CIV3701 ASSIGNMENT 2 WITH QUALITY ANSWERS SEMESTER 2 2024
1. Structural Analysis
Question: Calculate the reactions at the supports for a simply supported beam with a span of 6
meters, subjected to a uniformly distributed load (UDL) of 4 kN/m over the entire length.
Answer:
● Total Load: 4 kN/m×6 m=24 kN4 \, \text{kN/m} \times 6 \, \text{m} = 24 \,
\text{kN}4kN/m×6m=24kN
● Reactions at the Supports: Since it's a simply supported beam with symmetrical
loading,
○ Reaction at A (RA) = Reaction at B (RB) = 24 kN2=12 kN\frac{24 \, \text{kN}}{2} =
12 \, \text{kN}224kN=12kN.
2. Material Properties
Question: A steel bar with a cross-sectional area of 50 mm² is subjected to a tensile force of
100 kN. Calculate the stress in the bar.
Answer:
● Stress (σ\sigmaσ) = Force (F) / Area (A)
● σ=100 kN50 mm2=100×103 N50×10−6 m2=2000 MPa\sigma = \frac{100 \, \
text{kN}}{50 \, \text{mm}^2} = \frac{100 \times 10^3 \, \text{N}}{50 \
times 10^{-6} \, \text{m}^2} = 2000 \, \text{MPa}σ=50mm2100kN
=50×10−6m2100×103N=2000MPa
3. Design of Concrete Structures
Question: A rectangular reinforced concrete beam has a breadth of 300 mm and an effective
depth of 500 mm. It is reinforced with 4 bars of 16 mm diameter. Calculate the area of steel
reinforcement provided.
Answer:
● Area of one bar: Area=π×(d2)2=π×(16 mm2)2=201.06 mm2\text{Area} = \pi \times \left(\
frac{d}{2}\right)^2 = \pi \times \left(\frac{16 \, \text{mm}}{2}\right)^2 = 201.06 \, \
text{mm}^2Area=π×(2d)2=π×(216mm)2=201.06mm2
● Total Area of Steel: Area=4×201.06 mm2=804.24 mm2\text{Area} = 4 \times 201.06 \, \
text{mm}^2 = 804.24 \, \text{mm}^2Area=4×201.06mm2=804.24mm2
4. Soil Mechanics
Question: Determine the bearing capacity of a soil with a cohesion value of 20 kN/m² and an
angle of internal friction of 30° for a square footing 1.5 m by 1.5 m.
Answer:
, ● Assumptions:
○ Factor of safety = 3
○ NqN_qNq and NcN_cNc values can be taken from standard bearing capacity
factors tables.
● Calculations:
○ Using Terzaghi's bearing capacity equation:
qnet=c×Nc+γ×Df×NqFactor of Safetyq_{net} = \frac{c \times N_c + \gamma \
times D_f \times N_q}{\text{Factor of Safety}}qnet=Factor of Safetyc×Nc+γ×Df
×Nq
Where:
■ ccc = 20 kN/m²,
■ NcN_cNc and NqN_qNq are bearing capacity factors for cohesion and
depth (from standard tables),
■ DfD_fDf is the depth of the foundation.
○ Final Answer: The specific values would require looking up the factors and
inputting them.
5. Fluid Mechanics
Question: Water flows through a pipe of diameter 300 mm at a rate of 0.3 m³/s. Calculate the
velocity of flow in the pipe.
Answer:
● Flow Rate (Q): Q=A×vQ = A \times vQ=A×v
● Area (A) of Pipe: A=π×d24=π×(0.3 m)24=0.0707 m2A = \frac{\pi \times d^2}{4} = \frac{\
pi \times (0.3 \, \text{m})^2}{4} = 0.0707 \, \text{m}^2A=4π×d2=4π×(0.3m)2=0.0707m2
● Velocity (v): v=QA=0.3 m3/s0.0707 m2=4.24 m/sv = \frac{Q}{A} = \frac{0.3 \, \text{m}^3/\
text{s}}{0.0707 \, \text{m}^2} = 4.24 \, \text{m/s}v=AQ=0.0707m20.3m3/s=4.24m/s
6. Surveying
Question: A horizontal distance of 100 m is measured along a slope with an angle of 5°.
Calculate the corresponding horizontal distance.
Answer:
● Horizontal Distance (L): L=D×cos(θ)L = D \times \cos(\theta)L=D×cos(θ)
● L=100×cos(5°)=99.6 mL = 100 \times \cos(5°) = 99.6 \, \text{m}L=100×cos(5°)=99.6m
7. Hydrology
Question: A catchment area of 500 hectares receives a rainfall of 100 mm over 24 hours. If
60% of the rainfall infiltrates into the ground, calculate the total runoff.
Answer:
, ● Rainfall Volume: Volume=Area×Rainfall Depth\text{Volume} = \text{Area} \times \
text{Rainfall Depth}Volume=Area×Rainfall Depth
● Area=500×104 m2\text{Area} = 500 \times 10^4 \, \text{m}^2Area=500×104m2 (since 1
hectare = 10,000 m²)
● Rainfall Depth: Rainfall Depth=0.1 m\text{Rainfall Depth} = 0.1 \, \text{m}Rainfall
Depth=0.1m
● Total Rainfall Volume: Volume=500×104×0.1=5×106 m3\text{Volume} = 500 \times
10^4 \times 0.1 = 5 \times 10^6 \, \text{m}^3Volume=500×104×0.1=5×106m3
● Runoff Volume: Runoff=40% of total rainfall volume=0.4×5×106=2×106 m3\text{Runoff}
= 40\% \, \text{of total rainfall volume} = 0.4 \times 5 \times 10^6 = 2 \times 10^6 \, \
text{m}^3Runoff=40%of total rainfall volume=0.4×5×106=2×106m3
8. Concrete Mix Design
Question: Design a concrete mix with the following data:
● Cement: 350 kg/m³
● Water: 175 kg/m³
● Coarse Aggregate: 1150 kg/m³
● Fine Aggregate: 750 kg/m³
● Water-Cement Ratio: 0.5
Determine the mix ratio by weight.
Answer:
● Cement Ratio: 1 part
● Water Ratio: 175350=0.5\frac{175}{350} = 0.5350175=0.5 parts
● Coarse Aggregate Ratio: 1150350=3.29\frac{1150}{350} = 3.293501150=3.29 parts
● Fine Aggregate Ratio: 750350=2.14\frac{750}{350} = 2.14350750=2.14 parts
Mix Ratio: 1:2.14:3.29:0.5 (Cement
Aggregate
Aggregate
)
9. Reinforced Concrete Design
Question: A rectangular beam has a width of 300 mm and an effective depth of 600 mm. It is
subjected to a bending moment of 150 kNm. Calculate the required area of tensile
reinforcement using M20 concrete and Fe415 steel.
Answer:
● Step 1: Calculate the moment of resistance.
MR=fy⋅As⋅d(1−Asfc⋅b⋅d)Partial safety factor for steelM_R = \frac{f_y \cdot A_s
\cdot d \left( 1 - \frac{A_s}{f_c \cdot b \cdot d} \right)}{\text{Partial safety
factor for steel}}MR=Partial safety factor for steelfy⋅As⋅d(1−fc⋅b⋅dAs)
1. Structural Analysis
Question: Calculate the reactions at the supports for a simply supported beam with a span of 6
meters, subjected to a uniformly distributed load (UDL) of 4 kN/m over the entire length.
Answer:
● Total Load: 4 kN/m×6 m=24 kN4 \, \text{kN/m} \times 6 \, \text{m} = 24 \,
\text{kN}4kN/m×6m=24kN
● Reactions at the Supports: Since it's a simply supported beam with symmetrical
loading,
○ Reaction at A (RA) = Reaction at B (RB) = 24 kN2=12 kN\frac{24 \, \text{kN}}{2} =
12 \, \text{kN}224kN=12kN.
2. Material Properties
Question: A steel bar with a cross-sectional area of 50 mm² is subjected to a tensile force of
100 kN. Calculate the stress in the bar.
Answer:
● Stress (σ\sigmaσ) = Force (F) / Area (A)
● σ=100 kN50 mm2=100×103 N50×10−6 m2=2000 MPa\sigma = \frac{100 \, \
text{kN}}{50 \, \text{mm}^2} = \frac{100 \times 10^3 \, \text{N}}{50 \
times 10^{-6} \, \text{m}^2} = 2000 \, \text{MPa}σ=50mm2100kN
=50×10−6m2100×103N=2000MPa
3. Design of Concrete Structures
Question: A rectangular reinforced concrete beam has a breadth of 300 mm and an effective
depth of 500 mm. It is reinforced with 4 bars of 16 mm diameter. Calculate the area of steel
reinforcement provided.
Answer:
● Area of one bar: Area=π×(d2)2=π×(16 mm2)2=201.06 mm2\text{Area} = \pi \times \left(\
frac{d}{2}\right)^2 = \pi \times \left(\frac{16 \, \text{mm}}{2}\right)^2 = 201.06 \, \
text{mm}^2Area=π×(2d)2=π×(216mm)2=201.06mm2
● Total Area of Steel: Area=4×201.06 mm2=804.24 mm2\text{Area} = 4 \times 201.06 \, \
text{mm}^2 = 804.24 \, \text{mm}^2Area=4×201.06mm2=804.24mm2
4. Soil Mechanics
Question: Determine the bearing capacity of a soil with a cohesion value of 20 kN/m² and an
angle of internal friction of 30° for a square footing 1.5 m by 1.5 m.
Answer:
, ● Assumptions:
○ Factor of safety = 3
○ NqN_qNq and NcN_cNc values can be taken from standard bearing capacity
factors tables.
● Calculations:
○ Using Terzaghi's bearing capacity equation:
qnet=c×Nc+γ×Df×NqFactor of Safetyq_{net} = \frac{c \times N_c + \gamma \
times D_f \times N_q}{\text{Factor of Safety}}qnet=Factor of Safetyc×Nc+γ×Df
×Nq
Where:
■ ccc = 20 kN/m²,
■ NcN_cNc and NqN_qNq are bearing capacity factors for cohesion and
depth (from standard tables),
■ DfD_fDf is the depth of the foundation.
○ Final Answer: The specific values would require looking up the factors and
inputting them.
5. Fluid Mechanics
Question: Water flows through a pipe of diameter 300 mm at a rate of 0.3 m³/s. Calculate the
velocity of flow in the pipe.
Answer:
● Flow Rate (Q): Q=A×vQ = A \times vQ=A×v
● Area (A) of Pipe: A=π×d24=π×(0.3 m)24=0.0707 m2A = \frac{\pi \times d^2}{4} = \frac{\
pi \times (0.3 \, \text{m})^2}{4} = 0.0707 \, \text{m}^2A=4π×d2=4π×(0.3m)2=0.0707m2
● Velocity (v): v=QA=0.3 m3/s0.0707 m2=4.24 m/sv = \frac{Q}{A} = \frac{0.3 \, \text{m}^3/\
text{s}}{0.0707 \, \text{m}^2} = 4.24 \, \text{m/s}v=AQ=0.0707m20.3m3/s=4.24m/s
6. Surveying
Question: A horizontal distance of 100 m is measured along a slope with an angle of 5°.
Calculate the corresponding horizontal distance.
Answer:
● Horizontal Distance (L): L=D×cos(θ)L = D \times \cos(\theta)L=D×cos(θ)
● L=100×cos(5°)=99.6 mL = 100 \times \cos(5°) = 99.6 \, \text{m}L=100×cos(5°)=99.6m
7. Hydrology
Question: A catchment area of 500 hectares receives a rainfall of 100 mm over 24 hours. If
60% of the rainfall infiltrates into the ground, calculate the total runoff.
Answer:
, ● Rainfall Volume: Volume=Area×Rainfall Depth\text{Volume} = \text{Area} \times \
text{Rainfall Depth}Volume=Area×Rainfall Depth
● Area=500×104 m2\text{Area} = 500 \times 10^4 \, \text{m}^2Area=500×104m2 (since 1
hectare = 10,000 m²)
● Rainfall Depth: Rainfall Depth=0.1 m\text{Rainfall Depth} = 0.1 \, \text{m}Rainfall
Depth=0.1m
● Total Rainfall Volume: Volume=500×104×0.1=5×106 m3\text{Volume} = 500 \times
10^4 \times 0.1 = 5 \times 10^6 \, \text{m}^3Volume=500×104×0.1=5×106m3
● Runoff Volume: Runoff=40% of total rainfall volume=0.4×5×106=2×106 m3\text{Runoff}
= 40\% \, \text{of total rainfall volume} = 0.4 \times 5 \times 10^6 = 2 \times 10^6 \, \
text{m}^3Runoff=40%of total rainfall volume=0.4×5×106=2×106m3
8. Concrete Mix Design
Question: Design a concrete mix with the following data:
● Cement: 350 kg/m³
● Water: 175 kg/m³
● Coarse Aggregate: 1150 kg/m³
● Fine Aggregate: 750 kg/m³
● Water-Cement Ratio: 0.5
Determine the mix ratio by weight.
Answer:
● Cement Ratio: 1 part
● Water Ratio: 175350=0.5\frac{175}{350} = 0.5350175=0.5 parts
● Coarse Aggregate Ratio: 1150350=3.29\frac{1150}{350} = 3.293501150=3.29 parts
● Fine Aggregate Ratio: 750350=2.14\frac{750}{350} = 2.14350750=2.14 parts
Mix Ratio: 1:2.14:3.29:0.5 (Cement
Aggregate
Aggregate
)
9. Reinforced Concrete Design
Question: A rectangular beam has a width of 300 mm and an effective depth of 600 mm. It is
subjected to a bending moment of 150 kNm. Calculate the required area of tensile
reinforcement using M20 concrete and Fe415 steel.
Answer:
● Step 1: Calculate the moment of resistance.
MR=fy⋅As⋅d(1−Asfc⋅b⋅d)Partial safety factor for steelM_R = \frac{f_y \cdot A_s
\cdot d \left( 1 - \frac{A_s}{f_c \cdot b \cdot d} \right)}{\text{Partial safety
factor for steel}}MR=Partial safety factor for steelfy⋅As⋅d(1−fc⋅b⋅dAs)
