Determine the maximum uniformly distributed load which can be applied over the entire length of the beam if the shearing stress is limited to 1.6 MPa. Answer must be in kN/m.
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- If a twisting moment of 1100 N-m is impressed upon a 4.4 cm ø shaft;1. What is the maximum shearing stress (in MPa) developed? 2. What is the angle of twist (in rad) in a 150 cm length of the shaft?Gsteel = 85 GPa.a. Calculate the moment of inertia (I) of the given cross–section b. Calculate the first moment of area (Q) at certain depth by using 25 mm increments. c. Calculate and draw the horizontal shear stress distribution of the beam at 0.2 m from the left supportA simply supported beam carries a concentrated load of 31 kN at 7 meters from the left support (1 meters from the right support). Use EI = 80,000 kN-m2. Solve for the slope at the left support in degrees.
- Note : point load = 12.289KN/m *3 = 36.867KN . 1) calculate the reaction forces 2) Calculate and draw the Bending moment digram, and what is the value of maximium bending moment ? 3) calculate and draw the shear force digram and what is the value of the maximum shear force ?For the beam shown in figure below, draw the shear and moment diagrams: 1- Due to applied loads (without support settlement) 2- Due to support settlement of 60 mm downward at support B (without loads). Take the stifiness E1 as 12000 kN.m”. sou i The solution is according to Slope deflection methodThe T-shaped beam shown above is supporting a concentrated load P at its free end. The beam has an allowable bending stress of ?????? = 250 MPa and an allowable shear stress of ?????? = 100 MPa. a) Determine the distance to the neutral axis (?̅), second moment of area (?), and the section modulus (?) of the cross-section.b) Draw the shear force diagram (SFD) and bending moment diagram (BMD) of the beam. On your diagrams, express the values of shear and moment in terms of the applied load P.c) Determine the maximum value of P such that bending failure will not occur.d) Determine the maximum value of P such that shear failure will not occur.e) Based on your answers to (c) and (d), what is the maximum load P that can be applied to the beam? Is this beam bending or shear governed?
- Consider a 130-mm-long segment of a simply supported beam. The internal bending moments on the left and right sides of the segment are 55 kN-m and 60 kN-m, respectively. The cross-sectional dimensions of the flanged shape are shown in the accompanying figure. Assume b1 = 165 mm, b2 = 40 mm, b3 = 290 mm, d1 = 65 mm, d2 = 240 mm, d3 = 65 mm. Determine the maximum horizontal shear stress in this segment of the beam.Compute the value of EIy at 2.5 m from the left support.Consider a 155-mm-long segment of a simply supported beam. The internal bending moments on the left and right sides of the segment are 75 kN-m and 79 kN-m, respectively. The cross-sectional dimensions of the flanged shape are shown in the accompanying figure. Assume b1 = 135 mm, b2= 50 mm, b3= 205 mm, d1= 65 mm, d2= 180 mm, dz= 65 mm. Determine the maximum horizontal shear stress in this segment of the beam.
- As shown in the figure, aluminum rod AB and steel rod BC, hinged to rigid supports, are pinned together at B to carry a vertical load P = 6000 lb. If aluminum rod has a cross-sectional area of 0.60 in.² and E = 10 × 103and steel rod has 0.50 in.² and E = 29 × 106 each rod and the horizontal and vertical displacements of point B. Assume α = 60° and θ = 30°. ksi, psi, compute the elongation ofA cantilever beam carries a trapezoidaldistributed load (see figure). Let wB = 2.5 kN/m,wA = 5.0 kN/m, and L = 2.5 m. The beam has amodulus E = 45GPa and a rectangular cross sectionwith width b = 200 mm and depth h = 300 mm.Use the method of superposition and Cases 1 and8 in Table H-1 to calculate the def lection and rotationat B.(a) Compute the horizontal displacement of joint B produced by the 240-kN load in Figure P8.10. For all bars, area = 2400 mm2 and E = 200 GPa. (b) Assuming that no load acts, determine the horizontal displacement of joint B if support A moves 20 mm to the right and 30 mm down and support E moves downward 36 mm.