EN/m on a 6.0m e beam is 400 m n is 500 mm. Use ps - 10 mm diam and fy - 1146 A 114 6 N

Q: How much service live load, in kN/m, can be supported? The member weight is the only dead load. The…

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Q: Select an A992 W-shape for beam AB of the floor system shown in Figure. In addition to the weight of…

A: Solution; Given that; Beam spacing=5ft, Span Length=30 ft, L.L=80psf, Partial Load=20psf

Q: For the following designs, use A36 steel for the angles and A992 steel for the beam and column. Use…

A: Consider free body diagram;

Q: DETERMINE THE FORCE (b) FOR MEMBER AC GIVEN THE DATA BELOW F-599 tb F-1.283 lb "Note for…

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Q: *6-124 The member has a brass core bonded to a steel cas- ing. If a couple moment of 8 kN m is…

A: Step 1: Step 2: Step 3: Step 4:

Q: 2.5 The steel propeller shaft ABCD carries the axial loads shown in Fig. (a). Determine the change…

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Q: Kindly please get that acting force on member BF

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Q: For the inverse T-section beam that loaded as shown in figure, find the maximum tensile and…

A: A double overhanging beam is given we need to find out maximum tensile and compressive bending…

Q: H.W -4: For the beam of sections and detail shown in Fig. find the maximum live load (PL) that the…

A: at section a-a b = 350 mmh = 600 mmfc' = 35 MPafy = 420 MPaArea of steel at bottom = 5- dia 25…

Q: A bar AE is in equilibrium under the action of the five forces shown. Determine T (kN). F1 88 kN F2-…

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Q: RD RA R. = 40 KN VD1² VD2 = M B1 82 2 m Mg2 MB1 hinge t-4m. Vpz KN KN 2 kN/m KN KN KN KN kNm kNm kNm…

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Q: 4.8 Member AB is the beam under consideration. As shown in the illustration of the loading…

A: in the question, the free body  diagram is provided also the load of 500-lb/ft uniformly distributed…

Q: HW Draw S.t,B.M and elasticaiagams for the beam Shown beloi by using method of equations? low 4 KW…

A: Introduction : The sum of all transverse forces applied on either side of a portion of a beam…

Q: Find solution steel structure 

A: The image you sent is a part of a (civil engineering) design problem. The text specifies a member AB…

Q: 4 4a 4b 4c The designed axially loaded column is W 4 x 13 with the following data: kg/m |d=105.66 mm…

A: For W 4x13  ry = 25.4 mm  Ag = 2471 mm2

Q: a. Calculate the critical slenderness ratio of the W section if the column is braced at midpoint of…

A: Given, L = 6m or 6000 mm DL = 900 kN  LL = 400 kN  Fy = 345 MPa

Q: For the beam shown below: a) Calculate the elastic curve equation Ely Mo A = R³L 6 Mox³L + cx + C₂…

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Q: solve force in members DE, EK, and KJ

A: ASKED: To determine the force in members DE, EK and KJ.

Q: The beam shown in the figure has continuous lateral support of both flanges. The uniform load is a…

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Q: The beam is loaded as shown. which of the following nearly gives the maximum momen of the beam in…

A: Concept:- If the loading is symmetrical then load applied on the beam will be equally divided…

Q: F2-5. The force F 450 lb acts on the frame. Resolve this force into components acting along members…

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Q: B2,8m RC Beem Sec B-B Sec A-A C1 (0,8x0.8)m Sec A-A7 Sec Sec 8-B Sec A-A Y-Y 143.9* Sec Sec A-A Sec…

A: Radius of the beam = 3.165 m Circumference= Total length of three beams = 2*π*3.185= 20.012 m Length…

Q: assume EI is constant. 2 k/ft What is MAB in k-ft? 180 24ft -108 Select the correct response: B ₁12k…

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Q: Determine the values of S and Z and shape factor about horizontal x-axis. 14-in 2.5-in 2.5-in 1/2…

A: y = A1y + Ay2 +Ay3 A1 + A2 + A3⇒y = .5×6×3 + 16×.5×6.25 + .5×6×3.5×6 + 16×.5 + .5×6⇒y = 4.8571 in…

Q: F4-35 F4-36. Replace the loading shown by an equivalent single resultant force and specify the x and…

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Q: -L/2- -L/2- B WO Identify the equation of the elastic curve for the cantilever beam AB.

A: A cantilever beam is given of span L and free at left end and fixed at right end.To determine the…

Q: View Policies Current Attempt in Progress The simply supported beam consists of a W460 x 82…

A: Given:- P=62 KNw=28 KN/mLAB=LBC=2.9 mLCD=1.9 mE=200 GPaI=370×106  mm4 To find:- To calculate Vc at…

Q: Design the beam shown below according to the ACI 318 requirements. The beam is to have a uniform…

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Q: Compute the maximum deflection for the beam loaded as shown below 400 Ib 400 Ib 160 Ib/ft 100 Ib/ft…

A: To find Maximum deflection for the beam

Q: Compound axial member ABC has a uniform diameter of d= 1.50 in. • Segment (1) is an aluminum [E1 =…

A: step 1-

Q: Select an A992 (Es= 345 MPa and Fu= 450 MPa) W-shape for the floor beam AB of the floor system…

A: Given Data: The thickness of the floor is, t=125 mm The unit weight of concrete is, wc=24 kN/m3 The…

Q: For the beam(300x800mm) to carried the loads shown in the figure below,design the steel…

A: Given values: Size of beam: (Width b * Depth D) =  300 * 800 mm Span = 9 m f'c = 25 MPa fy = 414 MPa…

Q: The beam shown in Figure has lateral support only at the ends. The uniform load is a superimposed…

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Q: Q14: A singly RC beam has effective dimension of 250 mm x 450 mm. It is reinforced with 3 nos 20 mm…

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Q: 3.5 A simply supported beam is loaded as shown in the given figure. The bending moment at Ewould be…

A: Given: A simply supported beam figure,

Q: Use Singularity functions to solve for the deflection at the middle of the beam

A: When we write function of w0 it assumes to be acting over the entire length , hence to compensate…

Q: 8= Pb²t 2²- 46€1 8₁=8₁482 MEN -SHE AR NOMENT LOADED WITH CROSS SECTION AS SHOWN USING THE B. BEAM…

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Q: Tabulate the result of your design. In 2 columns, list the tension members on the 1st column and the…

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Q: Find solution steel structure

A: The text in the image asks you to design member AB within the frame shown in figure (2). The applied…

Q: 94-Det simply supported boam shown in fiy. use Pey = loo sin 4t %3D Pt) %3D m = 50 per unit length C…

A: The deflection of the simply supported beam with udl is to be calculated or standard expression to…

Q: 100kn 2okn 25kW 50kw A 30° 300 し 3m 3m 3m 20kN 2okn 20ka Solve for FAB, FCH, FeD, FOF inticate units…

A: WELL ,  WE CAN DETERMINE THE FORCES IN THE MEMBERS USING METHOD OF JOINTS.

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.