10a-Bending_Stresses_in_Beams

b) A 1.25 m long cable has a diameter 3.50 mm with a Young's Modulus, E, of 9.75 x 10⁹ N/m². When the wire is placed under tension, it experiences a stress of 202.52 x 106 N/m², the length of the cable extends by 36.35 mm. Calculate the force that the cable experiences under tension and the strain energy density (U/V) due to deformation. Give your answers in newtons (N) to 2 decimal places for the force; and in joules per cubic metre (J/m³) for the strain energy density to 2 decimal places. Assume the cable is solid and the material is homogeneous. c) Figure Q1c shows a bracket on rollers that allow the bracket to move along a beam in the horizontal direction. A force F₁ of 2625 N acts at point A where the angle between F, and the x-axis is 100. If force F₂ acts at an angle a = 400 between F2 and the y-axis: (i) Determine the magnitude of F2 necessary to maintain bracket on horizontal equilibrium. (ii) Determine the vertical force at A acting along the y-axis. Give your answers to the…

The steel I-beam in the drawing has a weight of 1.20 x 103 N and is being lifted at a constant velocity. What is the tension in each cable attached to its ends? 70.0° 70.0 I-beam Units XX> XIX XIXIX

some value P, and the final deflection in the direction of the load is A, the load does work U, =PA on the structure. By conservation of energy, this work is stored in the elastic deformation energy of the structure, U¡ . For a truss, the internal strain energy is the sum of the axial strain energies in all the elements. This gives the equation N²L PA =E 2AE where N is the internal axial force in each member.

Elastic Properties of Materials 4. A cylindrical bar of steel 10 mm (0.4 in.) in diameter is to be deformed elastically by application of a force along the bar axis. Using the data in Table given below, determine the force that will produce an elastic reduction of 3 × 10¯³mm (1.2 × 10-4 in.) in the diameter. Metal Alloy Aluminum Brass Copper Magnesium Nickel Steel Titanium Tungsten Modulus of Elasticity GPa 69 97 110 45 207 207 107 407 106 psi 10 14 16 6.5 30 30 15.5 59 Shear Modulus GPa 25 37 46 17 76 83 45 160 106 psi 3.6 5.4 6.7 2.5 11.0 12.0 6.5 23.2 Poisson's Ratio 0.33 0.34 0.34 0.29 0.31 0.30 0.34 0.28

A bar with varying cross section is subjected to various forces as shown below. Compute: a) Stresses in each section b) strain in each section c) total extension of the bar. Take E = 2.1x105 N/mm² 400 mm² P1=10kN A 1000 mm B 800 mm mm² P₂ P3=45KN+ 1500 mm 600 mm² 800 mm → P4-35kN

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

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

(2) lubäi The difference between shear strain and linear strain Linear Strain is change in length/original length while shear strain is the angle of deformation linear strain and shear strain are the same and equal to the change in length / original O length None O linear strain and shear strain are the same and equal to the change in angle under shear force (2) ibäi A concentrated load is one which acts at a point on a beam O varies uniformly over the whole length of a beam

REVIEW OF STRENGTH OF MATERIALS 6-in D. 300 lbf 50 Ibf 75 lbf 450lbf D 9 in D.C 6 in 8 in B 8 in For the loading shown, the force which makes the shaft to bend is the sum of the belt pulls. The force which makes the shaft to rotate is the difference between the two belt pulls. At any given point, find the Resultant moment by the Pythagorean theorem MRes = M + My *** The Direction of the Resultant Moment is e = tan Mxy ** -1 To be done: (a) Draw the Shear and Moment Diagram on the xz and on the xy planes. Solve for the resultant moments at points B, and C. (b) Find the location where the Resultant moment is maximum. (c) Find the maximum Bending Moment and its location.

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?

Question 3 Determine the tensile stress in each cable given that all the cables have a cross-sectional area of 4cm?. Take mass of cylinder = 60kg; and acceleration due to gravity, g = 9.81 ms². A 45° B D

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

materials mechanics A bar of circular cross section is made of an isotropic, linearly elastic material. The material has an elastic limit of 360 MPa. The bar is subjected to two equal and opposite bending moments M2=12 kNm and two equal and opposite torsional moments M1=16 kNm. Determine the minimum diameter of the bar for a safe design using the following factor of safety: F.S=3.

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

An elevator weighs 1000 lbs and is supported by a 5/26 inch diameter cable, 1500 feet long. When the elevator carries a 1500 Ib, the cable elongates 6 inches more. What is the modulus of elasticity of the cable? A 609x10^7 psi B) 460x10^7 psi 5.87x10^7 psi D 231x10^7 psi

P4 Alyear 1 4 N-1 N End of Year Interest = i%/year %3D

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

Question 7 Determine the maximum shear force in kN in the figure below. 10 kN/m 25 kN • m A E B C D 1 m 1 m 3 m 2 m

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

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…

For the stress-strain curves below, which statements are correct? Stress (MPa) 60 50 40 30 20 10 0 0 A B C. 1 2 3 4 Strain 5 6 7 8 a. The mechanical behavior shown in C is that of an elastomer with low elastic modulus b. While the deformation in polymer A is fully reversible until breakage, polymer B and C only display reversible deformation until 0.1X and 5x elongations respectively The mechanical behavior shown in B is that of an elastomer with low elasticity d. Polymer B and C have an elastic behavior whereas A has a plastic behavior e. The polymer with the necking behavior has the highest strength Of. The polymer with brittle nature is the one that possesses the highest elastic modulus

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) ?

2) The stress-strain diagram for an alloy specimen having an original diameter of 0.5 in. and a gage length of 2 in. is shown in the figure below. TENSILE STRESS (MPA) 450 400 350 300 250 200 150 100 50 Part A 0 0 STRESS VS. STRAIN 0.002 0.004 STRAIN 0.006 Determine the modulus of resilience. Show units. Part B Determine the modulus of elasticity of the material. Part C Estimate the Yield Strength of the material. 0.008 0.01

Question 8 Determine the maximum shear force in kN in the figure below. 30 kN 140 kN • m A D B 2 m 3 m 2 m

Question 1 The wedge shaped bar shown in Figure Q1 is to be analysed using a single one dimensional quadratic element as shown. The bar has a rectangular cross section with a constant width and a linearly varying depth. 500 k = 10E 2 100 3 Figure Q1. +1] a) Calculate the strain shape function matrix, [B], for the element. b) Using the strain shape function matrix derived in (a), show that the stiffness matrix for the element is given by the equation: I 255+ ALL DIMENSIONS IN mm (3-5) d h 100- with units of N/mm Where: E = Young's modulus of the material 10 c) Show how the strain shape function matrix may be used to determine the stress from the nodal displacement for this type of element. Do not attempt to calculate the stress. LEGION Question 2 Figure Q2 shows a two-dimensional quadratic element that has been distorted such that one Q Search

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

Question 1 Draw the shear force and bending moment diagrams by inspection and express the bending moment and shear force by using singularity function 20 kN 20 kN/m A °C 4 m 2 m

It's mechanical engineering question. A bar of rectangular section A × B = 1 × 2 cm and length L = 200 cm is subjected to the action of a load P = 2000 kgf, experiencing an axial elongation of Δl = 1 mm and a lateral contraction Δa = −1.45 × 10−4 cm. Determine: a) Elastic modulus of the material b) Poisson's ratio c) Elongation Δb d) Dimensions of the cross and longitudinal sections of the bar when subjected to a load P = 150 000 kgf

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