Q3// Find spacing required for stirrups 08mm at zone A and zone B forthe beam shown with the factored loads, use, fé = 25MPA, fy = 300MPA.%3D%3DPu = 42kNPu = 14kN65mmWu = 40kN/m435mm1.2 m1 m250mm-Vu = 144kN96KN54KNZone A14kNZone B

Q3// Find spacing required for stirrups Ø8mm at zone A and zone B for the beam shown with the factored loads, use, f = 25MPa, fy = 300MP%. %3D Pu = 42kN Pu = 14kN 65mm Wu 40kN/m f111111| 435mm 1.2 m 1m 250mm- Vu = 144kN %3D 96kN 54KN Zone A 14KN Zone B

Q3// Find spacing required for stirrups Ø8mm at zone A and zone B for the beam shown with the factored loads, use, f = 25MPa, fy = 300MP%. %3D Pu = 42kN Pu = 14kN 65mm Wu 40kN/m f111111| 435mm 1.2 m 1m 250mm- Vu = 144kN %3D 96kN 54KN Zone A 14KN Zone B

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

A simply-supported T-beam is subjected to two pairs of distributed loads: pair 1 (wD1, wL1)and pair 2 (wD2, wL2). The geometric, material and loading properties are all given on thetable above. bf=1368mm bw=408mm t=248mm fc'=25.2MPa fy=336MPa d=918mm d'=92mm Lbeam=12.2m wD1=42.2kn/m wL1=50.2kn/m wD2=170.2kn/m wL2=193.3kn/m1. Solve for the height of the compression stress block using pair 1. 2. Solve for the theoretical steel area using pair 1. 3. Using Es = 0.004, solve for the theoretical steel area using pair 2. 4. Using Es= 0.005, solve for the theoretical steel area using pair 2.
A W 250 x 49 steel column 6.0m long must support an axial load of 140kN plus an eccentric load p applied on the minor axis at a point 178mm from major axis. Determine the value of P using AISC Specs, with Fb = 138Mpa, Fy = 250Mpa, and k = 1.0. Properties of W 250 x 49: A = 650mm2, Sx = 572 x 103mm3, ry = 49.20mm, rx = 106mm. Using AISC Column-Interaction formula: fa/Fa + fb/Fb = 1.0 where, fa = actual axial compressive stress = F/A, fb = actual compressive bending stress = M/Sx, Fa = allowable axial compressive stress, Fb = allowable compressive bending stress.
Help, FEM: Given a differential eqn : D2 +3xD-6y =0. Find y(0.2) by Galeskin method.
At-beam has a width of flange equal to 700mm and a thickness of 100mm. It has an effective depth of 450mm and the width of the web section is 350mm. It is reinforced at the bottom with a steel area of 2925sq.mm. fc'=17.24MPa, fy=413.7MPa. The depth of the compression block is _mm. A. 136.00mmB. 146.00mmC. 126.00mmD. 156.00mm
Steel Design Given: S = 2m L = 10m Superimposed dead load = 5 kPa Live load = 3.6 kPa Properties of Beam BFA section = W 460 x 97 kg/mArea (A) = 12320 mm^2Depth (d) = 465 mmFlange Width (bf) = 193 mmFlange Thickness (tf) = 19mmWeb Thickness (tw) = 11 mm Moment of InertiaIx = 445 x 10^6 mm^4Iy = 23 x 10^6 mm^4Yield Strength, Fy = 248 MPar (radius of gyration of a section compromising the compression flange plus 1/3 of the compression web area of compression flange = 50 mm. Consider bending about the x-axis. 1. The compression flanges are laterally supported only at midspan. With this condition, what is the permissible flexural stress if Cb = 1.0?
A W410 x 46 beam section consisting of A 36 steel Fy = 248 MPa, Fu = 400 MPa has twoholes for 22mm diameter bolts in its tension flange. Fb = 0.66Fy. Assume diameter of hole is3mm greater than bolt diameter.Properties of W410 x 46A = 5,890 mm2d = 403 mmbf = 140 mmtf = 11.2 mmtw = 7.0 mmw = 46.2 kg/mSx = 773 x 103 mm3Ix = 156 x 106 mm4a. Determine the effective tension flange area.b. Determine the amount of moment of inertia to be reduced due to the effect of holes.c. Determine the safe concentrated load that it could support at the midspan besides its ownweight if it has a simple span of 8m.
FOLLOW INSTRUCTIONS AND WITH FBD FOR UPVOTE THANKS. DO AMM ONLY. DO AMM ONLY
16 A short rectangular column 300 mm on one side and 400 mm on the other side is reinforced with 8-16-mm-diameter longitudinal bars equally distributed to all sides. Use f’c = 21 MPa and fy = 420 MPa. Calculate the required spacing (mm) of laterial ties 10-mm-diameter. A. 300 B. 250 C. 450 D. none of the above
Situation 6: A W18x97 simply supported beam carries a uniformly distributed load. The properties of the section relevant to the problem are as follow: Use A50 steel and Cb = 1.14d = 472 mmbf = 282 mmtf = 22.1 mmtw = 13.6 mmkdes=32.3 mmry = 67.3 mmlx = 728 x 10^6 mm^4ly = 83.7 x 10^6 mm^4Sx= 3080 x 10^3 mm^3Zx =3460 x 10^3 mm^3 J = 2440 x 10^3 mm^3Cw= 4240 x 10^9 mm^6 Which of the following nearly gives the Design Strength Moment (LRFD) if the span of the beam is 9 m?a) 666 kN-mb) 367 kN-mc) 781 kN-md) 381 kN-mNEED ASAP
P= 388 452 ND= 32 mm σ= ______________ksiASTM DESIGNATION: _____________
Situation 23: Given beam data and use 2010 NSCP: bw = 300mm d = 600mm fc’ = 21 MPa (λ=0) Stirrups = 2-legs 10mm dia. ϕ=0.75 The required spacing of the stirrups if Vu=52kN. The required spacing of the stirrups if Vu=105kN. The required spacing of the stirrups if Vu=275kN. The required spacing of the stirrups if Vu=535kN.
A 400x12 mm steel plate has a staggered bolt hole arranged as shown. Diameter of bolt is taken as 29mm. Use A36 steel. Consider that the member is being tensioned on both sides. 1.) Calculate the distance "b" in millimeters so that the net area in path 1-2-3-4 is equal to the net area of path 1-2-4. 2.) Using the calculated "b" from the previous problem, solve for the critical net area in mm^2 3.) Using the calculated "b" from the previous problem, solve for the design tensile strength in KN.