2. Design the compression member in Problem 1 using W14 and A-50 steel. Use ASD with PDL= 990 KN AND PLL= 680 KN. 米 mu 2.5m 3.5m 3.5m

Q: 8. A W12 × 65 of A572 Grade 60 (414 Mpa) steel is used as a compression member. It is 7.9m long,…

A: Given, W12 x 65 Fy = 414 MPa DL = 600 kN  LL = 1420 kN

Q: Check the beam shown for compliance with the AISCS. Lateral support is provided only at the ends,…

A: Load=70 Sin       =70*(4/5)       =56 KN       =12.58 kipsDead Load=30% of 12.58=3.78 kipsLive…

Q: An HSS 10 x 6 x 5/16 with Fy = 46 ksi is used as a column. The length is 16 feet. Both ends are…

A: HSS 10 x 6 x 5/16 Fy = 46 ksi L = 16 ft

Q: 3#9 12" 21" 3" tc = 4 ksi Grade 60 steel. NWC. M = 60 ft-k The sectim is cracked Computer the…

A:

Q: A W610 x 82 beam of A 36 steel has end reaction of 302 kN and is supported on a plate with a length…

A: Yield Strength :Yield strength of an element is a maximum stress an element or component can bear…

Q: SOLVE FOR THE FORCE IN ALL MEMBERS OF THE TRUSS. SHOW SOLUTIONS CLEARLY USING METHOD OF JOINTS AND…

A:

Q: Calculate the development length at simply supported edge in mm from the following information: M25…

A: Side way clear spacing is 50 mm,M25 and Fe 415 steelbd =1.4 MPaV = 30 KN and M = 16.076 KN-md = 120…

Q: a. Nominal moment capacity of the overhanging portion of the section. b. Depth of the compression…

A: Solution: Givendata: bf = 750 mm tf = 50 mm d = 750 mm bw = 350 mm Mu = 816 kN.m fc' = 20.7 Mpa fy =…

Q: Use LRFD method to find maximum distributed live load (W₂) that can be applied on the concrete slab…

A: Built-up column:Section: 4 L6X6X1Effective length, KL = 8 ftA992 grade of steel;fy = 50 ksifu = 65…

Q: 3-3 A rectangular section with width b=200mm and depth h=450mm is reinforc with 4 16 of HRB400. The…

A: Width (b) = 200 mmDepth (h) = 450 mmReinforcement of concrete = 4 -16 diameter HRB400Design Moment…

Q: Use "Unit Load method" & Mueller-Breslau Principle Please Influence Lines for: MA VA MB V8 0.5L 0.5L

A:

Q: A compression member is built up from a W14x90 and a W10x49, both of A992 steel. a. Compute rx and…

A: We need to first obtain the Moment of inertia about X and Y axis We can determine the moment of…

Braced cuts

The braced cut is a method of providing lateral support to the soil in foundation trenches, to prevent the trenches from deformation by the caving in of soil and resisting it from moving laterally. During shallow excavation in firm soil, generally, slopes are provided to resist the caving in of soil. During shallow excavation in soft soil and deep excavation, the chances of caving in of soil increase. Hence, it becomes necessary to make a provision to resist the deformation of trenches due to caving in of soil. The braced cut is the method used for such conditions. The excavation using braced cuts is a bit uneconomical as compared to the traditional open-cut excavation method but provides better safety than the cut excavation method.