1 kN/m 1m
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- .2 A ligmio.irc ii supported by two vorlical beams consistins: of thin-walled, tapered circular lubes (see ligure part at. for purposes of this analysis, each beam may be represented as a cantilever AB of length L = 8.0 m subjected to a lateral load P = 2.4 kN at the free end. The tubes have a constant thickness ; = 10.0 mm and average diameters dA = 90 mm and dB = 270 mm at ends A and B, re s pec lively. Because the thickness is small compared to the diameters, the moment of inerlia at any cross section may be obtained from the formula / = jrrf3;/8 (see Case 22, Appendix E); therefore, the section modulus mav be obtained from the formula S = trdhlA. (a) At what dislance A from the free end docs the maximum bending stress occur? What is the magnitude trllul of the maximum bending stress? What is the ratio of the maximum stress to the largest stress (b) Repeat part (a) if concentrated load P is applied upward at A and downward uniform load q {-x) = 2PIL is applied over the entire beam as shown in the figure part b What is the ratio of the maximum stress to the stress at the location of maximum moment?The beam safely supports shear forces and bending moments of 2kN and 6.5 kN-m respectively. Based on this criterion, can it be safely subjected to the loads F = 1kN and C = 1.6 kN-m? Use the integration methodA beam of 8m length rests on two supports one at the right end and the other 2m from the left end. The beam carries a udl of 15kN/m over the entire span and a concentrated load of 80kN at the middle of its span. Draw SFD and BMD . Mark all salient values and locate point of contra-flexure. Ans: RB = 120kN, MC = 172.5kNm, x1 = 0.343m from B. answer must tally with those given
- PRIOR ANSWERS: Reaction Forces: By = 76.809 Cy = 15.94 Shear Forces: V (× = 10.5- ft (i.e., just to the left of support B)) = -36.75 kips V ( x = 10.5+ ft (i.e., just to the right of support B)) = 40.059 kips V (x = 25.5 ft) = -12.441 kips V (x = 26.5- ft (i.e.. just to the left of support C))= = -15.941 kips Bending moment: M (x = 10.5 ft (i.e., at support B)) = -192.9375 kips-ft M (x = 25.5 ft) = 14.1975 kips-ftA shallow foundation supported by a silty sand is shown in Figure 7.5. Given:Length: L = 2 mWidth: B = 1 mDepth of foundation: Df = 1 mThickness of foundation: t = 0.23 mLoad per unit area: qo = 190 kN/m2Ef = 15 x 106 kN/m2The silty sand has the following properties:H = 2 mμs = 0.4Eo = 9000 kN/m2k = 500 kN/m2/mUsing Eq. (7.17), estimate the elastic settlement of the foundation.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 beam has a bending moment of 3 kN-m applied to a section with a hollow circular cross-section of external diameter 3.4 cm and internal diameter 2.4 cm . The modulus of elasticity for the material is 210 x 109 N/m2. Calculate the radius of curvature and maximum bending stress. Also, calculate the stress at the point at 0.6 cm from the neutral axis Solution: (i) The moment of inertia = ii) The radius of curvature is (iii) The maximum bending stress is iv) The bending stress at the point 0.6 cm from the neutral axis is
- A beam with a circular cross section of radius c is subjected to pure bending, and it is made of a material with an elastic, perfectly plastic stress–strain curve, Eq. 12.1. Show that the moment M is related to εc, the strain at y = c, by where (a) applies for (εc ≤ εo), and (b) applies for (εc ≥ εo), with εo = σo/E being the yield strain, and α = εc/εo.Uniformly distributed load on a simple support beam with a length of 6 metersMaximum that occurs on the cross section when q is appliedIf bending stress is 337.5MPa,How many kN/m is the equal distribution load q?(However, the cross section of the beam is wide x height = 40mm x)100 mm.)Q: Shear of Thin-Walled Beams (open section 1. The figure below shows the cross section of a thin, singly symmetrical I-section. Show that the distance & of the shear center from the vertical web is given by d (1+12p) where p=d /h., The thickness is taken negligibly small in comparison with the other dimensions and fj < I,
- The beam W24x104 beam is constructed of A-36 steel. AC has a cross-sectional area of 1 square inch and is constructed of the same material. w = 30 lb/in, L1 = 8 ft, and L2 = 1 ft. Find reactions at A and B, and FAC Draw shear force and bending moment How does a answer change if AC is 6061-T6?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)…