5-17. A 24 ft axially loaded W12 x 96 column that has the bracing and end support condi- tions shown in the figure. (Ans. 1023.3 k LRFD; 680.4 k ASD) 10 ft 24 ft 14 ft х-х ахis У-у ахis FIGURE P5-17

Q: Determine the force in each member of the loaded truss. 6kN- Dybala SKN C 201 F E

A:

Q: Phe beam below 40KN Draw the Sf and BM digean 5m 5m 5m

A: To draw the SFD and BMD

Q: What is the reaction force at support C of this beam? A EL/2- -L/2 Multiple Choice (5wL 6P)/ 16 (6wL…

A: Given data : Point load=PUniformly distributed load =wTotal length=L

Q: e to sustain the loading applied to it, and that the des: al components are the bar BD and the pin…

A:

Q: *16-52 The simply supported beam carries a uniform load of 30 kN/m. Code restrictions, due to a…

A: Consider the free body diagram of the beam shown below. Consider the equilibrium moment about point…

Q: R12-5. The beam is constructed from four boards glued together at their seams. If the glue can…

A:

Q: Check the adequacy of column AB for applied live load. Fy= 350M

A: fy = 350 MPaCompression member:W 360 x 51A = 6450 mm2 rx = 148 mmry = 38.8 mmd = 355 mmb = 171 mmt =…

Q: F3-6. Determine the force in each member of the truss and state whether it is in tension or…

A: Determine the force in each member of the truss and state whether it is in tension or compression.

Q: 12 kN P N. 4 m 12 kN M K L. F 4 m 32 kN H. 6 m DX E - 10 m -- tom 10 m --| 16 m 16 m

A:

Q: Use the method of joint to determine the Force in all members of next trusses

A: The truss with loading diagram is shown below.

Q: Q-7b, An RC square column of size 600 mm is constructed to carry an unfactored load of 3000 kN. The…

A: An RC square column of size 600 mm,Unfactored load P=3000 kN,The ends of the column are restrained…

Q: pin at C is FS, = 3.3. (a) Determine the required thickness, 1, of the rectangular bar BD, whose…

A:

Q: Member-AB is_____. options:   a two-force member   a multi-force member   a zero-force member…

A: Note :- According to Bartleby guidelines, Since you have posted a question with multiple sub-parts,…

Q: 2-17. The 150-lb force of Fig. P2-17 is the resultant of the two forces shown and the couple and two…

A:

Q: What does the moment diagram look like 分 1 24 P →>> H An L

A:

Q: 12. Determine the axial forces in the indicated members of the following trusses using the optimal…

A: Solution

Q: Given: L1 = 8m S1 = 2.5 m L2 = 8 m S2 = 3 m L3 = 8 m L4 = 8 m Total dead load = 4.6 kPa Live load =…

A:

Q: L 3886 A D В

A: In order to draw the influence line for the member CJ, let us cut a section xx' from the member CJ…

Q: Date No. Cheek Deflection for Simply supported Girder Bridga :- Span L-20 DL from Section o.w weight…

A: To check the deflection for the simply supported girder bridge as described in the provided image,…

Q: Design (LRFD) a column to support the loads: Dead load = 378 kN Live load = 622 kN The 6.096 m-long…

A: Given, DL = 378 kN  LL = 622 kN  L = 6.096 m  Fy = 345 MPa

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.