Identify types of each slab panel and label the shear force coefficient for loading distribution according to Table 3.15 BS8110. (*all unit in m) B1 A B 5 0.5 3 5.5 2.5 3 6.5 6.5 5

Q: Two wooden planks, each 18 mm thick and 250 mm wide, are joined by the glued mortise joint shown.…

A: Given Thickness 18 mmLoad  19 kNCut d = 30 mmTo find out Maximum ave shear stress

Q: A square footing 2 m x 0.36 m thick supports a 300 mm square column at its center. Column loads are…

A:

Q: A square footing 2.3 m x 0.35 m thick supports a 300 mm square column at its center. Column loads…

A: The thickness of footing is designed based on the  Bending moment acting along the critical…

Q: Point A is located at coordinates (2.5, 1.5, -10) Point B is located at coordinates (2, 1,-5) Point…

A: Answer  We know that the increase in stress under a corner point of rectangle is given by  ∆σz =…

Q: For the figure shown. Design the upper leftmost corner slab. The dead load pressure is 5.19kpa and…

A:

Q: 5.1 Use the approximate bending moment equation, Mu = 4 As d to calculate the bending moment…

A: Answer  We are given  total h = 6 in concrete cover = 0.75 inch f'c = 4000 psi fy = 60000 psi…

Q: Two wood boards, each 18 mm thick, are joined by the glued finger joint shown. The finger joint will…

A: Asked: To determine the shortest allowable length of the cuts in the glued finger joint.   Given:…

Q: square footing 2.2 m x 0.31 m thick supports a 300 mm square column at its center. Column loads are…

A:

Q: SITUATION 1 For the figure shown, the diameter of holes is 20mm. Steel plate is made up of 10…

A: Calculating the diameter of the bolt hole on plate: dh=db+1.5   =20+1.5   =21.5 mm

Q: The lumped section shown has an internal shear force of 3000 lb. All lumped areas are 0.5 in² and…

A: internal shear force=3000lblumped area=0.5in2web thickness=0.050"

Methods to Determine Vertical Stress

The load applied to the soil is transferred to the underground. The presence of water at soil pores and solid grains above the soil are the primary components that are responsible for the effective load transfer. Due to the load acting, the internal component of the soil or the underground formations undergoes deformations or a tectonic shift. The amount of deformation depends upon the amount of load acting on the soil surface, which indeed, depends on the weight of everything present above the soil. Due to this, the underground soil element develops internal stresses which try to resist the tectonic changes.