02. A simply supported beam of 6.75 m span is to support a uniformly distributed load of 40 kN/m (inclusive of self weight). Design the beam making use of MB 300(or) MB 400 sections which are available in the stores. If required flange plates may be used. The beam may be assumed to be laterally supported. For given sections fbc = fr = 165 MPa fs = 100 MPa E = 2x10° MPa For ISMB 300: I = 89.90×10ʻ mm Z = 59.90x10ʻ mm? Br= 140 mm , tw = 7.7 mm For ISMB 400: L= 205×10°mm Z, = 102×10ʻmm? Br=140 mm, tw= 8.9 mm CSScanned with CamSceer

02. A simply supported beam of 6.75 m span is to support a uniformly distributed load of 40 kN/m (inclusive of self weight). Design the beam making use of MB 300(or) MB 400 sections which are available in the stores. If required flange plates may be used. The beam may be assumed to be laterally supported. For given sections fbc = fr = 165 MPa fs = 100 MPa E = 2x10° MPa For ISMB 300: I = 89.90×10ʻ mm Z = 59.90x10ʻ mm? Br= 140 mm , tw = 7.7 mm For ISMB 400: L= 205×10°mm Z, = 102×10ʻmm? Br=140 mm, tw= 8.9 mm CSScanned with CamSceer

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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note: SN is equal to 18, topload will now be 118kN
A W36x210 can be used for Transfer Girder ‘A’ as shown in the elevation next page.Assume the LRFD loads on the ‘Transfer Girder’ are as follows:Pu= 200 kips from the column above. Assume also that the loads from self-weight of thewu = 2.0 kip/ft (girder self weight and beams framing into this transfer girder are equivalentto a uniformly distributed load). These loads are shown in the last figure.a) Check whether the W36x210 ‘Transfer Girder’ is adequate for bending and shear.Assume the unbraced length Lb =20 ft.b) Suppose that service loads, w and P, are both 50% dead load and 50% live load. Does theW36x210 satisfy a service live load deflection limit of L/480?
6.9-2 Use A992 steel and select a structural tee shape for the top chord of the truss shown in Figure P6.9-2. This is the truss of Example 3.15. The trusses are spaced at 25 feet on centers and support W6 X 12 purlins at the joints and midway between the joints. The other pertinent data are summarized as follows: Metal deck: 2 psf Built-up roof: 5 psf Snow: 18 psf of horizontal roof projection a. Use LRFD. b. Use ASD. 46.6' 3' 3' 12' 3' 3' 2 @ -12' = 24- 2 @ 42' -12' = 24' 90 FIGURE P6.9-2
W 14 x 142 is used as a column having a length of 8.48 m. long. It is hinged at the r end and fixed at the lower end but there is a lateral bracing perpendicular to minor axis of the W section at a point 4.78 m. above the bottom support. It is assumed to be pinned connected at the bracing point. Using A 36 steel Fy= 248 MPa the NSCP Specificatioris. E, = 200000 MPa. Properties of W 14 x 142 A = 26967.69 mm2 d= 374.65 mm bf= 393.70 mm tf= 27.00 mm tw= 17.27 mm4 Ix= 695.11 x 106 mm4 Iy = 274.71 x 106 mm4 rx = 160.53 mm rY = 100.84 mm Assume: Kx = 0.80 for 8.48 m. length KY = 1.9 for the length on the upper support KY = 0.80 for the length on the bottom support Compute the slenderness ratio along y-axis. Please show clear solution thank you
6-1.2-1: Find lightest W14 for L = 20', P = 600k E = 29,000ksi, Fb = 30ksi and Fv = 20ksi. 30. CONTINUOUS BEAM - TWO EQUAL SPANS-CONCENTRATED LOAD AT CENTER OF ONE SPAN Total Equiv. Uniform Load 13 p 8 P 2 R,= V₁.. 32 R R₂ R₁ R₂= V₁ + V₂. 11 p 16 V₁ R= V₁ V₂ Mmax (at point of load). M. (at support R)..... .......... Amax (at 0.480 / from R.).... M Imax Shear Moment M₁ V₂ ********* ***************** 11 = 11 P 32 - 19 p 32 13 64 = 3. Pl PI 32 0.015 P³ El
Solve by LRFD (AISC ) masnual
A W 16 x50 is used as a beans to carry a uniform we and dead load including own weight of 37 kN/m and an axial tension load of T acting through the centroid, of the member. It has a simple span of 5.7 m. The compression flange of the member is laterally supported against local buckling.Use A 36 steel, Fy = 248 MPa.Properties of W 16x 50A = 9483.85 mm^2d= 412.75 mmbf= 179.65 mmtf= 15 95 mmtw= 9.65 mmSx= 1324 × 10^3 mm^3Sy = 172 x 10^3 mm^3 Questions:Determine the Safe axial load T that the beam could carry Hint Answer: 432.69 KN
Solve it.
A bracket of length 1.5m is rigidly attached to the simply-supported beam at its end. The beam is subjected to the combined effect of M,, M, and compressive force P caused by the factored loads as shown. There are lateral supports only at the ends. The member is made of HE300A section and steel grade is S275 (Fy-275MPA, Fu= 430 MPa). The loads acting are determined from LRFD combinations. Determine whether the member can safely carry the applied loads. Multiply first-order moment diagrams by a factor of 1.2 to account for second-order effects. No need to check shear limit state. You do not need to additionally consider the beam self-weight. E Use the following simple expression to calculate the value of L:: L, = T*rts 0.7*Fy y P = 400kN 1.5 m F = 150kN 2 т 4 т X: Lateral support НЕЗ00А
steel structure (AISC ) masnual
A W420x85 steel beam is fully restrained with a uniformly distributed live load of 25 kN/m. The beam has a span of 10m. Use A36 steel. Properties of W40 x 85 A = 10839 mm2 %D bf = 180 mm tf = 18 mm %3D tw = = 11 mm Ix = 310 x 106 mm4 %3D d = 420 mm (Note: Since this is a built-up section, for checking of web slenderness, use h = d - 2t;) %3D 1) Calculate the design shear strength of the section. Check if the section is adequate in shear. USE LRFD and U = 1.2D + 1.6L. %3D 2) Is the maximum deflection at the midspan of the beam due to live load adequate? 3) Calculate the maximum uniform live load that can be carried by the beam.