The deflection of a uniform beam subject to a linearly increasing distributed load can be computed as
Given that
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- 3) A point mass is attached to one end of a beam levered against a spike which acts as the pivot. The point mass weighs 250N and the beam has a uniform mass of 10kg and is 8m long. Parts a and b Part c Fapplied 3 of 6mov 1m 8m Krist 0 = 60° Scale 250 N Scale a. If a scale placed beneath the point mass reads 250N, what is the magnitude and direction for the force at the pivot point? Direction: North, 90° from x-axis Magnitude to k* 250N = 2500 N b. For the same scenario, how far must the pivot point be located relative to the left edge of the beam? (Hint, both the boxes weight and the force from the scale act at the farthest right edge of the beam.) www.acta 0672nos onl c. The pivot is then moved to be 6m from the left edge of the beam which now makes a 60 angle with the floor. How much downward force would be needed at the left-most edge of the beam to reduce the reading on the scale to zero? (Hint: the beam is still in static eq. when the scale reads zero.)arrow_forwardQUESTION 2 Question 2 A cross-section of a beam is shown in Figure Q2. If the shear force in this section is V = 125 KN, determine the value and the location of the maximum shear stress in the section. In Figure Q2, a = 30 mm and the origin of the coordinate system is at centroid of the cross section. 7 y= Z= A a AY S= 20 4a mm; mm; O Figure Q2 Answer The vertical coordinate (y-coordinate; the y-axis serves as the axis of symmetry of the cross- section.) and horizontal coordinate (z-coordinate) of the location where the maximum shear stress occurs in the section are ← a The vertical distance from the location where the maximum shear stress occurs in the section to the bottom side (AB cross section can be calculated as Distance = mm (units: mm) 3a Second moment of area The second moment of area employed in the equation to calculate maximum shear stress can be calculated as I₂ = a (units: mm²) Shear stress The second moment of area employed in the equation to calculate maximum shear…arrow_forwardThe bending dstributon in a beam as a function of distance 'x' is moment given by M= (5x2 + 20x 7) N-m. the shear force at x=2 m is 10 N 40 N 20 Narrow_forward
- Q6- The intervertebral disc as shown below is subjected to four parallel loadings (F,,F2, F3, F4). If F, = 10N and F = 5N, determine F, and F, if the resultant force acts through point c (x--2cm and y-3cm). %3D F2 = 10N F. 5N F2 Bcm 2cm 5cm 10cm 10cm 4cm 6cm 12cm 6cm ch Address DELLarrow_forward9:10 IM N 10 36 | zain IQ 22222.jpg > SAMPLE PROBLEM 2/5 Calculate the magnitude of the moment about the base point O of the 600-N force in five different ways. 600 N Solution. 2 m 600 N F- 600 cos 40 4m F= 600 sin 40arrow_forwardThe bending moment diagram and cross-section for a beam are shown below. Use the elastic flexural formula ox = -Mzy/l, to determine the required strength of the beam to resist the compressive stresses. Circle the correct answer for the compressive strength in ksi: 2.3, 2.5, 27.5, 30, 37.7 Show your work below. Bending Moment Diagram Beam Cross-Section 4" +50 2' (K-Fe) 2" 2" - 40 2"arrow_forward
- Use any geometric method to solve the response of the beam. Determine the maximum deflection (in mm) of the given beam. Indicate only the magnitude of the deflection. Use E=200GPA and l=5x107 mm4. P = 10kN w = 25 kN/m A В 下 5 m 2 marrow_forwardENGINEERING MECHANICS STATICS Example 10: Determine the resultant of the tensions of the wires shown in figure below. n' 12b 18lb 171b 3' 3arrow_forwardThe centre of mass of a non-uniform rod is not necessarily at the rod's mid-point. 4. A non-uniform beam AB of length 60 m and mass 5N rests on a pivot at the mid-point C. When a mass of 2N rests on the beam 15 m from B, the system is in equilibrium. Find the distance between A and the centre of mass of the beam. R B 15m 50N y 15m 20Narrow_forward
- For a uniformly loaded span of a cantilever beam attached to a wall at x = 0 with the free end at x = L, the formula for the vertical displacement from y = 0 under the loaded condition with y the coordinate in the direction opposite that of the load can be written as follows: Y= -(X4 – 4X³ + 6X²) where Y is the vertical displacement, X = x/L, and L is the length of the beam. The formula was put into dimensionless form to answer the following question: What is the shape of the deflection curve when the beam is in its loaded condition and how does it compare with its unloaded perfectly horizontal orientation? The answer is provided graphically in Figure Q4. Figure Q4 shows the vertical deflection of a uniformly loaded cantilever beam and its comparison with the unloaded perfectly horizontal orientation. Write a script to get the same figure as Figure Q4 by solving the following question. 1 · Unloaded cantilever beam 0.5 Uniformly loaded beam -0.5 -1E > -1.5 -2- -2.5 -3 -3.5 0.5 1 1.5…arrow_forward1. A simply supported beam with length of 5 Meters is loaded with a counterclockwise couple of 5 KN.M at I Meter from the left support and a point load of 5 KN at 3 Meters from the left support. The beam is an I-Section with the following dimensions: bf = 250 MM tf = 16 MM tw = 10MM d = 350 MM Determine the MaxiMUM tensile and compressive bending stress developed in the beam. 2. A simply supported beam with a length of 4M is loaded with a uniform distributed load (w). The beam has a rectangular hollow section with the following dimensions: Outer Base = 150 MM Outer Depth = 200 MM Inner Base = 100 MM Inner Depth = 150 mm Determine the maximum uniformly distributed load which can be applied over the entire length of the beam if the bending stress is limited to 8 Mpa.arrow_forwardYou are required to calculate the second moment of area about the x-axis for the figure below. a mm b mm a mm a mm One method for calculating the second moment of area is to break up the shape into three parts as shown below. Calculate the second moment of area of segment 2 if: a = 30.67 mm b = 21.14 mm |_x = mm^4 a mm ww earrow_forward
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