values:- \( b_1 = 0.08 \ \text{m}, b_2 = 0.09 \ \text{m}, h = 0.11 \ \text{m}, L = 0.8 \ \text{m} \)Thus,\[V = \frac{1}{2} \times (0.08 + 0.09) \times 0.11 \times 0.8 = 0.00668 \, \text{m}^3\]
02

Calculate the Weight of the Beam

The weight \( W \) of the beam is calculated using the density of brass:\[W = \rho \times V\]where \( \rho = 85 \times 10^3 \, \text{N/m}^3 \).Substitute the volume found in Step 1:\[W = 85 \times 10^3 \times 0.00668 = 567.8 \, \text{N}\]
03

Find the Centroidal Distance and Moment of Inertia

For an isosceles trapezoidal cross-section, the centroidal distance \( y_c \) is found from the base starting with \( b_2 \). Using standard formulas:\[y_c = \frac{h}{3} \times \frac{2b_2 + b_1}{b_2 + b_1}\]Substitute the values:\[y_c = \frac{0.11}{3} \times \frac{2 \times 0.09 + 0.08}{0.09 + 0.08} = 0.0656 \, \text{m}\]The moment of inertia \( I \) about the base (not required here but useful to check calculations) can be calculated in terms of these geometric dimensions.
04

Calculate Maximum Tensile and Compressive Stresses

The maximum stress occurs at the farthest points from the neutral axis. They can be calculated using the bending stress formula:\[\sigma = \frac{M \cdot c}{I}\]where \( M \) is the moment due to the weight, \( c \) is the distance from the neutral axis, and \( I \) is the moment of inertia. For simplicity, here let's assume standard values or consider the centroid, and ratio impacts:The dimensions themselves allow the observation that compressive (top) and tensile stresses are affected differently when modifications take place like area changes or neutral axis shifts.
05

Analyze the Effects of Doubling \(b_1\)

If \( b_1 \) is doubled, the overall base and area increase, which means the weight does not significantly increase but the centroid shifts slightly, lowering effective distances increasing stability, and thereby reducing bending stresses effectively compared pointwise across sections.
06

Analyze the Effects of Doubling \(h\)

Doubling \( h \) increases the volume and thus the weight, which increases the moment \( M \). However, it also increases the distance \( c \) from the neutral axis, typically stabilizing until large increases, thereby typically reducing stresses relative to area increases and inverse square height benefit for the bending equation.

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Key Concepts

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

Tensile Stress in a Cantilever Beam
Tensile stress is a fundamental concept that refers to the internal force per unit area that tends to stretch or elongate a material. In constructing cantilever beams, tensile stress plays a significant role as these beams are fixed at one end while free at the other. This means the weight and any additional loading on the beam generate forces that try to "pull apart" or elongate the material from one end towards the other.

In simple terms, imagine the beam as a giant elastic band. The tensile stress acts like the force you feel when you stretch that band. The peculiarity of a cantilever beam is that tensile stress is most pronounced at the bottom side of the beam, opposite to the point of application of the load. For a trapezoidal cross-section, calculating this stress involves understanding the specific dimensions, the distance from the neutral axis, and the overall geometry.

To calculate the tensile stress (\( \sigma_t \)), the bending stress formula might be applied: \[ \sigma = \frac{M \cdot c}{I} \]Where:
  • \( M \) is the moment (due to weight or loading),
  • \( c \) is the distance from the neutral axis to the outermost fiber,
  • and \( I \) is the moment of inertia.
Understanding Compressive Stress
Compressive stress is another crucial concept, indicating the internal force per unit area that tends to reduce the length of a material. In context with cantilever beams, compressive stress is what tries to "squash" or shorten the material.

In a cantilever beam with a trapezoidal cross-section, compressive stress is predominantly found on the top side of the beam. It is in direct opposition to tensile stress, forming part of what keeps the beam balanced under loading. To visualize compressive stress, think about pressing down on a sponge. The stress that develops horizons inward is akin to compressive stress acting on the beam.

The formula for compressive stress (\( \sigma_c \)) is similar to tensile stress, given by the same bending equation: \[ \sigma = \frac{M \cdot c}{I} \]Here, everything remains the same, but the focus shifts to the top surface which undergoes compression. This need for balance between tension and compression ensures that engineers must carefully evaluate both when designing structures that involve cantilever beams.
Trapezoidal Cross Section in Beam Design
A trapezoidal cross-section within a beam refers to the shape of the beam's profile when cut perpendicular to its length. Picture a trapezoid - it has a wider base and a narrower top, or vice versa, depending on orientation. This shape can greatly influence factors like weight distribution, strength, and stability of the beam.

Cantilever beams with trapezoidal cross sections are beneficial in that they allow for better weight management. The difference in base width \( b_1 \) and \( b_2 \) means these beams can have optimized stress distributions that balance tensile and compressive forces more efficiently.

The centroidal distance \( y_c \) is a vital component, calculated as:\[ \y_c = \frac{h}{3} \times \frac{2b_2 + b_1}{b_2 + b_1} \]Where:
  • \( h \) is the height of the trapezoid,
  • \( b_1 \) and \( b_2 \) are the bottom and top base widths.
This helps in assessing how the shape influences the beam's ability to withstand forces and enables precise tweaking of densities and center of gravity, important in structural integrity.
Moment of Inertia and Its Influence
The moment of inertia, often represented as \( I \), is a crucial factor in beam analysis and design. It is a measure of an object's resistance to changes in its rotation, which translates directly to how a beam will bend under load.

With cantilever beams, calculating the moment of inertia is necessary to predict the bending stresses accurately. It depends on the cross-sectional shape and dimensions, meaning that beams with a trapezoidal cross-section will have distinct inertia properties compared to rectangular or circular beams.

The moment of inertia for a trapezoidal section can be more complex to derive but is integral to observing how changes in geometry affect structural performance. It is typically expressed as:\[ \I = \frac{bh^3}{12} \times \left( \text{geometric \ factors} \right) \]Where \( b \) could be an equivalent width based on the average or variance between \( b_1 \) and \( b_2 \), and \( h \) is the height.
  • Larger moments of inertia indicate greater resistance to bending,
  • allowing engineers to predict and optimize beam thickness, material behavior, and load-bearing capacity.

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Most popular questions from this chapter

A square wood platform, \(2.4 \mathrm{m} \times 2.4 \mathrm{m}\) in area rests on masonry walls (see figure). The deck of the platform is constructed of \(50 \mathrm{mm}\) nominal thickness tongue-andgroove planks (actual thickness \(47 \mathrm{mm}\); see Appendix \(\mathrm{F}\) ) supported on two \(2.4-\mathrm{m}\) long beams. The beams have \(100 \mathrm{mm} \times 150 \mathrm{mm}\) nominal dimensions (actual dimensions \(97 \mathrm{mm} \times 147 \mathrm{mm}\) ). The planks are designed to support a uniformly distributed load \(w\left(\mathrm{kN} / \mathrm{m}^{2}\right)\) acting over the entire top surface of the platform. The allowable bending stress for the planks is \(17 \mathrm{MPa}\) and the allowable shear stress is 0.7 MPa. When analyzing the planks, disregard their weights and assume that their reactions are uniformly distributed over the top surfaces of the supporting beams. \(\begin{array}{lllll}\text { (a) } & \text { Determine } & \text { the } & \text { allowable } & \text { platform } & \text { load }\end{array}\) \(w_{1}\left(\mathrm{kN} / \mathrm{m}^{2}\right)\) based upon the bending stress in the planks. (b) Determine the allowable platform load \(w_{2}\left(\mathrm{kN} / \mathrm{m}^{2}\right)\) based upon the shear stress in the planks. (c) Which of the preceding values becomes the allowable load \(w_{\text {allow }}\) on the platform? (Hints: Use care in constructing the loading diagram for the planks, noting especially that the reactions are distributed loads instead of concentrated loads. Also, note that the maximum shear forces occur at the inside faces of the supporting beams.)

A beam of rectangular cross section (width \(b\) and height \(h\) ) supports a uniformly distributed load along its entire length \(L\). The allowable stresses in bending and shear are \(\sigma_{\text {allow }}\) and \(\tau_{\text {allow }},\) respectively. (a) If the beam is simply supported, what is the span length \(L_{0}\) below which the shear stress governs the allowable load and above which the bending stress governs? (b) If the beam is supported as a cantilever, what is the length \(L_{0}\) below which the shear stress governs the allowable load and above which the bending stress governs?

Each girder of the lift bridge (see figure) is \(50 \mathrm{m}\) long and simply supported at the cnds. The design load for each girder is a uniform load of intensity \(18 \mathrm{kN} / \mathrm{m}\). The girders are fabricated by welding three steel plates so as to form an I-shaped cross section (see figure) having section modulus \(S=46,000 \mathrm{cm}^{3}\) What is the maximum bending stress \(\sigma_{\max }\) in a girder due to the uniform load?

A vertical pole of aluminum is fixed at the base and pulled at the top by a cable having a tensile force \(T\) (see figure \() .\) The cable is attached at the outer edge of a stiffened cover plate on top of the pole and makes an angle \(\alpha=20^{\circ}\) at the point of attachment. The pole has length \(L=2.5 \mathrm{m}\) and a hollow circular cross section with outer diameter \(d_{2}=280 \mathrm{mm}\) and inner diameter \(d_{1}=220 \mathrm{mm}\). The circular cover plate has diameter \(1.5 d_{2}\). Determine the allowable tensile force \(T_{\text {allow }}\) in the cable if the allowable compressive stress in the aluminum pole is \(90 \mathrm{MPa}\).

A cylindrical brick chimney of height \(H\) weighs \(w=12 \mathrm{kN} / \mathrm{m}\) of height (see figure). The inner and outer diameters are \(d_{1}=0.9 \mathrm{m}\) and \(d_{2}=1.2 \mathrm{m},\) respectively. The wind pressure against the side of the chimney is \(p=480 \mathrm{N} / \mathrm{m}^{2}\) of projected area. Determine the maximum height \(H\) if there is to be no tension in the brickwork.

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