Tag Archives: Mechanics

Homework 1 in MEE 322: Structural Mechanics | Two-Dimensional Stress Analysis


Problem 1

The state of stress at a point is given in the figure. Find τ and τxy directly using force equilibrium. Do NOT use the stress transformation equations.

Problem 2

The state of stress at a point is given in the figure. Find σ and τxy using stress transformation equations.

Problem 3

The state of stress at a point is given in the figure. Find the principal stresses, principal directions and the maximum shear stress using

(a) Eigenvalue problem approach

(b) Stress transformation equations

Problem 4

The shaft shown in the figure has a gear at B with a force of 2098 N in -y and 6456 N in +z applied at its tip. The force along z produces a torque that drives the component attached at C, which produces an equal and opposite torque to that produced at the gear as well as forces on the shaft of 6000 N along +y and +z. The bearing at A can be considered a spherical hinge, whereas the bearing at D can be considered a planar hinge in the y-z plane.

(b) Draw bending moment and torsion diagrams for the shaft and show diagrams of the cross-section of the shaft where the critical points occur, i.e., the locations where the maximum normal stresses due to bending and maximum shear stress due to torsion coincide. Indicate the internal reactions (bending and torsion moments) in this diagram, as well as the locations of the critical points.

(b) If the diameter of the shaft is 33 mm, find the stresses at the critical point and use them to find the maximum shear stress at that location as well as the maximum and minimum principal stresses. Note: the bending normal stress can be taken as σx, while the torsion shear stress can be assumed to be τxy for the effects of this calculation, all other stresses can be assumed to be zero.

(c) It is known that the material of the shaft is such that it will fail if the maximum shear stress reaches 300 MPa. Is the shaft safe? If so, calculate the factor of safety.


Purchase the solution to this homework now

Book-Covers-2

Homework 4 in MEE 322: Structural Mechanics

This is the complete solution manual to the problems in Homework 4 of ME322: Structural Mechanics. The 4 problems included in the manual are written above. The PDF document will be sent to your email address within 24 hours from the time of purchase. The email will be coming from homeworkhelp@engineering-math.org Do not forget to check on your spam folder. If you have not received your file in 24 hours, kindly send us an email at help@engineering-math.org

$20.00

Homework 3 in MEE 322 Structural Mechanics | Normal and Shear Stresses Under Combined Loading Part 2


Problem 1

Three forces act on the tip of a L-shaped rod with a cross-sectional radius of 0.5 in.

(a) Determine the normal and shear stress at points A and B and draw the stress cube at those points based on the given coordinate system.

(b) Determine the maximum normal stress on the cross-section and locate the point at which it occurs.

Problem 2

The simply supported solid shaft has a radius of 15 mm and is under static equilibrium. Pulley C has a diameter of 100 mm. The pulleys B and D have the same diameter as each other. The forces on pulley B are at an angle of 45 to the negative z-axis. The forces on pulley C and pulley D are in the z and -y direction. The shaft dimensions are in mm.

(a) Determine the maximum bending and torsional stresses in the shaft.

(b) Locate the point(s) on the cross-section where the bending stress is maximum.

Problem 3

The structural part of a setup to measure net belt tensions in pulleys is shown in the figure. The belt tensions at both sides of the pulley at B (radius 10 cm) are P and F=0.1*P along z and a reaction force is measured from the pulley at C (radius 2 cm), which is connected to a load cell at E with an axial member parallel to x. Pulleys are rigidly attached to rod AD, which is made with a ductile steel rod 60 cm long and 1.27 cm in diameter. Length AB=0.20 m, and length DC=0.15 m. There is a spherical hinge at A and a plane hinge at D. The latter constrains motion in the x-z plane only.

(a) Draw bending moment and torsion diagrams for this structure as functions of the unknown tension P and use them to draw a diagram of the critical section showing internal loads (bending and torsion moments) and the critical points.

(b) Use your results from part a to determine the maximum normal stress due to bending and the maximum shear stress due to torsion in terms of the unknown tension P. Calculate the maximum value that P can have if only bending stresses are considered (with σallow = 350 MPa) and then if only torsion stresses are considered (with τallow = 175 MPa).


Purchase the Complete Solution to this Homework Now

Book-Covers

Homework #2 in MEE 322 Structural Mechanics

This is the complete solution manual to the problems in Homework 3 of ME322: Structural Mechanics. The 3 problems included in the manual are written above. The PDF document will be sent to your email address within 24 hours from the time of purchase. The email will be coming from homeworkhelp@engineering-math.org Do not forget to check on your spam folder. If you have not received your file in 24 hours, kindly send us an email at help@engineering-math.org

$20.00

Homework 1 in MEE 322: Structural Mechanics | Internal Reactions in 3D and 2D Bending


Problem 1

The structure shown in Fig. 1.1 is made with a steel bar that has a rectangular cross-section 5 cm tall and 10 cm wide.

Fig 1.1. Frame supporting point and distributed load

Given the geometry, loads and supports in the structure draw the bending moment diagrams for segments A-C and D-E of the structure. Then, use the diagrams to find the critical section and calculate the maximum bending stress in segment AC.

Problem 2

The torsion rod with variable cross-section shown in Fig. 2.1 is clamped at A and carries point torques at B (4 kN.m), C (8 kN.m) and D (unknown value T) with the senses indicated in the figure. It is known that the diameter of AB is 25 mm, the diameter of BC is 50 mm and the diameter of CD is 20 mm. Furthermore, 6LAB = 6LBC = 5LCD = 1200 mm. If the shear modulus of the material is 80 GPa, find:

a) The value of T that would make the angle of twist at point C with respect to A equal to zero.

b) The maximum value of T that can applied with the sense shown such that failure does not occur for an allowable shear stress of 1.1 GPa. Can the condition from part a be achieved without failure?

Fig. 2.1: Torsion rod with variable cross-section.

Problem 3

The cantilever beam shown in Fig. 3.1 has a rectangular cross-section with h = 120 mm and b = 80 mm. The 12 kN and 10 kN forces act parallel to the x and z axis, respectively, and pass through the centroid of the beam cross-section at the locations they act. The 10 kN force acts at the free end of the cantilever, whereas the 12 kN force acts 250 mm away from the free end. The cross-section ABCD is 750 mm away from the free end.

a) Determine the magnitude and location of the maximum tensile and compressive bending stress at the cross-section ABCD and indicate the neutral axis on it.

b) If an additional force, F, is applied on the beam parallel to the z axis at a point 500 mm away from the free end, what should be its magnitude and direction to make the bending stress at point B zero?

Fig. 3.1: Cantilever beam with loads along two axes.

Purchase the Solution to this Homework Now

Homework 1 in MEE 322: Structural Mechanics

This is the complete solution manual to the problems in Homework 1 of ME322: Structural Mechanics. The 3 problems included in the manual are written above. The PDF document will be sent to your email address within 24 hours from the time of purchase. The email will be coming from homeworkhelp@engineering-math.org Do not forget to check on your spam folder. If you have not received your file in 24 hours, kindly send us an email at help@engineering-math.org

$20.00


Advertisements
Advertisements

Mechanics of Deformable Bodies


Hibbeler Statics 14E P1.11 — Representing Measurements with SI units Having Appropriate Prefix


Represent each of the following with SI units having an appropriate prefix: (a) 8653 ms, (b) 8368 N, (c) 0.893 kg.

Statics of Rigid Bodies 14th Edition by RC Hibbeler, Problem 1-11


Solution:

Part A

\begin{align*}
8653 \ \text{ms} & = 8653 \ \left( 10^{-3} \right) \ \text{s} \\
& =8.653 \ \text{s}
\end{align*}

Part B

\begin{align*}
8368 \ \text{N} & = 8.368\times 10^3 \ \text{N}\\
& = 8.368 \ \text{kN}\\
\end{align*}

Part C

\begin{align*}
0.893 \  \text{kg} & = 0.893\times 10^3 \ \text{g} \\
& =893 \ \text{g}
\end{align*}

Advertisements

Mechanics of Materials: An Integrated Learning System 4th Edition by Timothy A. Philpot Complete Solution Manual



Advertisements
Advertisements

Arizona State University| PHY 121: Univ Physics I: Mechanics|Homework 1-1| Problem 2

Learning Goal:

To practice Tactics Box 1.2 Vector Subtraction.
Vector subtraction has some similarities to the subtraction of two scalar quantities. With numbers, subtraction is the same as the addition of a negative number. For example, 53 is the same as 5+(3). Similarly, A⃗ B⃗ =A⃗ +(B⃗ ). We can use the rules for vector addition and the fact that B⃗ is a vector opposite in direction to B⃗  to form rules for vector subtraction, as explained in this tactics box.
The figure shows two vectors. Vector A is on the left and is directed upwards to the right. Vector B is on the right and is directed downwards to the right.

To subtract B⃗  from A⃗  (Figure 1), perform these steps:

  1. Draw A⃗ .  The figure shows vector A that is directed upwards to the right.
  2. Draw B⃗  and place its tail at the tip of A⃗ .The figure shows two vectors. Vector A is on the left and is directed upwards to the right. Vector minus B is on the right and is directed upwards to the left. The tip of vector A and the tail of vector minus B are at the same point.
  3. Draw an arrow from the tail of A⃗ to the tip of B⃗ . This is vector A⃗ B⃗ .     The figure shows two vectors. Vector A is on the left and is directed upwards to the right. Vector minus B is on the right and is directed upwards to the left. The tip of vector A and the tail of vector minus B are at the same point. Vector A minus B starts at the tail of vector A and ends at the tip of vector minus B.
Part A
Draw vector C⃗ =A⃗ B⃗  by following the steps in the tactics box above. When drawing B⃗ , keep in mind that it has the same magnitude as B⃗  but opposite direction.
Part B
Draw vector F⃗ =D⃗ E⃗ by following the steps in the tactics box above. When drawing E⃗ , keep in mind that it has the same magnitude as E⃗  but opposite direction.
ANSWERS:
Part A
3
Part B
4

Arizona State University| PHY 121: Univ Physics I: Mechanics|Homework 1-1| Problem 1

Learning Goal:
To practice Tactics Box 1.1 Vector Addition.
Vector addition obeys rules that are different from those for the addition of two scalar quantities. When you add two vectors, their directions, as well as their magnitudes, must be taken into account. This Tactics Box explains how to add vectors graphically. 
To add B⃗  to A⃗  (Figure 1), perform these steps:
  1. Draw A⃗ .  The figure shows vector A that is directed upwards to the right.
  2. Place the tail of B⃗  at the tip of A⃗ .  The figure shows two vectors. Vector A is on the left and is directed upwards to the right. Vector B is on the right and is directed downwards to the right. The tip of vector A and the tail of vector B are at the same point.
  3. Draw an arrow from the tail of A⃗  to the tip of B⃗ . This is vector A⃗ +B⃗ .  The figure shows three vectors. Vector A is on the left and is directed upwards to the right. Vector B is on the right and is directed downwards to the right. The tip of vector A and the tail of vector B are at the same point. Vector A plus B starts at the tail of vector A and ends at the tip of vector B.
Part A
Create the vector R⃗ =A⃗ +B⃗  by following the steps in the Tactics Box above. When moving vector B⃗ , keep in mind that its direction should remain unchanged.
The figure shows two vectors. Vector A is on the left and is directed upwards to the right. Vector B is on the right and is directed downwards to the right.
Part B
Create the vector R⃗ =C⃗ +D⃗  by following the steps in the Tactics Box above. When moving vector D⃗ , keep in mind that its direction should remain unchanged. The location, orientation, and length of your vectors will be graded.
ANSWERS:
Part A
1
Part B
2

Expressing the Density of Water in SI Units


Water has a density of 1.94 slug/ft³. What is the density expressed in SI units? Express the answer to three significant figures.

Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 1-19
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-17


Solution:

\begin{align*}
\rho _w & =\left(\frac{1.94\:\text{slug}}{1\:\text{ft}^3}\right)\left(\frac{14.59\:\text{kg}}{1\:\text{slug}}\right)\left(\frac{1\:\text{ft}^3}{0.3048^3\:\text{m}^3}\right) \\
& =\left(\frac{1.94\:\text{slug}}{1\:\text{ft}^3}\right)\left(\frac{14.59\:\text{kg}}{1\:\text{slug}}\right)\left(\frac{1\:\text{ft}^3}{0.3048^3\:\text{m}^3}\right) \\
& =999.6\:\frac{\text{kg}}{\text{m}^3}\\
& =1.00\:\text{Mg/m}^3\\
\end{align*}

Advertisements
Advertisements

Representing combinations of units in correct SI Form


Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) 0.000431 kg, (b) 35.3(10³) N, and (c) 0.00532 km.

Statics of Rigid Bodies 14th Edition by RC Hibbeler, Problem 1-7


Solution:

Part A

\begin{align*}
0.000 \ 431 \ \text{kg} & = 0.431 \times 10^{-3} \ \text{kg}\\
& = 0.431\times 10^{-3} \times 10^3 \ \text{g} \\
& = 0.431 \ \text{g}
\end{align*}

Part B

\begin{align*}
35.3\left( 10^3 \right) \ \text{N} & = 35.3 \ \text{kN}
\end{align*}

Part C

\begin{align*}
0.00532 \ \text{km} & = 0.00532 \times 10^3 \ \text{m} \\
& = 5.32 \times 10^{-3} \times 10^3 \ \text{m} \\
& = 5.32 \ \text{m}
\end{align*}