Check the factor of safety against shear failure (bearing capacity) for the rectangular footing shown in Fig. (F.S= 2.5) and calculate consolidation settlement of the foundation 4 3 m W.T Fine Silty 6 m 1m V = 1000 kN Ycon. = 24 kN/m³ Ys = 19 kN/m³ $=25° G₁=2.7 Y₁ = 18 kN/m³ C₁ = 60 kN/m² G₁ = 2.65 G. s 7777 X1 m Dense Sand y= 20 kN/m³ =35° 10 m 6 m

Q: 3 Given: T- shape foundation as shown in figure is loaded with a uniform load of 120 kN/m².…

A: T shaped foundation loaded with uniform load q = 120 kPa

Q: For the rigid shallow foundation (2*4m) shown in Fig, calculate Immediate settlement the center of…

A: Given: net pressure, qo = 100 kPa Poisson ratio, μ=0.3 B = 2 m L = 4 m H = 3.5+0.5=4 m

Q: Figure Q1 shows two parallel strip footings 3m wide and 5m apart (measured from centre to centre),…

A: Given: The width of the footing is 3 m. The distance between footings is 5 m. The compressive…

Q: Circular foundation diameter B = 2m G.S. 1.0т Sand q = 150 kN/m² 0.5m W.T. Sand 0.5m Normally…

A: As per our company guidelines, we are supposed to answer only first three sub-parts. Kindly re-post…

Q: 2 ft 24 ft 24 ft 24 ft 2 ft Problem 4 B E F G 3 ft DL=190 kip DL-100 kip DL=180 kip LL = 60 kip LL=…

A: Given data: Size of the mat = 76 ft by 96 ft Column size = 24 in. by 24 in. qall(net)=1.5 kip/ft2…

Q: The 2.5 m x 3.5 m rectangular spread footing placed at a depth of 1.5 m supports a vertical load of…

A: To find:Factor of safety

Q: = Figure 2 shows a rectangular shallow foundation. The foundation measures 1.5 m x3 m (B x L) in…

A: Average effective stress on the clay layer due to applied foundation load is 24 KN/m2.Determine the…

Q: 1. Find the increase of pressure at point A due to loading q=400 kN/m² on the given rectangular…

A: Given data in question  Intensity 400 kN/m2 Dimensions 8×4 m To find out  Increase in pressure at A

Q: 3. A column foundation (Figure 1) is 3 m x 2 m in plane. Given: D;= 2m, o=25°, c = 50 kN/m². Using…

A: Since you have asked multiple questions, we will solve the first question for you. If you want any…

Q: A square column foundation with base of 1.40 m is shown in the figure. It carries an axial load of…

A:

Q: 3,500 kN/m2; and µs = 0.35. (Useful formula and table can be found from Appendix)

A: 5 images are attached to solve the following question

Q: 9.6 A mat foundation is shown in Figure P9.6. The design considerations are L = 12 m, B = 10 m, Dy =…

A: Length of foundation= 12 m Width of foundation= 10 m Depth of foundation= 2.2 m

Q: 7.7 Refer to Figure P7.7. Using the procedure outlined in Section 7.10, determine the average stress…

A: The given figure is shown below:

Q: 8.4 A rectangular foundation is shown in Figure P8.2, given B= 2 m, L=4m q= 240 kN/m², H= 6m, and D;…

A:

Q: Circular foundation diameter B = 2m G.S. 1.0m Sand q = 150 kN/m² y = 17 kNhm² 0.5m W.T. Sand 0.5m 7…

A: Given data-

Load and resistance factor design

Load and resistance factors design (LRFD), also known as Limit state design (LSD), refers to the design method used in building engineering. A boundary condition is a condition of a structure where it can no longer meet the relevant design requirements. The condition may refer to the load level or other actions on the structure, while the terms refer to the integrity of the structure, suitability for use, durability, or other design requirements. A structure designed for LRFD is limited to keeping track of all possible actions during its design time and is always ready for use, with the appropriate level of reliability in each boundary condition. LRFD-based construction codes clearly define the appropriate levels of reliability in their instructions.