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(I) A sign (mass \(1700 \mathrm{~kg}\) ) hangs from the end of a vertical steel girder with a cross-sectional area of \(0.012 \mathrm{~m}^{2} .\) (a) What is the stress within the girder? (b) What is the strain on the girder? \((c)\) If the girder is \(9.50 \mathrm{~m}\) long, how much is it lengthened? (Ignore the mass of the girder itself.)

Short Answer

Expert verified
(a) Stress: 1,389,750 N/m虏. (b) Strain: 6.95 脳 10鈦烩伓. (c) Length change: 0.066 mm.

Step by step solution

01

Calculate the Force on the Girder

The force experienced by the girder is due to the weight of the sign. We can calculate this using the formula for weight, where weight is mass times the acceleration due to gravity (assuming \( g = 9.81 \text{ m/s}^2 \)): \[F = m \cdot g = 1700 \text{ kg} \cdot 9.81 \text{ m/s}^2 = 16677 \text{ N}\]
02

Calculate the Stress on the Girder

Stress is defined as the force applied per unit area. Using the force calculated in Step 1 and the cross-sectional area given, we can find stress by:\[\text{Stress} = \frac{F}{A} = \frac{16677 \text{ N}}{0.012 \text{ m}^2} = 1,389,750 \text{ N/m}^2 \]
03

Calculate the Strain on the Girder

Strain is the ratio of the change in length to the original length. Before calculating the actual strain, note that it is dimensionless:\[\text{Strain} = \frac{\Delta L}{L} = \frac{\text{Stress}}{E}\]where \(E\) is the Young's Modulus for steel (approximately \( 2 \times 10^{11} \text{ N/m}^2 \)). So:\[\text{Strain} = \frac{1,389,750 \text{ N/m}^2}{2 \times 10^{11} \text{ N/m}^2} \approx 6.95 \times 10^{-6}\]
04

Calculate the Change in Length of the Girder

Finally, use the relation for strain to find \(\Delta L\), the change in length, where strain \(\epsilon = \frac{\Delta L}{L}\):\[\Delta L = \epsilon \cdot L = 6.95 \times 10^{-6} \cdot 9.50 \text{ m} = 6.60 \times 10^{-5} \text{ m}\]So the girder is lengthened by approximately 0.066 mm.

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Key Concepts

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

Young's Modulus
Young's Modulus is a fundamental property of materials that measures their stiffness. Consider it a way to determine how much a material will stretch or compress when a force is applied. The value is incredibly important in engineering and construction as it helps predict how structures will behave under stress. For steel, which is the material of our girder, Young's Modulus is approximately \( 2 \times 10^{11} \text{ N/m}^2 \). This large value tells us that steel is very stiff, so it won't stretch much under normal conditions.
To calculate strain, which is essentially how much a material deforms, we use the formula \( \text{Strain} = \frac{\text{Stress}}{E} \). Here, \( \text{Stress} \) is the force applied per unit area, and \( E \) is Young's Modulus. A higher Young's Modulus means less deformation under the same stress, highlighting the material's rigidity.
In summary, Young's Modulus is crucial in determining how much a material changes shape when a stress is applied. It's a key factor in ensuring that structures like bridges, buildings, and girders remain safe and reliable.
Cross-sectional Area
The cross-sectional area is the area of the cut surface of a 3D object when it's sliced, typically perpendicular to its length. Imagine cutting through a girder like slicing a loaf of bread; the face you see is the cross-sectional area. This value is crucial when calculating stress, as stress is the force distributed over the cross-sectional area
The formula for stress is \( \text{Stress} = \frac{F}{A} \), where \( F \) is force and \( A \) is the cross-sectional area. In the exercise, we see a girder with a cross-sectional area of \(0.012 \text{ m}^2\). This small area means that the concentrated force from the hanging sign will result in considerable stress. A larger cross-sectional area would distribute the same force more widely, resulting in less stress.
Understanding the cross-sectional area is vital in construction and engineering design. It affects how much load a material can support without failing. By selecting appropriate cross-sectional areas, engineers ensure structures can handle anticipated forces.
Force due to Gravity
Force due to gravity, commonly referred to as weight, is a force exerted by the Earth pulling objects downwards. It's calculated by multiplying an object's mass by the acceleration due to gravity (\(9.81 \text{ m/s}^2\) on Earth).
In the given exercise, the girder experiences a force owing to the weight of the sign it holds. The calculation \(F = 1700 \text{ kg} \times 9.81 \text{ m/s}^2 = 16677 \text{ N}\) shows the force due to gravity acting on the girder. This force is critical for calculating stress and, subsequently, the strain and potential deformation of the girder.
Understanding the force of gravity is crucial for accurately predicting how structures will hold up against their own weight, and additional loads, and ensure they remain stable and safe.

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Most popular questions from this chapter

(I) A marble column of cross-sectional area 1.4 \(\mathrm{m}^{2}\) supports a mass of \(25,000 \mathrm{kg}\) . (a) What is the stress within the column? \((b)\) What is the strain?

A \(23.0-\mathrm{kg}\) backpack is suspended midway between two trees by a light cord as in Fig. \(12-50 .\) A bear grabs the backpack and pulls vertically downward with a constant force, so that each section of cord makes an angle of \(27^{\circ}\) below the horizontal. Initially, without the bear pulling, the angle was \(15^{\circ}\); the tension in the cord with the bear pulling is double what it was when he was not. Calculate the force the bear is exerting on the backpack.

A uniform ladder of mass \(m\) and length \(\ell\) leans at an angle \(\theta\) against a wall, Fig. \(12-101 .\) The coefficients of static friction between ladder-ground and ladder-wall are \(\mu_{\mathrm{G}}\) and \(\mu_{\mathrm{W}},\) respectively. The ladder will be on the verge of slipping when both the static friction forces due to the ground and due to the wall take on their maximum values. (a) Show that the ladder will be stable if \(\theta \geq \theta_{\min },\) where the minimum angle \(\theta_{\min }\) is given by $$ \tan \theta_{\min }=\frac{1}{2 \mu_{\mathrm{G}}}\left(1-\mu_{\mathrm{G}} \mu_{\mathrm{W}}\right) $$ (b) "Leaning ladder problems" are often analyzed under the seemingly unrealistic assumption that the wall is frictionless (see Example \(12-6\) ). You wish to investigate the magnitude of error introduced by modeling the wall as frictionless, if in reality it is frictional. Using the relation found in part \((a)\), calculate the true value of \(\theta_{\min }\) for a frictional wall, taking \(\mu_{\mathrm{G}}=\mu_{\mathrm{W}}=0.40 .\) Then, determine the approximate value of \(\theta_{\min }\) for the "frictionless wall" model by taking \(\mu_{\mathrm{G}}=0.40\) and \(\mu_{\mathrm{W}}=0 .\) Finally, determine the percent deviation of the approximate value of \(\theta_{\min }\) from its true value.

A uniform \(6.0-\mathrm{m}\) -long ladder of mass 16.0 \(\mathrm{kg}\) leans against a smooth wall (so the force exerted by the wall, \(\overline{\mathbf{F}}_{\mathrm{W}},\) is perpendicular to the wall). The ladder makes an angle of \(20.0^{\circ}\) with the vertical wall, and the ground is rough. Determine the coefficient of static friction at the base of the ladder if the ladder is not to slip when a 76.0 -kg person stands three-fourths of the way up the ladder.

(III) A uniform ladder of mass \(m\) and length \(\ell\) leans at an angle \(\theta\) against a frictionless wall, Fig. \(12-67 .\) If the coefficient of static friction between the ladder and the ground is \(\mu_{\mathrm{s}},\) determine a formula for the minimum angle at which the ladder will not slip.

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