Square footing, B = 2 m, sandy soil, γ = 18 kN/m³, φ = 30°, Df = 1.5 m, c = 0. Find ultimate bearing capacity.

 Square footing, B = 2 m, sandy soil, γ = 18 kN/m³, φ = 30°, Df = 1.5 m, c = 0. Find ultimate bearing capacity.

Solution:

$$q_u = \gamma Df N_q + 0.4 \gamma B N_\gamma$$

• $N_q \approx 18.4, N_\gamma \approx 22.4$

$$q_u = 18 \cdot 1.5 \cdot 18.4 + 0.4 \cdot 18 \cdot 2 \cdot 22.4 = 858.7 \text{ kPa}$$

Answer: 0.859 MPa

Slump = 90 mm. Determine workability and suitability for beams/columns/slabs.

 Slump = 90 mm. Determine workability and suitability for beams/columns/slabs.

Solution:

• Slump 50–100 mm → Medium workability

• Medium workability suitable for beams, columns, slabs

Answer: Medium workability; suitable for beams, columns, slabs

Target mean strength f c k = 20 fck​=20 MPa, water/cement ratio = 0.5, water content = 200 kg/m³. Find cement content.

 Target mean strength 

$f_{ck} = 20$ MPa, water/cement ratio = 0.5, water content = 200 kg/m³. Find cement content.

Solution:

$$Cement = \frac{Water}{w/c} = 200/0.5 = 400 \text{ kg/m³}$$

Answer: 400 kg/m³

A simply supported beam, 6 m, carries central point load 12 kN. Find reactions and max bending moment.

 A simply supported beam, 6 m, carries central point load 12 kN. Find reactions and max bending moment.

Solution:

• Reactions: $R_A = R_B = 12/2 = 6 \text{ kN}$

• Max bending moment: $M_{max} = PL/4 = 12 \cdot 6/4 = 18 \text{ kNm}$

Answer: RA = RB = 6 kN, Mmax = 18 kNm

A simply supported beam, 8 m long, carries a uniformly distributed load of 10 kN/m over the entire span. Find: Reactions at supports Maximum bending moment

 A simply supported beam, 8 m long, carries a uniformly distributed load of 10 kN/m over the entire span. Find:

1. Reactions at supports

2. Maximum bending moment

Solution:

• Reaction at supports: $R_A = R_B = \frac{wL}{2} = \frac{10 \cdot 8}{2} = 40 \text{ kN}$

• Max bending moment (midspan): $M_{max} = \frac{wL^2}{8} = \frac{10 \cdot 8^2}{8} = 80 \text{ kNm}$

Answer:

• Reactions: 40 kN each

• Max bending moment: 80 kNm

slope distance D = 150 m measured at a slope of 12°. Find the horizontal distance.

 slope distance D = 150 m measured at a slope of 12°. Find the horizontal distance.

Solution:

Step 1: Horizontal distance formula

$$H = D \cos \theta$$ $$H = 150 \cos 12° \approx 150 \cdot 0.9781 \approx 146.7 m$$

✅ Answer: Horizontal distance = 146.7 m