Determine the force in each member of the loaded truss. All triangles are 3-4-5. Enter a positive number if the member is in tension, negative if in compression. 11 kN H 34KN 4 panels at 8 m- 20 KN D F E
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- For the given loads F1=42 KN, F2=48 KN, F3=45 KN, F4=32 KN (Assume internal forces are in tension) 1- Magnitude of reaction at K is (KN) :(a. 151 - b. 226.5 - c. 302 - d. 377.5 - e. 453 ) 2- internal force in member JH (KN) :(a. -327 - b. -218 - c. -436 - d. -545 - e. -654) 3- internal force in member IH (KN) : (a. -204.600222 - b. -136.400148 - c. -272.800296 - d. -341.00037 - e. -409.200444 ) 4- Magnitude of reaction at A is (KN) 5- internal force in member IK (KN) is 6- internal force in member GF (KN)The frame is used to support the wood deck in a residential dwelling. Sketch the loading that acts along the members BG and ABCD. Set b=4.5m, a=2.4m. Use Live Load=1.92 kPa.34 - A support is a fixed, B support is a sliding joint. loads M= 30 kNm, P1= 12 kN, P2= 13 kN, q1= 6 kN ⁄ m, q2= 7 kN ⁄ m , spans are a= 2 m, b= 3 m. The support reactions in the beam whose loading condition is given in the figure will be found. Accordingly, Ax = ?A) 22.25B) 0C) None.D) 39.25E) 21/2
- The following frame has pin connections at A and B. If F1 = 60lbs, F2 = 45 lbs, a = 4.3 ft., b = 3.0 ft., c = 4.9 ft., and has atriangular distributed load with a max loading of wpeak = 14 lb/ft,Determine the connection forces at C acting on member BCD .Prove that the truss is statically determinate and calculate the internal forces in each truss member using the method of joints or the method of sections for a load of F=350N with workings.Find the reactions to girder G-2 (A3-D3). Start with beams B-1 and B-3, then girder G-4. The decking has a uniform load of 110 lb/ft2: Assume all beams are weightless.
- The beam CBA is supported by a ball and socket support at A and two cables BD,BE as shown. It is loaded by two forces F=945 KN and P=315 KN.A short rectangular column 300 mm on one side and 400 mm on the other side. It is reinforced with 8-20-mm-diameter (28) longhitudinal bars equally distributed to the shorte sides of the column. Use f'c = 21 MPa and fy = 415 MPa. Calculate the required spacing of 10-mm-diameter ties, s (mm). Calculate the nominal axial strength of the column, Pn (kN). Calculate the maximum ultimate axial load the column can carry, Pu (kN)A W section steel purlin span 6.7 m between roof trusses on centers. The roof is assumed to support a dead load of 880 N/m2 of roof surface including self-weight and a live load of 816 N/m2 of horizontal roof surface projection. The slope of the roof truss is 1 vertical to 2 horizontal and the purlins are spaced 1 m on centers. Use A36 with Fy = 248 MPa. Assume all loads pass through the center of gravity of the section. Sag rods are to be placed at the middle thirds between trusses. Determine the ratio of the actual to the allowable bending stress. DO NOT ROUND-OFF DURING THE COURSE OF SOLVING. ANSWER IN FOUR DECIMAL PLACES. Properties of W Section A = 2,887 mm2 d = 192.5 mm bf = 100.7 mm tf = 10.82 mm Sx = 174,900 mm3 Sy = 36,133 mm3 tw = 6.1 mm Use the following allowable stresses (NSCP 2001): For bending about strong axis, Fbx=0.66 Fy For bending about weak axis, Fby=0.75Fy
- Find the max compression on member DI in kN. Put "0" if none. a= 4.5m b= 4.8m c= 50kN d= 68kNThe steel framework is used to support the reinforced stone concrete slab that is used for an office. The slab is 200 mm thick. Sketch the loading that acts along members BE and FED. Take a = 3 m, b = 4 m. Hint: See Tables 1.2 and 1.4.Identify the maximum shear and maximum moment. Answers: Vmax = 28.98 kN, Mmax = -85.25 kN-m. Kindly give me the solution of this please.