Problem No. 3 Using double integration method to determine the deflection at L and 1.5L for the beam shown below. E = 200 GPa 4g (106) mm4 w kN/m P1 kN P1 kN A, B /2-
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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 Z-section of Example D-7 is subjected to M = 5 kN · m, as shown. Determine the orientation of the neutral axis and calculate the maximum tensile stress c1and maximum compressive stress ocin the beam. Use the following numerical data: height; = 200 mm, width ft = 90 mm, constant thickness a = 15 mm, and B = 19.2e. Use = 32.6 × 106 mm4 and I2= 2.4 × 10e mm4 from Example D-7A 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.
- For the beam shown, use only singularity functions. V₁ = 45 lbf/in and V/₂ = 5 in. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. 0 1400 lbf Hinge JA B₁ C R₂ R₁ 4 in 4 in 2 in V1 V2 D R₂ + What is the value of the peak moment between points Cand D? The peak moment between points C and Dis Ibf.in.The shape of the beam made of U profile ensures safe loading. Find the length c of the overhangs so that it can move. Of the material Safety stress (ç) emn = 50 N / mm2 for tension and compression conditions, (ob) emn = 80 N / mm2. Support spacing of the beam L = 6 m, cross section its dimensions are a = %D 210 mm, b = 150 mm and wall thickness t = 40 mm. The loading condition is P = 30 kN and qo = 40 kN / m. P B (Kesit)Problem 4 For the beam shown below, find the value of EIY at 2m from R2. E= 18500MPa and; beam cross -section=200mmX275mm . Use Area-moment method w=15kN/m
- The cross-sectional dimensions of the beam are a = 4.8 in, b = 6.5 in, d=4.5 in, and t= 0.30 in. The internal bending moment about the z centroidal axis is Mz= -3.90 kip-ft. Answers both in psiConsider the beam in the picture below: 7kN/m 5kN/m P N/m Section 1 Section 2 Section 3 - L/3 L/3 L/3 %3D Take P = the last four digits of your student number in N/m. If P<250 N/m then take P = 30OON/m instead. Take L = the third digit of your student number, reading left to reight. If this value is zero then take L = 2 Assume: The reaction at the Pin = V pin 47000L+9PL )N 54 The reaction at the Roller = Vroller = 61000L+9PL 54 and that both reactions act vertically upwards. a) Find an expression for the internal moment for Section 1. Show all working and any relevant free body diagrams. b) What is the maximum magnitude of the internal moment for Section 1? Mark sure you prove that the value you calculate is the maximum. c) Find an expression for the internal moment for Section 2. Show all working and any relevant free body diagrams. d) What is the maximum magnitude of the internal moment for Section 2? Mark sure you prove that the value you calculate is the maximum. e) Find an…A beam is supported and loaded as shown. In the section of the beam between 'A' and 'B', the equation for the resisting bending moment was determined as: Mr = -6.2x²+50x. Determine the magnitude of maximum resisting bending moment (Mcmax) in the section. Note: Do NOT include units in your answer Answer: AY Fun
- Consider the following values in the given beam above: L1=6m L2 = 2 m L3 = 4 m L4=3m L5=2 m L6=1m W1 = 90 kN/m W2 = 30 kN/m P = 50 kN M = 60 kN-m Point E is an internal hinge W1 B L2 L3 2 m E W2 L5 L6 HA simply supported wood beam with a span of L = 24 ft supports a uniformly distributed load of wo = 350 lb/ft. The allowable bending stress of the wood is 1.45 ksi. If the aspect ratio of the solid rectangular wood beam is specified as h/b = 2.00, calculate the minimum width b that can be used for the beam. h B C Answer: b = i in.1. For the given beam cross section and its properties shown; (a) determine the section modulus of the beam, (b) compute the maximum flexural stress at section B., and (c) compute the maximum flexural stress at section C. 5kN 5 KN•m D B 60mm 200mm Im 2m 2m INA = 40x10 6 4 mm