How much weight can a 4-meter-tall steel (E = 200 GPa) column of cross section 20 x 60 mm with one fixed end support assuming a safety factor of 2? (P = 1) P
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A: Ra = 15 kNRb = 15 kNRc = 0 kNExplanation:Please see attach photo for the solutions thank you.
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A: Pc= 400 KN PB=310 KNE(STEEL)=200 GPa E(Concrete)=30 GPa A2=1850 mm2d1=530 mmL1=4.8 m L2=5.8 m
Q: Show step by step sol
A: P1= 981 PaV1 = 1.53 m/sExplanation:
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- The floor system of a residential building (LL=1.9KPa) consists of 125mm thick reinforced concrete slab (unit weight = 23.6 KN/m^3) resting on three 8 m spanned RC floor beams spaced at 2.5 meters on centers. Size of RC beam is 200mm x 450mm. NSCP 2015 Load factors: 1.2DL & 1.6LL. Assume simple supports. Calculate the ultimate/factored live load in KN/m. Express your answer in 1 decimal place.A simply supported beam, having the following properties: 5.5 m length, Mcr = 30 kN-m, Ig = 7000x10^6 mm^4, Icr = 3000x10^6 mm^4, uniformly distributed dead load = 5 kN/m, uniformly distributed live load = 7 kN/m, Ec = 21,000 MPa. a) Determine the value of Ma (DL+LL).Pls help me I need answer for this asap. Thank you!! The frame is used to support the wood deck in a residential building. Calculate and illustrate the magnitude loading in kN/m acting at beam DE. Set a = 1.5 and b =4. Dead Load = 0.54 kN/m2Live Load = 1.92 kN/m2
- A steel column W300 x 203 kg/m section is used as to support an aial load. A=25740mm^2 Tw = 20mm. Fy= 415 Mpa Tf = 32mm. Lu= 8.5 m d = 340mm. Ix=5.16*10^ 8 mm^4 Iy= 1.65*10^8mm^4 E = 200GPA k = 1.0 Sidesway is prevented. Determine the ultimate design capacity of Column using NSCP 2015. Use Phi= 0.9.A 6.20 meter long A36 steel column is subjected to an axial compressive load, P= 2500 kN and a lateral force, F as shown that caused bending about the strong axis of the column. Compute for the safe lateral force, F in kN. Use Fy = 250 Mpa Es = 200,000 Mpa, and Cm value equal to 1.0 155 mm 2500 kN 12 mm 3.10 m 250 mm tw- 10 mm F(kN) -12 mmThe steel framework shown below is used to support the reinforced concrete slab (unit weight of concrete =23.6KN/m3) that is used for an office use (minimum live load = 2.4KPa). The slab is 150 mm thick. Use a=3m, b=5m. Neglect the self-weight of the beam and girder. Assume all beams and girders have simple supports. NSCP 2015 Load factors: 1.2DL & 1.6LL. Calculate the total ultimate/factored liveload in Kilo Pascals? Express your answer in 2 decimal places
- Determine the pure axial load capacity (kN) of the given tied column. Express your answer in whole number. B = 650 mm db = 32 mm f'c = 32 MPa fy = 420 MPaA steel structural member is to be used as a column 5 m long. The column is to be pinned at both ends for bending about X-X axis and fixed at one end and pinned at the other for bending about Y-Y axis. The properties of the member section are as follows:E= 200 GPa, yield stress= 350 MPa, Ixx = 222 x 106 mm4, Iyy = 72.71x106 mm4 and A= 12.33 x 103 mm2Using Perry-Roberston equation, calculate the buckling stress and the corresponding loadA steel column W300 x 203 kg/m section is used as to support an aial load. A=25740mm^2 Tw = 20mm. Fy= 415 Mpa Tf = 32mm. Lu= 4m d = 340mm. Ix=5.16*10^ 8 mm^4 Iy= 1.65*10^8mm^4 E = 200GPA k = 1.0 Sidesway is prevented. Determine the ultimate design capacity of Column using NSCP 2015. Use Phi= 0.9. 7718.11kN C. 8575.88 kN 333.17 kN D. 20356.06kN
- A concrete circular column with a radius of 350mm has f'c=25 MPa and is braced against sidesway. The column is assumed to be pinned on both ends with an unsupported length is 6.1meters. The column is subjected to the following service loads: Axial Dead Load =350 kN Axial Live Load=480 kN Moment due to Dead Load on the top of the column = 125 kN-m Moment due to Dead Load on the bottom of the column = 60 kN-m Moment due to Live Load on the top of the column = 122 kN-m Moment due to Live Load on the bottom of the column = 75 kN-m. If the moments bent the column in a singular curve, determine the following: 1. Critical Load on the column, kN 2. Moment Magnifying factor 3. Design Moment of the column, kN-mA tubular steel strut 2 m long, both ends hinged, must have a buckling load of120 kN, based on the Euler formula. The outside diameter is 60 mm. Calculatethe thickness required. What will be the factor of safety according to Rankine's formula if oc = 350 MPa and Rankine's constant is 1/7500 for hinged ends? (E = 200 GPa.)A circular aluminum column 6 m long has a diameter of 200 mm. Determine the critical load if one end is recessed. E = 73 GPa. σY = 250 MPa.