10a-Bending_Stresses_in_Beams

You need the shear modulus for a particular material in order to calculate shear strain for a known applied shear shear stress. You can't find shear modulus for the material, but you can find elastic modulus (200 GPa) and Poisson's ratio (0.3). Which of the following could you use as the shear modulus for your calculations? 200 GPa O 60 GPa O 77 GPa O 154 GPa

B7

a) i) The numerical value of plastic section modulus of the solid rectangular section is mm^3 ii) The numerical value of plastic moment of the solid rectangular section is N.mm   b) The numerical value of ?t can be calculated as mm

Draw shear force and bending moment diagrams P= 1000 N .w = 250 N/cm 5 cm 20 cm Figure 1.2 (not to scale)

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.)

A Please enter your supplied question data (kN) P 300 KN Dimensions (mm) B 65 mm 45 mm (kNm) M 2.5 KN

vi) Using Macauley's method, determine the deflection of the member at the locations where the 25 kN, 20 kN and 50 kN point loads are applied. Comment on how the maximum deflection of the member could be determined. 1 Engineering Design & Analysis 2 25 kN Delta Strain 20 kN Gauge 5 kN/m 50 kN D=100 mm ! 5.75 kNm 8 kN/m 110.kN _7.5MIN/m2 1.5m l0.5m l0.5ml 0.75m 0.5m| t=15 mm 4m Ес 3 375x10-5 Ев — 550х10-6 Data: Material properties for steel: Maximum allowable tensile stress = 750 MN/m? Maximum allowable torsional shear stress = 375 MN/m? %3D Young's Modulus of elasticity = 205 GN/m² Modulus of rigidity = 81 GN/m? Poisson's ratio = 0.31 EA = 880x106 %3D Delta Strain Gauge Readings at Point X on Member

Could I get this question explained step for step thanks

A.) It is the boundary condition for fixed support Choices: Both choices Y=0 Y’=0 Either of the choices   B.) It is a type of simple stress acting parallel to an area Choices: Flexural stress Shear stress Axial stress Bearing stress   C.) For concentric row of bolts in a coupling connection, the shear deformations in the bolts is proportional to the: Choices Modulus of rigidity Diameter of bolt circle Radial distance Number of bolts   note: please help us in the word meaing sample problems thank you

Draw shear force and bending moment diagrams

2. answer B C and D including free body diagram

A copper wire has a radius of 3.5 mm. When forces of a certain equal magnitude but opposite directions are applied to the ends of the wire, the wire stretches by 5.0×10^-3 of its original length. Young's modulus for copper is 11×10^10 Pa What is the magnitude of the force on either end?

Element 1 Ук ** Answer in MPa. 2 cm Element 2 Assume the reaction forces at the origin were solved to be Rf = [325, 100, 0] N and = Rm [10, -2, 5] Nm. We will consider a hypothetical cut at the connection of the dolly to the axle. Z is positive out of the page. What is the value of normal stress in the x-direction Ox for element 2? Answer options: a) Ox < -300 b) -300 ≤ Ox < -100 c) -100 ≤ Ox < -200 d) 0 ≤ Ox < 100 e) 100 ≤ Ox < 300 f) 300 ≤ Ox

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HELLO SIR, COULD YOU SOLVE THIS TABLE TOPIC ABOUT STRENGTH OF MATERIALS LABORATORY ?BENDING TEST SECOND STAGE AL ESRAA Example: Copper Material Distance between anchor= 100 mm L = 1m h = 6.5mm, b = 25.5 mm I= bh/12 = 583.6 mm Y = h/2 = 6.5/2 Mass F (N) Moment 02 (N/m?)= E (N/m?) 1/R No. R (m) (gm) (N.m) (mm) (m) 1 100 0.981 0.0981 0.13 961.238 5.463*105 14.008*1010 1.04*103 2 200 1.962 0.1962 0.38 328.947 10.926*105 11.06 *1010 3 300 0.53 4 400 0.7 500 1.07 6. 600 1.29 7 700 1.49 8 800 1.7 9. 900 1.89 10 1000 2.16

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An elevator weighs 1000 lbs and is supported by a 5/16 inch diameter cable, 1500 ft long. When the elevator carries a passenger load of 1000 lbs, the cable elongates 6/7 inches more. What is the modulus elasticity of the cable? A 4.56 x108 Ib/in? B 5.48 x108 Ib/in? 5.48 x106 Ib/in? D 4.56 x106 Ib/in²

In mechanical resistance

In mechanical resistance

Learning Goal: To calculate the shear stress at a point in the web of an l- beam section subjected to a shear force. When a beam section is subjected to a shear load, a shear stress distribution is developed on the section. The distribution of the shear stress is not linear. Elasticity theory can be used to calculate the shear stress at any point. However, a simpler method can be used to calculate the average shear stress across the width of the section, a distance y above or below the neutral axis. The average VQ shear stress is given by T = Here V is the shear It force on the section, I is the moment of inertia of the entire section about the neutral axis, and ₺ is the width of the section at the distance y where the shear stress is being calculated. Q is the product of the area of the section above (or below) y and the distance from the neutral axis to the centroid of that area (Figure 1). In short, Qis the moment of the area about the neutral axis. Figure 1 of 1 An I-beam has a…

R10 O Reaction Points 84 cm Loading Points Area = 1594 cm^2 W/2 W/2 54 cm L1=27 cm Figure 2. Skateboard dimensions and loading conditions. Loading conditions Weight of a skateboarder, W 90 kg Load case: Calculate Bending and twisting moment, Ms, Mxy Use Tsai-Wu failure criterion for selection of relevant laminate design. Specifically, the core materials failure is free of any failure; therefore, it can be analyzed based on maximum strain failure criterion with =0.1. D3D20 cm

Question #1: Determine the shear, axial, and bending moment distributions for the loaded beam shown below. Write equations for axial and shear forces as well as bending moment as a function of distance along the beam from left to right. Also, sketch graphs of shear and axial forces and bending moments. 10 lb/ft -6 ft →→ 30 lb 20 lb-ft ∙3 ft- 3 ft-

1) A cylindrical rod of constant shear modulus, G is of length L and located between z=0 and z=L. It has a polar moment of inertia given by J=(Jo)/(1-Z/L) and t=-sin(πZ/2L)where Jo is a constant.The boundary conditions are dΦ(o)/dz =0 AND Φ(L)=0 Find Φ(z) and graph (1/GJo)Φ(Z),  and (1/GJo)(dΦ(Z)/dz) ?

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Vibration Engineering Mechanical Engineering

Please Provide correct solution, correct Method.   No Missing steps / calculations.   No Handwriting please

For the simply supported steel beam shown, determine the mid-span deflection and slope at both ends if the beam is fabricated from a W457 x 113 wide-flange section. Given: • L=7m • L₂ = 5.25 m F = 95 kN • q = 76 kN/m -L₂ L F q