The beam cross section shown below has been proposed for a short pedestrian bridge. The cross section will consist of two pipes that are welded to a rectangular web plate. Dimensions of the cross section are: h = 470 mmtw = 16 mmd = 120 mmt = 4.8 mm Additionally:• The area of each pipe is A = 1737 mm2.• The moment of inertia of the entire beam cross section about the z centroidal axis is IZ = 254790000 mm4. If the beam will be subjected to a shear force of V = 250 kN, determine the shear stress at point K, located at yK = 80 mm below the z centroidal axis. The beam cross section shown below has been proposed for a short pedestrian bridge. The cross section will consist of two pipes thatare welded to a rectangular web plate. Dimensions of the cross section are:h = 470 mmtw = 16 mmd = 120 mmt = 4.8 mmAdditionally:• The area of each pipe is A = 1737 mm².• The moment of inertia of the entire beam cross section about the z centroidal axis is lz=254790000 mm*.If the beam will be subjected to a shear force of V = 250 kN, determine the shear stress at point K, located at yk = 80 mm below the zcentroidal axis.HH.VKKAnswer:TK =NZMPaУнУк

Answered: Image /qna-images/answer/f1729258-f228-4d3a-89e3-d402aa47b1c5.jpg

The beam cross section shown below has been proposed for a short pedestrian bridge. The cross section will consist of two pipes that are welded to a rectangular web plate. Dimensions of the cross section are: h = 470 mm tw = 16 mm d = 120 mm t = 4.8 mm Additionally: • The area of each pipe is A = 1737 mm². • The moment of inertia of the entire beam cross section about the z centroidal axis is lz = 254790000 mm². If the beam will be subjected to a shear force of V = 250 kN, determine the shear stress at point K, located at yk = 80 mm below the z centroidal axis. H H.V K Answer: TK = N Z MPa Ун Ук h

The beam cross section shown below has been proposed for a short pedestrian bridge. The cross section will consist of two pipes that are welded to a rectangular web plate. Dimensions of the cross section are: h = 470 mm tw = 16 mm d = 120 mm t = 4.8 mm Additionally: • The area of each pipe is A = 1737 mm². • The moment of inertia of the entire beam cross section about the z centroidal axis is lz = 254790000 mm². If the beam will be subjected to a shear force of V = 250 kN, determine the shear stress at point K, located at yk = 80 mm below the z centroidal axis. H H.V K Answer: TK = N Z MPa Ун Ук h

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN:9781337093347

Author:Barry J. Goodno, James M. Gere

Publisher:Barry J. Goodno, James M. Gere

Chapter6: Stresses In Beams (advanced Topics)

Section: Chapter Questions

Problem 6.8.1P: A simple beam with a W 10 x 30 wide-flange cross section supports a uniform load of intensity q =...

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A simple beam with a W 10 x 30 wide-flange cross section supports a uniform load of intensity q = 3.0 kips/ft on a span of length L = 12 ft (sec figure). The dimensions of the cross section are q = 10.5 in., b = 5.81 in., t1= 0.510 in., and fw = 0.300 in. Calculate the maximum shear stress tjuly on cross section A—A located at distance d = 2.5 ft from the end of the beam. Calculate the shear stress rat point Bon the cross section. Point B is located at a distance a = 1.5 in. from the edge of the lower flange.
A simple beam ACE is constructed with square cross sections and a double taper (see figure). The depth of the beam at the supports is dAand at the midpoint is dc= 2d 4. Each half of the beam has length L. Thus, the depth and moment of inertia / at distance x from the left-hand end are, respectively, in which IAis the moment of inertia at end A of the beam. (These equations are valid for .x between 0 and L, that is, for the left-hand half of the beam.) Obtain equations for the slope and deflection of the left-hand half of the beam due to the uniform load. From the equations in part (a), obtain formulas for the angle of rotation 94at support A and the deflection Scat the midpoint.
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A C 200 x 17.1 channel section has an angle with equal legs attached as shown; the angle serves as a lintel beam. The combined steel section is subjected to a bending moment M having its vector directed along the z axis, as shown in the figure. The cent roi d C of the combined section is located at distances xtand ycfrom the centroid (C1) of the channel alone. Principal axes yl and yvare also shown in the figure and properties Ix1,Iy1and 0pare given. Find the orientation of the neutral axis and calculate the maximum tensile stress exand maximum compressive stress if the angle is an L 76 x 76 x 6.4 section and M = 3.5 kN - m. Use the following properties for principal axes for the combined section:/^, = 18.49 X 106 nrai4,/;| = 1.602 X 106 mm4, ep= 7.448*(CW),_r£ = 10.70 mm,andvf= 24.07 mm.
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The beam cross section shown below has been proposed for a short pedestrian bridge. The cross section will consist of two pipes that are welded to a rectangular web plate. Dimensions of the cross section are: h = 430 mm tw = 16 mm d = 110 mm t = 4.8 mm Additionally: • The area of each pipe is A = 1586 mm2. • The moment of inertia of the entire beam cross section about the z centroidal axis is IZ = 194710000 mm4. If the beam will be subjected to a shear force of V = 170 kN, determine the shear stress at point K, located at yK = 70 mm below the z centroidal axis.
N for Newton, m for meter, mm for millimeter, N/(mm^2) for Stress, mm^2 or m^2 for Area, mm^4 for Moment of inertia and Nm for bending moment. Use brackets if the power is MINUS for Example: 0.00125 N =1.25*10^(-3)N. A simply supported beam AB = 11 m has a hollow rectangular cross-section with 14 cm as width, 29 cm as depth and inner thickness as 1 cm is subjected to a point load of 6 N & 8 N acting at C and D respectively and a uniformly distributed load (UDL) of 8 N/m starts from mid-span and ends at the right support of the beam. Determine the maximum bending stress and the bending stress at 1 cm from the top. Take AC = 1 m & CD = 2 m. Solution: i) Reaction force at B = ii) Reaction Force at A = iii) The distance from B at which the shear Force value changes from "-" to "+" = iv) Maximum Bending Moment (Please write the Maximum bending moment valve in "Nm") = v) Moment of Inertia, I = vi) Maximum bending stress = vii) Bending stress at 1 cm from…
N for Newton, m for meter, mm for millimeter, N/(mm^2) for Stress, mm^2 or m^2 for Area, mm^4 for Moment of inertia and Nm for bending moment. Use brackets if the power is MINUS for Example: 0.00125 N =1.25*10^(-3)N. A simply supported beam AB = 11 m has a hollow rectangular cross-section with 14 cm as width, 29 cm as depth and inner thickness as 1 cm is subjected to a point load of 6 N & 8 N acting at C and D respectively and a uniformly distributed load (UDL) of 8 N/m starts from mid-span and ends at the right support of the beam. Determine the maximum bending stress and the bending stress at 1 cm from the top. Take AC = 1 m & CD = 2 m. Solution: i) Reaction force at B = ii) Reaction Force at A = iii) The distance from B at which the shear Force value changes from "-" to "+" = Answer and unit for part 3 iv) Maximum Bending Moment (Please write the Maximum bending moment valve in "Nm") = v) Moment of Inertia, I = vi) Maximum bending stress = vii)…
A beam having a tee-shaped cross section is subjected to equal 13 kN-m bending moments, as shown. Assume bf = 95 mm, tf = 25 mm, d = 165 mm, tw = 40 mm. The cross-sectional dimensions of the beam are also shown. Determine(a) the centroid location (measured upward from the bottom), the moment of inertia about the z axis, and the controlling section modulus about the z axis.(b) the bending stress at point H (positive if tensile and negative if compressive).(c) the maximum bending stress (positive if tensile and negative if compressive) produced in the cross section.