December 2008

Solution to Problem 255 Statically Indeterminate

Problem 255
Shown in Fig. P-255 is a section through a balcony. The total uniform load of 600 kN is supported by three rods of the same area and material. Compute the load in each rod. Assume the floor to be rigid, but note that it does not necessarily remain horizontal.
 

Figure P-255

 

Derivation of Quadratic Formula

The roots of a quadratic equation ax2 + bx + c = 0 is given by the quadratic formula
 

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

 

The derivation of this formula can be outlined as follows:

  1.   Divide both sides of the equation ax2 + bx + c = 0 by a.
  2.   Transpose the quantity c/a to the right side of the equation.
  3.   Complete the square by adding b2 / 4a2 to both sides of the equation.
  4.   Factor the left side and combine the right side.
  5.   Extract the square-root of both sides of the equation.
  6.   Solve for x by transporting the quantity b / 2a to the right side of the equation.
  7.   Combine the right side of the equation to get the quadratic formula.

See the derivation below.
 

Sum and Product of Roots

The quadratic formula
 

$x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

 

give the roots of a quadratic equation which may be real or imaginary. The ± sign in the radical indicates that
 

$x_1 = \dfrac{-b + \sqrt{b^2-4ac}}{2a}$   and   $x_2 = \dfrac{-b - \sqrt{b^2-4ac}}{2a}$

 

where x1 and x2 are the roots of the quadratic equation ax2 + bx + c = 0. The sum of roots x1 + x2 and the product of roots x1·x2 are common to problems involving quadratic equation.
 

Derivation of Pythagorean Theorem

Derivation of Pythagorean Theorem | Plane Trigonometry

Pythagorean Theorem
In any right triangle, the sum of the square of the two perpendicular sides is equal to the square of the longest side. For a right triangle with legs measures $a$ and $b$ and length of hypotenuse $c$, the theorem can be expressed in the form
 

$a^2 + b^2 = c^2$

 

Derivation of Sum of Arithmetic Progression

Arithmetic Progression, AP
Definition

Arithmetic Progression (also called arithmetic sequence), is a sequence of numbers such that the difference between any two consecutive terms is constant. Each term therefore in an arithmetic progression will increase or decrease at a constant value called the common difference, d.
 

Examples of arithmetic progression are:

  • 2, 5, 8, 11,... common difference = 3
  • 23, 19, 15, 11,... common difference = -4

Derivation of Sum of Finite and Infinite Geometric Progression

Geometric Progression, GP
Geometric progression (also known as geometric sequence) is a sequence of numbers where the ratio of any two adjacent terms is constant. The constant ratio is called the common ratio, r of geometric progression. Each term therefore in geometric progression is found by multiplying the previous one by r.
 

Eaxamples of GP:

  • 3, 6, 12, 24, … is a geometric progression with r = 2
  • 10, -5, 2.5, -1.25, … is a geometric progression with r = -1/2

 

Solution to Problem 256 Statically Indeterminate

Problem 256
Three rods, each of area 250 mm2, jointly support a 7.5 kN load, as shown in Fig. P-256. Assuming that there was no slack or stress in the rods before the load was applied, find the stress in each rod. Use Est = 200 GPa and Ebr = 83 GPa.
 

Figure P-256

 

Solution to Problem 257 Statically Indeterminate

Problem 257
Three bars AB, AC, and AD are pinned together as shown in Fig. P-257. Initially, the assembly is stress free. Horizontal movement of the joint at A is prevented by a short horizontal strut AE. Calculate the stress in each bar and the force in the strut AE when the assembly is used to support the load W = 10 kips. For each steel bar, A = 0.3 in.2 and E = 29 × 106 psi. For the aluminum bar, A = 0.6 in.2 and E = 10 × 106 psi.
 

Figure 257

 

Thermal Stress

Temperature changes cause the body to expand or contract. The amount δT, is given by
 

$\delta_T = \alpha L \, (T_f \, - \, T_i) = \alpha L \, \Delta T$

where α is the coefficient of thermal expansion in m/m°C, L is the length in meter, Ti and Tf are the initial and final temperatures, respectively in °C. For steel, α = 11.25 × 10-6 m/m°C.
 

If temperature deformation is permitted to occur freely, no load or stress will be induced in the structure. In some cases where temperature deformation is not permitted, an internal stress is created. The internal stress created is termed as thermal stress.
 

Relationship Between Arithmetic Mean, Harmonic Mean, and Geometric Mean of Two Numbers

For two numbers x and y, let x, a, y be a sequence of three numbers. If x, a, y is an arithmetic progression then 'a' is called arithmetic mean. If x, a, y is a geometric progression then 'a' is called geometric mean. If x, a, y form a harmonic progression then 'a' is called harmonic mean.
 

Let AM = arithmetic mean, GM = geometric mean, and HM = harmonic mean. The relationship between the three is given by the formula
 

$AM \times HM = GM^2$

 

Below is the derivation of this relationship.
 

Solution to Problem 261 Thermal Stress

Problem 261
A steel rod with a cross-sectional area of 0.25 in2 is stretched between two fixed points. The tensile load at 70°F is 1200 lb. What will be the stress at 0°F? At what temperature will the stress be zero? Assume α = 6.5 × 10-6 in/(in·°F) and E = 29 × 106 psi.
 

About

Mathalino.com is a compilation of solved engineering problems from different sources. This website will target the following engineering subjects:

  1. College Algebra
  2. Plane and Spherical Trigonometry
  3. Plane, Solid, and Analytic Geometry
  4. Differential and Integral Calculus
  5. Differential Equations
  6. Advance Engineering Mathematics
  7. Engineering Economy
  8. Engineering Mechanics
  9. Strength of Materials

Problems will be taken from different sources such as past board exams, text books, problems from different competitions, problems emailed by readers, and problems created by the contributors of this website.
 

Solution to Problem 262 Thermal Stress

Problem 262
A steel rod is stretched between two rigid walls and carries a tensile load of 5000 N at 20°C. If the allowable stress is not to exceed 130 MPa at -20°C, what is the minimum diameter of the rod? Assume α = 11.7 µm/(m·°C) and E = 200 GPa.
 

Solution to Problem 263 Thermal Stress

Problem 263
Steel railroad reels 10 m long are laid with a clearance of 3 mm at a temperature of 15°C. At what temperature will the rails just touch? What stress would be induced in the rails at that temperature if there were no initial clearance? Assume α = 11.7 µm/(m·°C) and E = 200 GPa.
 

Solution to Problem 264 Thermal Stress

Problem 264
A steel rod 3 feet long with a cross-sectional area of 0.25 in.2 is stretched between two fixed points. The tensile force is 1200 lb at 40°F. Using E = 29 × 106 psi and α = 6.5 × 10-6 in./(in.·°F), calculate (a) the temperature at which the stress in the bar will be 10 ksi; and (b) the temperature at which the stress will be zero.
 

Solution to Problem 265 Thermal Stress

Problem 265
A bronze bar 3 m long with a cross sectional area of 320 mm2 is placed between two rigid walls as shown in Fig. P-265. At a temperature of -20°C, the gap Δ = 2.5 mm. Find the temperature at which the compressive stress in the bar will be 35 MPa. Use α = 18.0 × 10-6 m/(m·°C) and E = 80 GPa.
 

Figure P-265

 

Solution to Problem 266 Thermal Stress

Problem 266
Calculate the increase in stress for each segment of the compound bar shown in Fig. P-266 if the temperature increases by 100°F. Assume that the supports are unyielding and that the bar is suitably braced against buckling.
 

Figure P-266

 

Solution to Problem 267 Thermal Stress

Problem 267
At a temperature of 80°C, a steel tire 12 mm thick and 90 mm wide that is to be shrunk onto a locomotive driving wheel 2 m in diameter just fits over the wheel, which is at a temperature of 25°C. Determine the contact pressure between the tire and wheel after the assembly cools to 25°C. Neglect the deformation of the wheel caused by the pressure of the tire. Assume α = 11.7 μm/(m·°C) and E = 200 GPa.
 

Solution to Problem 268 Thermal Stress

Problem 268
The rigid bar ABC in Fig. P-268 is pinned at B and attached to the two vertical rods. Initially, the bar is horizontal and the vertical rods are stress-free. Determine the stress in the aluminum rod if the temperature of the steel rod is decreased by 40°C. Neglect the weight of bar ABC.
 

268 Steel and aluminum rods

 

Solution to Problem 269 Thermal Stress

Problem 269
As shown in Fig. P-269, there is a gap between the aluminum bar and the rigid slab that is supported by two copper bars. At 10°C, Δ = 0.18 mm. Neglecting the mass of the slab, calculate the stress in each rod when the temperature in the assembly is increased to 95°C. For each copper bar, A = 500 mm2, E = 120 GPa, and α = 16.8 µm/(m·°C). For the aluminum bar, A = 400 mm2, E = 70 GPa, and α = 23.1 µm/(m·°C).
 

Figure P-269

 

Solution to Problem 270 Thermal Stress

Problem 270
A bronze sleeve is slipped over a steel bolt and held in place by a nut that is turned to produce an initial stress of 2000 psi in the bronze. For the steel bolt, A = 0.75 in2, E = 29 × 106 psi, and α = 6.5 × 10-6 in/(in·°F). For the bronze sleeve, A = 1.5 in2, E = 12 × 106 psi and α = 10.5 × 10-6 in/(in·°F). After a temperature rise of 100°F, find the final stress in each material.
 

Pages