Terracotta Jali Brick

 

6

438753-001

Terracotta Jali Brick

1. Ajit Vilas Karande 2. Ganesh Dhareshwar Lakade 3. Chetan Pitambar Pise 4. Amol Ashok Kamble

28/11/2024

221912

 

Hexagonal Brick

 

1

438588-001

Hexagonal Brick

1. Yashwant Prabhakar Pawar 2. Satyawan Dagadu Jagdale 3. Chetan Pitambar Pise 4. Chandrakant Mahadeo Deshmukh 5. Sachin Arun Ghadge 6. Ganesh Dhareshwar Lakade 7. Shriganesh Shantikumar Kadam 8. Jaywant Prabhakar Pawar 9. Amol Ashok Kamble

27/11/2024

221790

Problems on Engineering Mechanics

Problem: Newton’s Second Law

Problem Statement:
A 5 kg block is pulled by a 20 N force on a frictionless surface. Find its acceleration.

Solution:
Using Newton’s Second Law:

$$F = ma$$ $$20 = 5a$$ $$a = 4 \text{ m/s}^2$$

Answer: Acceleration = 4 m/s²

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Problem: Equilibrium of Forces

Problem Statement:
A 10 kg block is suspended by two ropes making 30° and 45° angles with the ceiling. Find the tension in each rope.

Solution:
Let $T_1$ and $T_2$ be the tensions in the two ropes.
Using equilibrium equations:

$$T_1 \cos 30^\circ + T_2 \cos 45^\circ = 98 \text{ N} (Weight of the block)$$ $$T_1 \sin 30^\circ = T_2 \sin 45^\circ$$

Solving these two equations gives:

$$T_1 = 113.14 \text{ N}, \quad T_2 = 80.21 \text{ N}$$

Answer: $T_1 = 113.14$ N, $T_2 = 80.21$ N

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Problem: Friction

Problem Statement:
A 15 kg box is placed on a rough surface with a coefficient of friction 0.4. Find the force required to just move the box.

Solution:
Friction force is given by:

$$F_f = \mu N$$ $$F_f = (0.4)(15 \times 9.81)$$ $$F_f = 58.86 \text{ N}$$

Answer: 58.86 N

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Problem: Centroid Calculation

Problem Statement:
Find the centroid of a right triangle with base 6 m and height 4 m, measured from the base.

Solution:
For a right triangle, the centroid is given by:

$$x_c = \frac{b}{3}, \quad y_c = \frac{h}{3}$$ $$x_c = \frac{6}{3} = 2 \text{ m}, \quad y_c = \frac{4}{3} = 1.33 \text{ m}$$

Answer: Centroid = (2 m, 1.33 m)

Problem: Pushing a Crate on a Rough Surface

Problem Statement:
A 20 kg wooden crate is pushed along a rough horizontal floor with a force of 100 N at an angle of 30° from the horizontal. If the coefficient of friction between the crate and the floor is 0.3, determine:

1. The normal force acting on the crate.
2. The acceleration of the crate.

Solution:

Step 1: Compute Normal Force

The weight of the crate:

$$W = mg = (20)(9.81) = 196.2 \text{ N}$$

The vertical component of the applied force reduces the normal force:

$$N = W - 100 \sin 30^\circ$$ $$N = 196.2 - 50 = 146.2 \text{ N}$$

Step 2: Compute Friction Force

$$F_f = \mu N = (0.3)(146.2) = 43.86 \text{ N}$$

Step 3: Compute Acceleration

The net horizontal force:

$$F_{\text{net}} = 100 \cos 30^\circ - 43.86$$ $$= 86.6 - 43.86 = 42.74 \text{ N}$$

Applying Newton’s Second Law:

$$F = ma$$ $$42.74 = (20)a$$ $$a = 2.14 \text{ m/s}^2$$

Answer:

1. Normal force = 146.2 N
2. Acceleration = 2.14 m/s²

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Problem: Beam with an Uneven Load

Problem Statement:
A 4-meter-long beam is simply supported at both ends. A 500 N force is applied 1 m from the left support, and a 200 N force is applied 3 m from the left support. Find the reaction forces at both supports.

Solution:
Using the sum of moments about the left support:

$$\sum M_A = 0$$

Let $R_B$ be the reaction at the right support:

$$(500)(1) + (200)(3) = R_B (4)$$ $$500 + 600 = 4 R_B$$ $$R_B = 275 \text{ N}$$

Using the sum of vertical forces:

$$R_A + R_B = 500 + 200$$ $$R_A + 275 = 700$$ $$R_A = 425 \text{ N}$$

Answer:
Reaction at left support $R_A = 425 N$
Reaction at right support $R_B = 275 N$

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Problem: Inclined Plane with a Pulley System

Problem Statement:
A 10 kg block is placed on a 30° incline and connected to a 5 kg hanging mass via a frictionless pulley. If the coefficient of friction between the block and the incline is 0.2, determine whether the system moves and find its acceleration.

Solution:

Step 1: Forces Acting on the 10 kg Block

• Gravity component along incline: $$F_{\text{gravity}} = mg \sin 30^\circ = (10)(9.81) \sin 30^\circ = 49.05 \text{ N}$$
• Normal force: $$N = mg \cos 30^\circ = (10)(9.81) \cos 30^\circ = 84.95 \text{ N}$$
• Friction force: $$F_f = \mu N = (0.2)(84.95) = 16.99 \text{ N}$$

Step 2: Forces Acting on the 5 kg Hanging Mass

$$F = mg = (5)(9.81) = 49.05 \text{ N}$$

Step 3: Determine Motion

The net force pulling the system:

$$F_{\text{net}} = 49.05 - (49.05 + 16.99)$$ $$F_{\text{net}} = -16.99 \text{ N}$$

Since the net force is negative, the block does not move; friction is preventing motion.

Answer: The system does not move.

Chute Spillway Discharge

HSWPE 12

Determine the discharge over a Chute spillway with ‘Ogee Crest’ using the following data: The length of the spillway is 250 m, Height of the spillway crest in front of at upstream approach channels is 10 m. The width of the approach channel is 250 m and the depth of water over the spillway crest is 5 m.

Answer :

Given
Length of spillway L = 250 m
Width of channel =250 m 
Cd = 2.2 for Ogee spillway design
h= 10m 
He =5m
h/He =10/5 =2

 from the graph 

C/Cd =1  

C/2.2 =1

C = 2.2

capacity of the Hydropower plant

HSWPE 10

Three generators each of capacity 9000 KW have been installed at a power station. During a certain period of load, the load on the plant varies from 14000 KW to 24000 KW. Determine: (6 mark)          Dec 2018

1. Total installed capacity

2. Load factor

3. Plant factor

4. Utilization factor

Answer :

HSWPE 11

A runoff river plant is installed on a river having a minimum flow of 15 cumecs. Head available at the plant is 16 m and the plant efficiency may be assumed as 80%. If the plant is used as a peak load plant operating for 6 hours daily, compute the firm capacity of the plant: (6 mark) May 2018
i.  Without pondage
ii. With pondage but allowing 8% of water to be lost in evaporation and other losses

Answer :

Spillway Discharge

HSWPE 6

Compute the discharge over an ogee weir with a coefficient of discharge equal to 2.4 and a head of 2 m. The length of the spillway is 100 m. The Weir crest is 8 m above the bottom of the approach channel having the same width as that of the spillway. (7 mark) 

Answer :

HSWPE 7

The Siphon spillway of the rectangular cross-section had the following dimensions at its throat. Height of throat = 1.5 m. Width of throat = 4 m. At the design flow, the tailwater elevation is 7 m below the summit of the siphon, and the headwater elevation is 2 m above the summit. Taking the coefficient of discharge as 0.6, determine the discharge capacity of the siphon. Also, determine the head that would be required on an ogee spillway 3.8 m long to discharge this flow if the coefficient of discharge is 2.25. (7 mark)

Answer :

HSWPE 8

A siphon spillway had the following cross-section at its throat. Height of the throat = 1.5 m. Width of throat = 4 m. At the design flow, the tail water elevation is 2 m above the summit.
i.   Taking a coefficient of discharge as 0.6, determine the capacity of the siphon.
ii. Determine the head that would be required on an ogee spillway 3.8 m long to discharge this flow, if coefficient of discharge is 2.25.
iii. What length of the ogee weir would be required to discharge the same flow with a head of 2.2 m on the crest? (7 mark)

Answer :

HSWPE 9

A saddle siphon spillway has the following data. Full reservoir level = 485 m, Level of the center of siphon outlet = 479.6 m, Highest flood level = 485.9 m, Highest flood discharge = 570 cumecs. If the dimensions of the throat of the siphon are: width = 4.2 m and height = 1.9 m, determine the number of siphon units required to pass the flood safely. The siphon is to discharge freely in the air. Assume the coefficient of discharge = 0.65. (6 mark)

Answer :

Slip Circle Methods

HSWPE 5

In order to find the factor of safety of downstream, slope during steady seepage, the section of the dam was drawn to scale 1 cm = 4 m. The following results were obtained on a critical slip circle.

Area of N-rectangle = 14.4 sqcm

Area of T-rectangle = 6.4 sqcm

Area of U-rectangle = 4.9 sqcm

Length of arc = 12.6 cm 

Laboratory tests have furnished values 260 for effective angle of friction and 19.5 kN/m2 for cohesion, unit weight of soil = 19 kN/m3. Determine the factor of safety of the slope.            (7 mark) 

Answer :

∴𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 𝑜𝑓 𝑠𝑙𝑜𝑝𝑒 𝑖𝑠 1.41

Seepage Analysis

HSWPE 4

A flow net was constructed for a homogeneous earth dam 52 m high and 2 m freeboard, and the following results were obtained. The number of potential drops = 25, Number of flow channels = 4. The dam has a horizontal filter of 40 m in length at its downstream end. Calculate the seepage discharge per meter length of the dam. Assume the dam material's permeability coefficient as 3x10^-3 cm/sec. (7 mark)

Answer: 

Given

K = 3x10^-3 cm/sec = 3x10^-5 m/sec

H = 52-2 = 50 m

Nf = 4

Nd = 25

There seepage discharge per meter length of the dam is 2.4×10^(−4) 𝑐𝑢𝑚𝑒𝑐𝑠/𝑚

HSWPE 3

 The figure shows the section of a gravity dam (non-overflow portion) built of concrete. Calculate (neglecting earthquake effects). (13 mark)       Dec 2016

i. The maximum vertical stresses at the toe and heel of the dam

ii. The major principal stresses at the toe of the dam

iii. Intensity of shear stress on a horizontal plane near the toe

Assume the unit weight of concrete = 23.5 kN/m2.

Allowable compressive stress in concrete = 2500 kN/m².

Allowable shear stress in concrete = 420 kN/m².

Assume that reservoir is full of water up to M.W.L.

ANSWER

For More Problems in Hydraulic Structure and Water Power Engineering