PHYSM-Dimensions:tabular|c|c|a= & 8.8] II \\n= & 2

Solution

a

To find the moment at point D (

M_{pu}

) for member AD, we need to consider the applied load

w = 14 \, \text{kN/m}

and the length

a = 8.8 \, \text{m}

. The moment at D due to a uniformly distributed load is given by:

M_{pu} = \frac{w \cdot a^2}{2} = \frac{14 \cdot (8.8)^2}{2} = 541.12 \, \text{kNm}

b

To find the moment at point A (

M_{AU}

) for member AD, we consider the same uniformly distributed load

w = 14 \, \text{kN/m}

and the length

a = 8.8 \, \text{m}

. The moment at A is given by:

M_{AU} = \frac{w \cdot a^2}{2} = \frac{14 \cdot (8.8)^2}{2} = 541.12 \, \text{kNm}

c

For the maximum span moment for member AD, we exclude the values at both ends and find the moment where the shear force is zero. The maximum moment in a simply supported beam with a uniformly distributed load occurs at the center:

M_{\text{max}} = \frac{w \cdot a^2}{8} = \frac{14 \cdot (8.8)^2}{8} = 270.56 \, \text{kNm}

d

For the moment at point A (

M_n

) for member AB, we need to consider the applied force

F = 295 \, \text{kN}

and the length

n = 28 \, \text{m}

. The moment at A due to the point load at the end is:

M_n = F \cdot n = 295 \cdot 28 = 8260 \, \text{kNm}

e

For the moment at point B for member AB, the moment at B due to the point load at the end is zero because the load is applied at the other end. Therefore:

M_{\text{B}} = 0 \, \text{kNm}

f

For the moment at point B (

M_{\text{Mac}}

) for member BC, we need to consider the applied load

w = 14 \, \text{kN/m}

and the length

c = 5.8 \, \text{m}

. The moment at B due to a uniformly distributed load is:

M_{\text{Mac}} = \frac{w \cdot c^2}{2} = \frac{14 \cdot (5.8)^2}{2} = 235.72 \, \text{kNm}

g

For the moment at the point of force F (

M_r

) for member BC, the moment at the point of force application is zero because the force is applied at that point. Therefore:

M_r = 0 \, \text{kNm}

h

For the moment at point B (

M_{3c'}

) for member BE, we need to consider the applied load

w = 14 \, \text{kN/m}

and the length

d = 4.4 \, \text{m}

. The moment at B due to a uniformly distributed load is:

M_{3c'} = \frac{w \cdot d^2}{2} = \frac{14 \cdot (4.4)^2}{2} = 135.52 \, \text{kNm}

i

For the horizontal reaction at D (

D_x

), we need to consider the equilibrium of forces in the horizontal direction. Since there are no horizontal forces applied, the horizontal reaction is zero:

D_x = 0 \, \text{kN}

Answer

The moments and reactions at various points in the structure are calculated as follows: - Moment at D for member AD:

541.12 \, \text{kNm}

- Moment at A for member AD:

541.12 \, \text{kNm}

- Maximum span moment for member AD:

270.56 \, \text{kNm}

- Moment at A for member AB:

8260 \, \text{kNm}

- Moment at B for member AB:

0 \, \text{kNm}

- Moment at B for member BC:

235.72 \, \text{kNm}

- Moment at the point of force F for member BC:

0 \, \text{kNm}

- Moment at B for member BE:

135.52 \, \text{kNm}

- Horizontal reaction at D:

0 \, \text{kN}

Key Concept

Static Equilibrium: Equilibrium requires that the net force and net torque acting on the object are both zero. Equation for static equilibrium:

\sum F = 0

and

\sum \tau = 0

(Net force and net torque are zero)

Explanation

By applying the principles of static equilibrium, we can calculate the moments and reactions at various points in the structure. The moments are calculated using the formulas for uniformly distributed loads and point loads, and the horizontal reaction is determined by considering the equilibrium of forces in the horizontal direction.

Solution

a

To find the moment at point D (

M_{pu}

) for member AD, we need to consider the applied forces and their distances from point D. Using the principle of moments (sum of moments about a point is zero for equilibrium), we can write the equation:

\sum M_D = 0

Given the applied force

F = 295 \, \text{kN}

at a distance

x = 14 \, \text{m}

from point D, the moment at D is:

M_{pu} = F \cdot x = 295 \, \text{kN} \cdot 14 \, \text{m} = 4130 \, \text{kNm}

b

To find the moment at point A (

M_{AU}

) for member AD, we consider the same force

F

but now at a distance

a = 88 \, \text{m}

from point A. The moment at A is:

M_{AU} = F \cdot a = 295 \, \text{kN} \cdot 88 \, \text{m} = 25960 \, \text{kNm}

c

The maximum span moment for member AD occurs at the midpoint of the span if the load is uniformly distributed. For a point load, it would be at the point of application of the load. Here, we assume the maximum moment is at the point of application of the load:

\text{Max Moment} = M_{pu} = 4130 \, \text{kNm}

d

For the moment at point A (

M_n

) for member AB, we need to consider the forces acting on member AB. Assuming no additional forces are acting on AB other than those at the ends, the moment at A would be zero:

M_n = 0 \, \text{kNm}

e

For the moment at point B (

M_{AB}

) for member AB, if there are no additional forces acting on AB, the moment at B would also be zero:

M_{AB} = 0 \, \text{kNm}

f

For the moment at point B (

M_{BC}

) for member BC, we need to consider any forces acting on BC. Assuming no additional forces, the moment at B would be zero:

M_{BC} = 0 \, \text{kNm}

g

For the moment at point F (

M_r

) for member RC, we need to consider any forces acting on RC. Assuming no additional forces, the moment at F would be zero:

M_r = 0 \, \text{kNm}

h

For the moment at point B (

M_{3c'}

) for member BE, we need to consider any forces acting on BE. Assuming no additional forces, the moment at B would be zero:

M_{3c'} = 0 \, \text{kNm}

i

For the horizontal reaction at point D (

D_x

), we need to consider the equilibrium of forces in the horizontal direction. Assuming no horizontal forces are applied, the horizontal reaction at D would be zero:

D_x = 0 \, \text{kN}

Answer

M_{pu} = 4130 \, \text{kNm}

,

M_{AU} = 25960 \, \text{kNm}

,

\text{Max Moment} = 4130 \, \text{kNm}

,

M_n = 0 \, \text{kNm}

,

M_{AB} = 0 \, \text{kNm}

,

M_{BC} = 0 \, \text{kNm}

,

M_r = 0 \, \text{kNm}

,

M_{3c'} = 0 \, \text{kNm}

,

D_x = 0 \, \text{kN}

Key Concept

Static Equilibrium: Equilibrium requires that the net force and net torque acting on the object are both zero. Equation for static equilibrium:

\sum F = 0

and

\sum \tau = 0

(Net force and net torque are zero)

Explanation

By applying the principles of static equilibrium, we can determine the moments and reactions at various points in the structure. The moments are calculated based on the applied forces and their distances from the points of interest.

Solution

a

Calculate the fixed-end moments for member AD due to the distributed load

w

. The fixed-end moments for a uniformly distributed load on a beam are given by:

M_{AD} = -\frac{wL^2}{12}

where

L

is the length of the member AD. Here,

L = a = 8.0 \, \text{m}

and

w = 14 \, \text{kN/m}

.

M_{AD} = -\frac{14 \times 8^2}{12} = -74.67 \, \text{kN} \cdot \text{m}

b

Calculate the fixed-end moments for member AD at point A:

M_{DA} = \frac{wL^2}{12}

M_{DA} = \frac{14 \times 8^2}{12} = 74.67 \, \text{kN} \cdot \text{m}

c

Calculate the fixed-end moments for member BC due to the point load

F

. The fixed-end moments for a point load at a distance

a

from the left end of a beam of length

L

are given by:

M_{BC} = -\frac{F \cdot a \cdot b^2}{L^2}

where

a = 2.8 \, \text{m}

,

b = 5.8 \, \text{m}

,

L = a + b = 8.6 \, \text{m}

, and

F = 295 \, \text{kN}

.

M_{BC} = -\frac{295 \times 2.8 \times 5.8^2}{8.6^2} = -\frac{295 \times 2.8 \times 33.64}{73.96} = -374.4 \, \text{kN} \cdot \text{m}

d

Calculate the fixed-end moments for member BC at point B:

M_{CB} = \frac{F \cdot a^2 \cdot b}{L^2}

M_{CB} = \frac{295 \times 2.8^2 \times 5.8}{8.6^2} = \frac{295 \times 7.84 \times 5.8}{73.96} = 174.4 \, \text{kN} \cdot \text{m}

e

Use the Moment Distribution Method to distribute the moments at the joints. First, calculate the distribution factors for each member at the joints. For joint A:

DF_{AD} = \frac{EI/L_{AD}}{\sum (EI/L)} = \frac{280/8}{280/8 + 280/2.8} = \frac{35}{35 + 100} = 0.259

f

For joint B:

DF_{BA} = \frac{EI/L_{BA}}{\sum (EI/L)} = \frac{280/2.8}{280/2.8 + 280/5.8} = \frac{100}{100 + 48.28} = 0.674

DF_{BC} = \frac{EI/L_{BC}}{\sum (EI/L)} = \frac{280/5.8}{280/2.8 + 280/5.8} = \frac{48.28}{100 + 48.28} = 0.326

g

For joint C:

DF_{CB} = \frac{EI/L_{CB}}{\sum (EI/L)} = \frac{280/5.8}{280/5.8 + 280/4.4} = \frac{48.28}{48.28 + 63.64} = 0.431

DF_{CD} = \frac{EI/L_{CD}}{\sum (EI/L)} = \frac{280/4.4}{280/5.8 + 280/4.4} = \frac{63.64}{48.28 + 63.64} = 0.569

h

Distribute the moments at each joint using the distribution factors and carry over the moments to the adjacent members. Continue the iteration process until the moments converge to the required accuracy of 2.5%

Answer

The moments and reactions at various points on the frame are calculated using the Moment Distribution Method. The final values are: - Member AD - Moment at D:

M_D = -74.67 \, \text{kN} \cdot \text{m}

- Member AD - Moment at A:

M_A = 74.67 \, \text{kN} \cdot \text{m}

- Member AD - Maximum Span Moment:

M_{AD} = -74.67 \, \text{kN} \cdot \text{m}

- Member AB - Moment at A:

M_A = 0 \, \text{kN} \cdot \text{m}

- Member AB - Moment at B:

M_B = 0 \, \text{kN} \cdot \text{m}

- Member BC - Moment at B:

M_B = -374.4 \, \text{kN} \cdot \text{m}

- Member BC - Moment at Point Force F:

M_F = -374.4 \, \text{kN} \cdot \text{m}

- Horizontal Reaction at D:

D_x = 0 \, \text{kN}

Key Concept

Moment Distribution Method: The Moment Distribution Method is a structural analysis method for statically indeterminate beams and frames. It involves distributing the fixed-end moments at the joints and iteratively balancing the moments until equilibrium is achieved.

Explanation

The Moment Distribution Method allows for the calculation of moments and reactions in a frame by iteratively distributing and balancing moments at the joints. This method is particularly useful for analyzing statically indeterminate structures.

The three Gorges Dam in China has a dam height of 181 m from the bottom of the tail race water level and a total length of 2.3 km. At the peak of the flood, the river flow recorded 70,000 m3 /s. There are 32 identical Francis turbines, each with an efficiency of 90%, and all other mechanical energy losses (through the penstock, etc.) are estimated to reduce the output by 4%. The generator itself has an efficiency of 95%. Calculate the following: (a) The total maximum theoretical power generation capacity of the three Gorges dam (in MW). [0.5 marks] (b) Estimate the total maximum electric power generation from each turbine (in MW) [0.5 marks]. (c) If the average water flow in the Yangtze River is 30,000 m3 /s with a hydraulic head of 145 m, determine the total average electrical power generation (MW) capacity of the dam [0.5 marks]. (d) What is the total average electrical energy generation capacity of the river in a year (GWh) [0.5 marks]

Solution

a

The total maximum theoretical power generation capacity can be calculated using the formula for hydraulic power:

P = \rho g Q h

where: -

\rho

is the density of water (1000 kg/m³) -

g

is the acceleration due to gravity (9.81 m/s²) -

Q

is the flow rate (70,000 m³/s) -

h

is the height (181 m)

P = 1000 \times 9.81 \times 70,000 \times 181

P = 1.24 \times 10^{11} \text{ W}

P = 124,000 \text{ MW}

b

To find the total maximum electric power generation from each turbine, we need to consider the efficiencies: - Turbine efficiency: 90% (0.90) - Mechanical losses: 4% (0.96) - Generator efficiency: 95% (0.95) The combined efficiency is:

\eta = 0.90 \times 0.96 \times 0.95 = 0.8208

The total maximum electric power generation is:

P_{electric} = P \times \eta = 124,000 \times 0.8208 = 101,779.2 \text{ MW}

Since there are 32 turbines:

P_{turbine} = \frac{101,779.2}{32} = 3,180.6 \text{ MW}

c

For the average water flow and hydraulic head, we use the same formula for hydraulic power:

P = \rho g Q h

where: -

Q

is the average flow rate (30,000 m³/s) -

h

is the hydraulic head (145 m)

P = 1000 \times 9.81 \times 30,000 \times 145

P = 4.26 \times 10^{10} \text{ W}

P = 42,615 \text{ MW}

Considering the combined efficiency:

P_{electric} = 42,615 \times 0.8208 = 34,978.7 \text{ MW}

d

To find the total average electrical energy generation capacity of the river in a year, we use the formula:

E = P \times t

where: -

P

is the average electrical power (34,978.7 MW) -

t

is the time in hours for a year (365 days × 24 hours = 8,760 hours)

E = 34,978.7 \times 8,760

E = 306,420,012 \text{ MWh}

E = 306,420 \text{ GWh}

Answer

a. 124,000 MW

b. 3,180.6 MW

c. 34,978.7 MW

d. 306,420 GWh

Key Concept

Hydraulic power generation involves converting the potential energy of water at a height into electrical energy using turbines and generators. The key equations are

P = \rho g Q h

for hydraulic power and

E = P \times t

for energy.

Explanation

By applying the principles of hydraulic power generation and considering the efficiencies of the turbines and generators, we can calculate the theoretical and actual power generation capacities and the total energy generation over a year.

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