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The allowable stress ( referred to as the yield stress ) for the bar shown in the figure below is given as sigma ymeld =250 MPa. What is the maximum force in N that the bar can support ? (the small arrows indicate the direction of the applied force ). If the picture isn't clear , the bar is 5mm thick and 10 cm wide .
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- Two similar channel sections from Figure Q3 (a) arranged back-to-back with 10 mm gap as shown in Figure Q3 (b). Calculate the moment of inertia, Iyy and I, of the double channel section given in Figure O3 (b).Locate the centroid of the plate area shown below: (All units are in mm). Ø200 y Ø150 300 300 Ø50 50 50- 300 +Part 2: Beam analysis The timber beam in Figure 2 is simply supported at points A and E. The cross-section of the beam changes from segment AC to CE, as shown below. The Young's modulus of timber may be taken as 12 GPa. The load, geometry and cross-sections of the beam will be instructed to each student in class. B A :B C D E 0.5L 0.5L 0.75L 0.75L bị 0.25hI b2 h2 NA hị Section B-B Section D-D Figure 2 (a) Determine the deflection of the beam at point D by both Virtual Work Method and Castigliano's Theorem. Compare and comment on the differences in these approaches. (b) Calculate the bending and shear stresses at point F on the Section B-B. (c) Develop and sketch the Mohr's circle of stresses corresponding to point F. Show all necessary details, including principal stresses and maximum shear stress.
- Determine the internal moments at B and C and the support reactions due to the loads and support settlements of 8 mm at B and 25 mm at C. Assume the supports at B and C are rollers and A and D are pins. El is constant. 12 kN/m 12 kN/m A B. 4 m 6 m- 4 m FRAMES WITHOUT SIDESWAY Use slope deflection method to determine the reactions for the frame shown. 36 kN/m A 4 m 4 m 4 m Solve Problem 6 due to the given loads and a support settlement of 5 mm at B. Determine the member end moments and reactions for the frames shown by using the slope- deflection method. 7 8. Note: Use the 100 kN · m acting at joint C for Equilibrium Equation at the joint, not on FEM's. 18 kN/m B 100 kN · m 5 m El = constant A E = 200 GPa I = 1,360 (106) mm4 5 m 5 mX is 193 Y is 93 B is 7 C is 9GIVEN: 240 mm 50 mm 400 mm 150 mm 150 mm 50 mm 50 mm please include table.
- Find ỹ shown in Figure below. - 3 cm 0.5 cm - 0.5 cm A 1 cm B 0.5 cm D 6 cm -Xc E 0.75 cm F 0.75 cm X. 0.75 cm-e -0.5 cm cmWhat is the product of inertia about the centroidal x-axis and y-axis of the body presented below? Show complete solutionDetermine the coordinates of the centroid of the lines shown with respect to the indicated X and Y axes. A. B. 2 60 mm ¥ … 80mm 5洲 80mm + 为那时 x
- 1) Compute the centroid y bar with respect to its base. 120mm 30 Mm 20Mm 100Mm 30MMDetermine the coordinates of centroid for the given figure shown. 80 mm 50 mm Y 50 mm. 150 mm XQuestion 6 Determine the coordinates of the centroid of the shaded area. (X,Y)= (Blank 1,Blank 2) mm Blank 1 Add your answer Blank 2 Add your answer 10 30 y 15 10 55 15 25 15. Dimensions in millimeters --x