6/4= v ViewBfound from12 ft1000 lb |Rg= 1160 lb(b)V = 1000–15x² = 0V (lb)2°1000x = 8.165 ftdM= 1000 – 15x? = 0dx8.165 ft(d)-1160x = 8.165 ft.305443M (lb • ft)the maximum bending3360moment isM.5(8.165)3 = 5443 lb- ft1000(8.165) -`max(e)7STUThi-Qar- Technical CollegeЕ.М. Е. Dept.Subject: Strength of MaterialsSecond ClassCollegeLecturer: Dr. Mohanad MousaН.WDraw shear force digam and bending moment digram for the following beam300 N500 N800 ND0.5 m*0.7 m0.8 m-Techalcal2.II

The given loading system is a cantilever beam loading. Shear force is the force which makes the…

Answered: Draw shear force digam and bending…

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Mechanical Engineering

Draw shear force digam and bending moment digram for the following beam 300 N 500 N 800 N to A B 0.5 m 0.7 m- 0.8 m

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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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-ft
A 16 ft steel line shaft has no bending action except its own weight. What power in hp can shaft deliver at a speed of 200 rpm. Consider that the torsional deflection wm not exceed 0.08o/ft of length.
3) The force, FT = 1 kN, and moment, MT = 0.5 kN-m, at the tip are caused by a wing tip vortexand a winglet, not shown. L = 12 m and the spar has an elastic modulus of E = 70 GPa and aPoisson’s ratio of n = 0.33. The mass of the wing is 4000 kg, and the weight of the engine is 107kN. Use 9.8 m/s 2 for the acceleration due to gravity.Consider the cross-section shown (you can look these up). Pay attention to the coordinate systemgiven in the drawing.d) where is the centroid?e) what is the area?f) Calculate the shear modulus
3) The force, FT = 1 kN, and moment, MT = 0.5 kN-m, at the tip are caused by a wing tip vortexand a winglet, not shown. L = 12 m and the spar has an elastic modulus of E = 70 GPa and aPoisson’s ratio of n = 0.33. The mass of the wing is 4000 kg, and the weight of the engine is 107kN. Use 9.8 m/s 2 for the acceleration due to gravity.Consider the cross-section shown (you can look these up). Pay attention to the coordinate systemgiven in the drawing.g) Compute each of these moments of inertia with the values b = 0.4 m, d =0.5 m , t =0.1 m.
3) The force, FT = 1 kN, and moment, MT = 0.5 kN-m, at the tip are caused by a wing tip vortexand a winglet, not shown. L = 12 m and the spar has an elastic modulus of E = 70 GPa and aPoisson’s ratio of n = 0.33. The mass of the wing is 4000 kg, and the weight of the engine is 107kN. Use 9.8 m/s 2 for the acceleration due to gravity.Consider the cross-section shown (you can look these up). Pay attention to the coordinate systemgiven in the drawing.a) what is the moment of inertia about y-axis in terms of the symbolic dimensions shown?b) what is the moment of inertia about z-axis in terms of the symbolic dimensions shown?c) what is the polar moment of inertia in terms of the symbolic dimensions shown (i.e.about the x-axis)?d) where is the centroid?e) what is the area?f) Calculate the shear modulusg) Compute each of these moments of inertia with the values b = 0.4 m, d =0.5 m , t =0.1 m.
A simple wooden beam, 150 mm wide by 250 mm deep and 7.5 long has a maximum deflection of 102 mm under a uniform load. If a third support is placed midway between the two supports to make the deflection equal to zero, E = 12411 MPa.1. What is the maximum bending stress? a. 6.75 MPab. 7.53 MPac. 8.34 MPad. 8.87 MPa2. What is the maximum shear stress? a. 0.45 MPab. 0.33 MPac. 0.56 MPad. 0.28 MPa
A load of 100 kN, followed by another load of 50 kN, at a distance of 10 metres, advances across a girder with a 100-metre span. Obtain an expression for the maximum bending moment at a section of the girder at a distance of z metres from an abutment. Please provide solutions. Answer is z(140-1.5z) for z< (100/3)m;(100-z)(1.5z-5) for z>(100/3) m.
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How can we determine the Slope and Displacement by the moment area method?
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A section of a rotating shaft is under an alternating bending moment of 70Nm and a torque of 35Nm. The yield strength of the material is 560 MPa and the ultimate tensile strength of the material is 700 MPa. The estimated actual endurance strength is assumed to be 210 MPa and the stress concentration factor is 1.5 for well rounded fillet. (Design Factor = 2) Determine the minimum diameter of the section of the shaft: D = mm