PA PL ¹½/2 x 6 3″ 2 PL ¾/16 x 3 P

Q: PROBLEM 9.9 9.10 In the bolted connection of Figure 9.12, assume that the bolts are in. in diameter…

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Q: app 3.0. Investigate all limit states. The tension m gusset plate is A36. The weld is a 1/8-inch fil…

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Q: 6. Design the tension plate attached to the wide-flange (W) section as well the welds, assuming the…

A: Step 1:

Q: Determine the design tensile strength, Pn, of a 3/8 x 6 plate tension member shown in the figure.…

A: 38×6 plateFu = Ultimate strength = 58 ksiFy = Yield strength = 36 ksi

Q: Fillet Welds Oriented Longitudinally and Transversely to the Load The two 4 x 4x3/8 angles shown in…

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Q: 7. Design the welds indicated to develop the full strength of the angles and minimize eccentricity.…

A: Problem Statement:Design the welds indicated to develop the full strength of the angles and minimize…

Q: Determine the design strength for C-shaped welds. USE E70 electrodes and 5/16" weld. 2011d

A: Given, E70 electrodes Weld thickness (w) = 5/16"

Q: can be decided as the Class b section. Determine the design value of axial load capacity.

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Q: A lap welded joint of A36 steel plates must hold a tensile load. The lower plate is .5 inches thick…

A: Lower plate tensile load = 0.5 inch thickWidth of lower plate = 7.7 inch wideUpper plate tensile…

Q: Compute the maximum acceptable tensile SERVICE LOAD that may act on a single tee section that is…

A: Answer In order to calculate the minimum acceptable tensile service load we need to first find the…

Q: Determine the required diameter of a plug weld using E60 electrodes on A-36 steel to develop the…

A: We have to determine the value of required diameter of a plug weld using E60 electrodes on A36 steel…

Q: 5. Select a pair of channels and design the fillet welds using SMAW process. The loading is 85% live…

A: Step 1:For A36 SteelGiven:Live Load (LL) = 170 kipsDead Load (DL) = 30 kipsDesign Load…

Q: Problem 3. An un-equal leg angle L8x4x½ made of ASTM A242 steel (Fy = 50 ksi, Fu = 70 ksi) is used…

A: Unequal leg angle, L 8 x 4 x 1/2" of ASTM 242 steel, Fy=50 ksi , Fu= 70ksi To Find: (a) Factored…

Q: Determine the size of the weld using E60 electrode. Dead Load = 38 kip and LL = 95 kips (Service…

A: Nominal strength of  weld is given by Rn = fnw*Anw fnw = Nominal weld strength fnw = 0.6*fEXX  fnw =…

Q: A C130x10 tension member is welded to a plate as shown in the figure. The channel is of Grade 350W…

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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.