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Problem 1 (10 Points): Determine the resultant internal loadings in the beam at...
Sep 22, 2024
Solution
a
To determine the resultant internal loadings at section D, we first need to calculate the reactions at the supports. The total distributed load on the beam is given by the formula W=q⋅(L1+L2)=2 kN/m⋅(5 m+8 m)=26 kNW = q \cdot (L_1 + L_2) = 2 \, \text{kN/m} \cdot (5 \, \text{m} + 8 \, \text{m}) = 26 \, \text{kN}. The point load FF is 14 kN. We can use the equilibrium equations to find the reactions at supports A and C
b
Using the moment about point A to find the reaction at C, we have: ∑MA=0⇒RC⋅(L1+L2)−W⋅(L1+L22)−F⋅(L1+L2−5)=0\sum M_A = 0 \Rightarrow R_C \cdot (L_1 + L_2) - W \cdot \left(\frac{L_1 + L_2}{2}\right) - F \cdot (L_1 + L_2 - 5) = 0. Substituting the values, we can solve for RCR_C
c
After finding RCR_C, we can use the vertical force equilibrium equation ∑Fy=0⇒RA+RC−W−F=0\sum F_y = 0 \Rightarrow R_A + R_C - W - F = 0 to find the reaction at support A
d
Now, to find the internal loadings at section D, we consider the left side of the section. The internal shear force VDV_D can be calculated as VD=RA−WleftV_D = R_A - W_{left}, where WleftW_{left} is the total load to the left of section D
e
For section E, just to the left of the point load F, the internal shear force VEV_E will be VE=VD−FV_E = V_D - F. The internal bending moment at section D can be calculated using MD=RA⋅d−Wleft⋅(L12)M_D = R_A \cdot d - W_{left} \cdot \left(\frac{L_1}{2}\right), and for section E, ME=MD−F⋅(L1+L2−5)M_E = M_D - F \cdot (L_1 + L_2 - 5)
Answer
Resultant internal loadings at section D and E can be calculated using equilibrium equations and considering the loads acting on the beam.
Key Concept
Static Equilibrium: The sum of forces and moments acting on a beam must be zero for it to be in equilibrium. Equations: ∑F=0\sum F = 0 and ∑M=0\sum M = 0.
Explanation
By applying the principles of static equilibrium, we can determine the reactions at the supports and subsequently find the internal loadings at the specified sections of the beam.
Solution
a
To determine the resultant internal loadings at cross section CC, we first need to calculate the reactions at the supports AA and BB. The total distributed load qq acting on the shaft is given as 700 kN/m700 \, \text{kN/m} over a length of L1+L2=0.6+0.6=1.2 mL_1 + L_2 = 0.6 + 0.6 = 1.2 \, \text{m}. The total distributed load can be calculated as: W=q⋅(L1+L2)=700 kN/m⋅1.2 m=840 kN.W = q \cdot (L_1 + L_2) = 700 \, \text{kN/m} \cdot 1.2 \, \text{m} = 840 \, \text{kN}.
b
Next, we need to find the reactions at the supports. The total load acting downwards is the sum of the distributed load and the concentrated load F=700 kNF = 700 \, \text{kN}. Therefore, the total downward load is: Wtotal=W+F=840 kN+700 kN=1540 kN.W_{total} = W + F = 840 \, \text{kN} + 700 \, \text{kN} = 1540 \, \text{kN}. We can use the equilibrium equations to find the reactions at supports AA and BB. Assuming the reaction at AA is RAR_A and at BB is RBR_B, we have: ∑Fy=0⇒RA+RB−1540=0.\sum F_y = 0 \Rightarrow R_A + R_B - 1540 = 0.
c
To find the moments about point AA, we can set up the moment equation: ∑MA=0⇒RB⋅(L1+L2)−W⋅(L1+L22)−F⋅L1=0.\sum M_A = 0 \Rightarrow R_B \cdot (L_1 + L_2) - W \cdot \left(\frac{L_1 + L_2}{2}\right) - F \cdot L_1 = 0. Substituting the values, we can solve for RBR_B. After finding RBR_B, we can substitute back to find RAR_A
d
Finally, the internal loadings at section CC can be determined by considering the equilibrium of the segment to the left of section CC. The internal shear force VCV_C and bending moment MCM_C can be calculated using the equations: VC=RA−Wdistributed⋅L1V_C = R_A - W_{distributed} \cdot L_1 and MC=RA⋅L1−Wdistributed2⋅L12−F⋅(L1−LdistancetoC).M_C = R_A \cdot L_1 - \frac{W_{distributed}}{2} \cdot L_1^2 - F \cdot (L_1 - L_{distance to C}).
Answer
[Insert final answer here]
Key Concept
Static Equilibrium: The sum of forces and the sum of moments acting on a body in static equilibrium must be zero. Equations: ∑F=0\sum F = 0 and ∑M=0\sum M = 0.
Explanation
By applying the principles of static equilibrium, we can determine the reactions at the supports and subsequently calculate the internal loadings at the cross section. This involves resolving the forces and moments acting on the shaft.
Solution
a
To determine the average normal stress at point D in the pipe AB, we first need to calculate the internal force acting at that section. The average normal stress is given by the formula: σD=FA\sigma_D = \frac{F}{A} where FF is the internal axial force and AA is the cross-sectional area of the pipe. The cross-sectional area of the pipe can be calculated as: A=π4(dout2−din2)A = \frac{\pi}{4} (d_{out}^2 - d_{in}^2)
b
For point E in the solid rod BC, the average normal stress is calculated similarly. The diameter of the rod is given as 15 mm, so the cross-sectional area is: AE=π4(15 mm)2A_E = \frac{\pi}{4} (15 \, \text{mm})^2. The average normal stress at point E is then given by: σE=FAE\sigma_E = \frac{F}{A_E} where FF is the internal axial force acting at that section
c
To represent the stress on a volume element located at points D and E, we can illustrate the stress vectors acting on the elements. At point D, the stress vector will act perpendicular to the cross-section of the pipe, while at point E, the stress vector will also act perpendicular to the cross-section of the rod
Answer
[Insert final answer here]
Key Concept
Average Normal Stress: The average normal stress in a material is defined as the internal force divided by the cross-sectional area over which the force acts. The formula is given by σ=FA\sigma = \frac{F}{A}. The cross-sectional area for hollow and solid sections can be calculated using the respective diameters.
Explanation
The average normal stress at points D and E is calculated using the forces acting on the shaft and the respective cross-sectional areas of the pipe and rod. This allows us to understand how the forces are distributed across the different sections of the built-up shaft.
Solution
a
To determine the largest intensity of the uniform loading w w that can be applied to the frame without exceeding the average normal stress σ \sigma and average shear stress τ \tau at section b-b, we first need to analyze the forces acting on the frame. The total load W W on the horizontal beam can be expressed as W=w⋅L W = w \cdot L , where L=4 m L = 4 \, \text{m} . The reactions at the supports can be calculated using static equilibrium equations: ∑Fy=0 \sum F_y = 0 and ∑MA=0 \sum M_A = 0
b
The average normal stress σ \sigma at section b-b can be calculated using the formula: σ=FA\sigma = \frac{F}{A} where F F is the axial force and A A is the cross-sectional area of member CB. Given that member CB has a square cross-section of a mm a \, \text{mm} on each side, the area A A can be expressed as A=a2 A = a^2 . The maximum allowable normal stress gives us the equation: σ=w⋅LA≤σmax\sigma = \frac{w \cdot L}{A} \leq \sigma_{max}
c
The average shear stress τ \tau at section b-b can be calculated using the formula: τ=VA\tau = \frac{V}{A} where V V is the shear force at the section. The shear force can be determined from the loading conditions and is equal to half the total load on the beam. Thus, we have: τ=w⋅L2A≤τmax\tau = \frac{\frac{w \cdot L}{2}}{A} \leq \tau_{max}. By substituting A=a2 A = a^2 into the equations for σ \sigma and τ \tau , we can solve for the maximum w w that satisfies both conditions
Answer
[Insert final answer here]
Key Concept
Static Equilibrium: The conditions for static equilibrium require that the sum of forces and the sum of moments acting on a structure are both zero. This is essential for analyzing structures under loads. Equations: ∑F=0\sum F = 0 and ∑M=0\sum M = 0. Average Normal Stress: σ=FA\sigma = \frac{F}{A} and Average Shear Stress: τ=VA\tau = \frac{V}{A}.
Explanation
To find the maximum uniform loading w w that can be applied without exceeding the stress limits, we analyze the forces and calculate the stresses at the critical section using the principles of static equilibrium. This ensures that the structure remains safe under the applied loads.
Solution
a
To determine the displacement of the center C, we need to integrate the normal strain expression ϵx=ksin⁡(πLx)\epsilon_{x} = k \sin\left(\frac{\pi}{L} x\right) over the length of the rod from 0 to L. The displacement uu at point C (which is at x=L2x = \frac{L}{2}) can be found using the relationship between strain and displacement: u=∫0xϵx dxu = \int_0^{x} \epsilon_{x} \, dx. Thus, we have: uC=∫0L2ksin⁡(πLx) dxu_C = \int_0^{\frac{L}{2}} k \sin\left(\frac{\pi}{L} x\right) \, dx. Evaluating this integral gives us the displacement at point C
b
The average normal strain in the entire rod can be calculated by integrating the strain over the entire length of the rod and then dividing by the length. The average strain ϵavg\epsilon_{avg} is given by: ϵavg=1L∫0Lϵx dx=1L∫0Lksin⁡(πLx) dx\epsilon_{avg} = \frac{1}{L} \int_0^{L} \epsilon_{x} \, dx = \frac{1}{L} \int_0^{L} k \sin\left(\frac{\pi}{L} x\right) \, dx. Evaluating this integral will yield the average normal strain in the rod
Answer
Displacement at C: uC=2kLπu_C = \frac{2kL}{\pi}; Average normal strain: ϵavg=2kπ\epsilon_{avg} = \frac{2k}{\pi}
Key Concept
Strain and Displacement: The relationship between strain and displacement is fundamental in mechanics of materials. Strain is defined as the change in length per unit length, and displacement can be found by integrating the strain over a length. The average strain is calculated by integrating the strain over the entire length and normalizing it by the length.
Explanation
By integrating the given strain function, we can find both the displacement at the center of the rod and the average strain across its length. This approach utilizes the fundamental definitions of strain and displacement in mechanics.
Solution
a
To determine the elongation of the square hollow bar when subjected to an axial force of P=100 kN P = 100 \, \text{kN} , we first need to calculate the cross-sectional area A A of the hollow bar. The outer dimensions are 50 mm 50 \, \text{mm} and the wall thickness is 5 mm 5 \, \text{mm} . The inner dimensions are 50−2×5=40 mm 50 - 2 \times 5 = 40 \, \text{mm} . The cross-sectional area is given by: A=π4(dout2−din2)=π4((50 mm)2−(40 mm)2)=π4(2500−1600)=π4×900 mm2=706.86 mm2. A = \frac{\pi}{4} (d_{out}^2 - d_{in}^2) = \frac{\pi}{4} ((50 \, \text{mm})^2 - (40 \, \text{mm})^2) = \frac{\pi}{4} (2500 - 1600) = \frac{\pi}{4} \times 900 \, \text{mm}^2 = 706.86 \, \text{mm}^2. Next, we convert this area to m2 \text{m}^2 : A=706.86×10−6 m2. A = 706.86 \times 10^{-6} \, \text{m}^2. Now, we calculate the stress σ \sigma using the formula: σ=PA=100×103 N706.86×10−6 m2=141.56 MPa. \sigma = \frac{P}{A} = \frac{100 \times 10^3 \, \text{N}}{706.86 \times 10^{-6} \, \text{m}^2} = 141.56 \, \text{MPa}.
b
Since the stress is within the elastic limit (up to 500 MPa), we can use Young's modulus Y Y to find the elongation ΔL \Delta L . Assuming Y Y for the metal alloy is approximately 200 GPa 200 \, \text{GPa} (or 200×103 MPa 200 \times 10^3 \, \text{MPa} ), we can calculate the elongation using the formula: ΔL=σLY=141.56 MPa×600×10−3 m200×103 MPa=0.423 mm. \Delta L = \frac{\sigma L}{Y} = \frac{141.56 \, \text{MPa} \times 600 \times 10^{-3} \, \text{m}}{200 \times 10^3 \, \text{MPa}} = 0.423 \, \text{mm}.
c
For the second part, when the axial force is increased to P=360 kN P = 360 \, \text{kN} , we again calculate the new stress: σnew=360×103 N706.86×10−6 m2=509.25 MPa. \sigma_{new} = \frac{360 \times 10^3 \, \text{N}}{706.86 \times 10^{-6} \, \text{m}^2} = 509.25 \, \text{MPa}. This stress exceeds the elastic limit, so we need to find the permanent elongation. The yield point is at 250 MPa 250 \, \text{MPa} and the corresponding strain is 0.00125 mm/mm 0.00125 \, \text{mm/mm} . The permanent elongation can be calculated as: ΔLpermanent=ΔLelastic+ΔLplastic=250 MPa×600×10−3 m200×103 MPa+(plastic strain)×L. \Delta L_{permanent} = \Delta L_{elastic} + \Delta L_{plastic} = \frac{250 \, \text{MPa} \times 600 \times 10^{-3} \, \text{m}}{200 \times 10^3 \, \text{MPa}} + \text{(plastic strain)} \times L. The plastic strain can be calculated from the stress above the yield point, which is 509.25−250=259.25 MPa 509.25 - 250 = 259.25 \, \text{MPa} . The total elongation will include both elastic and plastic components
Answer
The elongation of the bar under 100 kN is approximately 0.423 mm 0.423 \, \text{mm} and the permanent elongation after increasing the load to 360 kN will depend on the plastic deformation calculated from the stress-strain curve.
Key Concept
Young's Modulus: Young's modulus is a measure of the stiffness of a material, defined as the ratio of stress to strain. The formula is given by: Y=σϵ Y = \frac{\sigma}{\epsilon} where σ \sigma is stress and ϵ \epsilon is strain.
Explanation
The elongation of the bar is calculated using the stress and Young's modulus, while the permanent elongation considers the plastic deformation beyond the yield point.
Solution
a
First, we need to calculate the cross-sectional area A A of the steel specimen. The original diameter is d=12.5 mm d = 12.5 \, \text{mm} , so the area is given by the formula: A=πd24=π(12.5 mm)24≈122.72 mm2 A = \frac{\pi d^2}{4} = \frac{\pi (12.5 \, \text{mm})^2}{4} \approx 122.72 \, \text{mm}^2
b
Next, we convert the load from kN to N for consistency in units. The load values from the table are multiplied by 1000 1000 to convert to N. The stress σ \sigma can be calculated using the formula: σ=FA \sigma = \frac{F}{A} where F F is the load in N and A A is the area in mm2 \text{mm}^2 . For example, for the first load of 11.1 kN 11.1 \, \text{kN} : σ1=11.1×1000 N122.72 mm2≈90.4 MPa \sigma_1 = \frac{11.1 \times 1000 \, \text{N}}{122.72 \, \text{mm}^2} \approx 90.4 \, \text{MPa}
c
We repeat this calculation for each load value in the table to obtain the corresponding stress values. Then, we calculate the strain ϵ \epsilon using the formula: ϵ=ΔLL0 \epsilon = \frac{\Delta L}{L_0} where ΔL \Delta L is the elongation and L0 L_0 is the original gage length of 50 mm 50 \, \text{mm} . For the first elongation of 0.0175 mm 0.0175 \, \text{mm} : ϵ1=0.0175 mm50 mm=0.00035 \epsilon_1 = \frac{0.0175 \, \text{mm}}{50 \, \text{mm}} = 0.00035
d
After calculating stress and strain for all data points, we can plot the stress-strain diagram. The modulus of toughness can be approximated as the area under the stress-strain curve up to the point of fracture. This can be calculated using numerical integration methods or by estimating the area geometrically if the curve is linear or piecewise linear
Answer
The modulus of toughness can be determined from the area under the stress-strain curve, which requires plotting the calculated stress and strain values.
Key Concept
Stress and Strain: Stress is defined as the force per unit area, and strain is the deformation per unit length. The relationship between stress and strain is crucial in determining material properties such as Young's modulus and modulus of toughness. The modulus of toughness is the total area under the stress-strain curve up to the fracture point.
Explanation
By calculating stress and strain from the provided load and elongation data, we can plot the stress-strain diagram and determine the modulus of toughness, which represents the energy absorbed by the material before failure.
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