Statistics-lab™可以为您提供olemiss.edu CE310 Structural Mechanics结构力学课程的代写代考和辅导服务！
CE310 Structural Mechanics课程简介
Dynamics of particles is the study of the motion of particles under the influence of various forces. This field of study is important in understanding the behavior of structures subjected to loads, such as civil engineering structures.
Rigid body mechanics in a plane is the study of the motion and forces of rigid bodies, which are objects that maintain their shape even when subjected to external forces. In civil engineering, this field of study is important in understanding the behavior of structures that are made up of interconnected rigid bodies.
PREREQUISITES
Applications to civil engineering involve the use of principles from dynamics and rigid body mechanics to design and analyze civil engineering structures such as buildings, bridges, and tunnels. These principles are used to determine the loads that a structure will experience and to ensure that the structure is designed to withstand those loads.
Structural systems and loads refer to the different types of structures that can be used in civil engineering, as well as the types of loads that these structures can be subjected to. Understanding these concepts is essential for designing and analyzing civil engineering structures.
Analysis of statically determinate trusses, beams, and frames involves applying principles from rigid body mechanics to analyze these types of structures. Trusses, beams, and frames are commonly used in civil engineering and understanding how they behave under different loads is important in ensuring their safety and reliability.
CE310 Structural Mechanics HELP（EXAM HELP， ONLINE TUTOR）
Calculate the maximum shear stress, angle of twist and strain energy of the following: (Assuming that the shaft is $2 \mathrm{~m}$ long, the applied torque is $10 \mathrm{kNm}$ and $\mathrm{G}=80 \mathrm{kN} / \mathrm{mm}^2$ )
(a) a solid circular shaft of $50 \mathrm{~mm}$ in diameter.
(b) A hollow circular shaft with outer diameter of $100 \mathrm{~mm}$ and thickness of $3 \mathrm{~mm}$.
(c) A hollow square tube of outer dimension of $80 \mathrm{~mm}$ square and $5 \mathrm{~mm}$ thickness. Use Bredt-Batho Theory in your analysis.
(a) For a solid circular shaft of $50 \mathrm{~mm}$ in diameter, the cross-sectional area is $A = \frac{\pi d^2}{4} = \frac{\pi (50 \mathrm{~mm})^2}{4} = 1963.5 \mathrm{~mm}^2$. The maximum shear stress is given by $ \tau_{max} = \frac{16T}{\pi d^3} = \frac{16(10 \mathrm{kNm})}{\pi (50 \mathrm{~mm})^3} = 8.13 \mathrm{~MPa}$. The angle of twist can be calculated using $\theta = \frac{TL}{GJ}$, where $J = \frac{\pi d^4}{32}$ is the polar moment of inertia. Thus, $\theta = \frac{10 \mathrm{kNm} \times 2000 \mathrm{~mm}}{(80 \mathrm{kN}/\mathrm{mm}^2) \times \frac{\pi (50 \mathrm{~mm})^4}{32}} = 0.0091 \mathrm{~rad}$. The strain energy can be calculated using $U = \frac{1}{2} \frac{T^2}{GJ} = \frac{1}{2} \frac{(10 \mathrm{kNm})^2}{(80 \mathrm{kN}/\mathrm{mm}^2) \times \frac{\pi (50 \mathrm{~mm})^4}{32}} = 361.6 \mathrm{~J}$.
(b) For a hollow circular shaft with an outer diameter of $100 \mathrm{~mm}$ and a thickness of $3 \mathrm{~mm}$, the inner diameter is $d_1 = 100 \mathrm{~mm} – 2(3 \mathrm{~mm}) = 94 \mathrm{~mm}$. The cross-sectional area is $A = \frac{\pi}{4} (d_2^2 – d_1^2) = \frac{\pi}{4} ((100 \mathrm{~mm})^2 – (94 \mathrm{~mm})^2) = 1810.9 \mathrm{~mm}^2$. The maximum shear stress is given by $ \tau_{max} = \frac{16T}{\pi d^3} = \frac{16(10 \mathrm{kNm})}{\pi ((100 \mathrm{~mm})^3 – (94 \mathrm{~mm})^3)} = 16.68 \mathrm{~MPa}$. The angle of twist can be calculated using $\theta = \frac{TL}{GJ}$, where $J = \frac{\pi}{32}(d_2^4 – d_1^4)$ is the polar moment of inertia. Thus, $\theta = \frac{10 \mathrm{kNm} \times 2000 \mathrm{~mm}}{(80 \mathrm{kN}/\mathrm{mm}^2) \times \frac{\pi}{32}((100 \mathrm{~mm})^4 – (94 \mathrm{~mm})^4)} = 0.0078 \mathrm{~rad}$. The strain energy can be calculated using $U = \frac{1}{2} \frac{T^2}{GJ} = \frac{1}{2} \frac{(10 \mathrm{kNm})^2}{(80 \mathrm{kN}/\mathrm{mm}^2) \times \frac{\pi}{32}((100 \
Question 3. Figure 2 shows an idealised diagram of a model bridge. Beam $\mathrm{BC}$ is freely supported at its ends on two similar cantilever beams $\mathrm{AB}$ and $\mathrm{CD}$. The cross-sections of all the beams are similar and is shown in Figure 2(a).
The allowable bending stresses for the material of the model are $80 \mathrm{MPa}$ in compression and $160 \mathrm{MPa}$ in tension. Find the maximum uniformly distributed load (w kN/m) that can be allowed on span BC so that every part of the model is within the allowable bending stresses. Note that the location of centroid and $I_{x x}$ are given on the Figure.
First, we need to find the maximum bending moment that can be allowed on span BC. Since the beam is symmetric, we can assume that the maximum bending moment occurs at the middle of the span. The maximum bending moment can be calculated using:
$$M_{max} = \frac{wL^2}{8}$$
where $w$ is the uniformly distributed load, and $L$ is the length of span BC.
Substituting $w = 1 \mathrm{kN/m}$ and $L = 4 \mathrm{m}$, we get:
$$M_{max} = \frac{(1 \mathrm{kN/m})(4 \mathrm{m})^2}{8} = 8 \mathrm{kNm}$$
Next, we need to check whether this bending moment exceeds the allowable bending stresses in the material. We can do this by calculating the bending stress at the top and bottom fibers of span BC.
For the top fiber, the maximum tensile stress occurs when the beam is fully loaded. The distance from the centroid to the top fiber is $y_t = 105 \mathrm{~mm} + 25 \mathrm{~mm} = 130 \mathrm{~mm}$. The maximum tensile stress can be calculated using:
$$\sigma_{max,t} = \frac{My_t}{I_{xx}}$$
Substituting $M_{max} = 8 \mathrm{kNm}$, $y_t = 130 \mathrm{~mm}$, and $I_{xx} = 1.08 \times 10^8 \mathrm{~mm}^4$, we get:
$$\sigma_{max,t} = \frac{(8 \mathrm{kNm})(130 \mathrm{~mm})}{1.08 \times 10^8 \mathrm{~mm}^4} = 0.098 \mathrm{~MPa}$$
Since this value is less than the allowable tensile stress of $160 \mathrm{MPa}$, the top fiber is within the allowable stress limits.
For the bottom fiber, the maximum compressive stress occurs when the beam is fully loaded. The distance from the centroid to the bottom fiber is $y_b = 105 \mathrm{~mm} – 25 \mathrm{~mm} = 80 \mathrm{~mm}$. The maximum compressive stress can be calculated using:
$$\sigma_{max,c} = \frac{My_b}{I_{xx}}$$
Substituting $M_{max} = 8 \mathrm{kNm}$, $y_b = 80 \mathrm{~mm}$, and $I_{xx} = 1.08 \times 10^8 \mathrm{~mm}^4$, we get:
$$\sigma_{max,c} = \frac{(8 \mathrm{kNm})(80 \mathrm{~mm})}{1.08 \times 10^8 \mathrm{~mm}^4} = 0.059 \mathrm{~MPa}$$
Since this value is less than the allowable compressive stress of $80 \mathrm{MPa}$, the bottom fiber is also within the allowable stress limits.
Therefore, the maximum uniformly distributed load that can be allowed on span BC is $w = \frac{M_{max} \times I_{xx}}{y_t^2 – y_b^2}$. Substituting $M_{max} = 8 \mathrm{kNm}$, $y_t = 130 \mathrm{~mm}$, $y_b = 80 \mathrm{~mm}$, and $I_{xx} = 1.08 \times
Textbooks
• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.
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