Concept explainers
A = 2300 mm2, I = 9.5(106) mm4.
R14–1
Answer to Problem 14.1RP
The total axial and bending strain energy in the A992 steel beam is
Explanation of Solution
Given information:
The cross-sectional area of the beam is
Moment of inertia of the beam is
Assumption:
The modulus of elasticity or Young’s modulus of theA992 steelis
Explanation:
Determine the reactions:
Entire beam:
Show the free body diagram of the entire beam as in Figure 1.
Moment about the point A:
Determine the vertical reaction at point B by taking moment about point A.
Along the vertical direction:
Determine the vertical reaction at point B by resolving the vertical component of forces.
Along the horizontal direction:
Determine the horizontal reaction at point A by resolving the horizontal component of force.
Show the calculation of reaction as follows:
Solve Equation (1).
Substitute 7.5kN for
Solve Equation (3).
Region
Show the free-body diagram of the section as in Figure 2.
Moment about the section:
Determine the moment at section by taking moment about the section.
Along the horizontal direction:
Determine the normal axial force at the section by resolving the horizontal component of forces.
Show the calculation of values as follows:
Substitute 7.5kN for
Substitute 15 kN for
Strain energy due to axial load:
Determine the strain energy of a bar of constant cross-sectional area A and constant internal axial load N using the equation.
Here, N is the axial load, L is the length of beam, E is Young’s modulus or modulus of elasticity, and A is cross-sectional area of the beam.
Substitute 15 kN for N, 10 m for L,
Strain energy due to Bending:
Determine the strain energy in the beam due to bending using the equation.
Here, M is the moment in the beam and I is the moment of inertia of the beam.
Substitute 10 m for L,
Total strain energy:
Determine the total strain energy by adding the strain energy due to axial load and the strain energy due to bending.
Substitute 2.4456 J for
Thus, the total axial and bending strain energy in the A992 steel beam is
Want to see more full solutions like this?
Chapter 14 Solutions
EBK MECHANICS OF MATERIALS
Additional Engineering Textbook Solutions
Management Information Systems: Managing The Digital Firm (16th Edition)
Database Concepts (8th Edition)
BASIC BIOMECHANICS
Java: An Introduction to Problem Solving and Programming (8th Edition)
Elementary Surveying: An Introduction To Geomatics (15th Edition)
Electric Circuits. (11th Edition)
- Calculate the load that will make point A move to the left by 6mm, E=228GPa. The diameters of the rods are as shown in fig. below. 2P- PA 80mm B 200mm 2P 0.9m 1.3m.arrow_forwardIf the rods are made from a square section with the dimension as shown. Calculate the load that will make point A move to the left by 6mm, E=228GPa. 2P- P A 80mm B 200mm 2P 0.9m 1.3marrow_forward3. 9. 10. The centrifugal tension in belts (a) increases power transmitted (b) decreases power transmitted (c) have no effect on the power transmitted (d) increases power transmitted upto a certain speed and then decreases When the belt is stationary, it is subjected to some tension, known as initial tension. The value of this tension is equal to the (a) tension in the tight side of the belt (b) tension in the slack side of the belt (c) sum of the tensions in the tight side and slack side of the belt (d) average tension of the tight side and slack side of the belt The relation between the pitch of the chain (p) and pitch circle diameter of the sprocket (d) is given by 60° (a) p=d sin (c) p=d sin (120° T where T Number of teeth on the sprocket. 90° (b) p=d sin T 180° (d) p=d sin Tarrow_forward
- OBJECTIVE TYPE QUESTIONS 1. The maximum fluctuation of energy is the 2. (a) sum of maximum and minimum energies (b) difference between the maximum and minimum energies (c) ratio of the maximum energy and minimum energy (d) ratio of the mean resisting torque to the work done per cycle In a turning moment diagram, the variations of energy above and below the mean resisting torque line is called (a) fluctuation of energy (b) maximum fluctuation of energy (c) coefficient of fluctuation of energy (d) none of the above Chapter 16: Turning Moment Diagrams and Flywheel 611 The ratio of the maximum fluctuation of speed to the mean speed is called 3. (a) fluctuation of speed (c) coefficient of fluctuation of speed 4. (b) maximum fluctuation of speed (a) none of these The ratio of the maximum fluctuation of energy to the.......... is called coefficient of fluctuation of energy. (a) minimum fluctuation of energy (b) work done per cycle The maximum fluctuation of energy in a flywheel is equal to 5.…arrow_forwardOBJECTIVE TYPE QUESTIONS 1. The velocity ratio of two pulleys connected by an open belt or crossed belt is 2. (a) directly proportional to their diameters (b) inversely proportional to their diameters (c) directly proportional to the square of their diameters (d) inversely proportional to the square of their diameters Two pulleys of diameters d, and d, and at distance x apart are connected by means of an open belt drive. The length of the belt is (a)(d+d₁)+2x+ (d₁+d₂)² 4x (b)(d₁-d₂)+2x+ (d₁-d₂)² 4x (c)(d₁+d₂)+ +2x+ (d₁-d₂)² 4x (d)(d-d₂)+2x+ (d₁ +d₂)² 4x 3. In a cone pulley, if the sum of radii of the pulleys on the driving and driven shafts is constant, then (a) open belt drive is recommended (b) cross belt drive is recommended (c) both open belt drive and cross belt drive are recommended (d) the drive is recommended depending upon the torque transmitted Due to slip of the belt, the velocity ratio of the belt drive 4. (a) decreases 5. (b) increases (c) does not change When two pulleys…arrow_forwardQ3: (10 MARKS) A piston with a weight of 29.4 N is supported by a spring and dashpot. A dashpot of damping coefficient c = 275 N.s/m acts in parallel with the spring of stiffness k = 2400 N/m. A fluctuating pressure p = 960 sin 30t N/m² acts on the piston, whose top surface area is 0.05 m². Determine the steady-state displacement as a function of time and the maximum force transmitted to the base. P=Po sin cot Warrow_forward
- 9. Design a spur gear drive required to transmit 45 kW at a pinion speed of 800 r.p.m. The velocity ratio is 3.5 : 1. The teeth are 20° full-depth involute with 18 teeth on the pinion. Both the pinion and gear are made of steel with a maximum safe static stress of 180 MPa. Assume a safe stress of 40 MPa for the material of the shaft and key. 10. Design a pair of spur gears with stub teeth to transmit 55 kW from a 175 mm pinion running at 2500 r.p.m. to a gear running at 1500 r.p.m. Both the gears are made of steel having B.H.N. 260. Approximate the pitch by means of Lewis equation and then adjust the dimensions to keep within the limits set by the dynamic load and wear equation.arrow_forward7. A motor shaft rotating at 1440 r.p.m. has to transmit 15 kW to a low speed shaft rotating at 500 r.p.m. The teeth are 20° involute with 25 teeth on the pinion. Both the pinion and gear are made of cast iron with a maximum safe stress of 56 MPa. A safe stress of 35 MPa may be taken for the shaft on which the gear is mounted. Design and sketch the spur gear drive to suit the above conditions. The starting torque may be assumed as 1,25 times the running torque. Ruins 20 LW at 100 nm to another shaft running approxiarrow_forward6. A two stage reduction drive is to be designed to transmit 2 kW; the input speed being 960 r.p.m. and overall reduction ratio being 9. The drive consists of straight tooth spur gears only, the shafts being spaced 200 mm apart, the input and output shafts being co-axial.arrow_forward
- 2 A metal block of mass m = 10 kg is sliding along a frictionless surface with an initial speed Vo, as indicated below. The block then slides above an electromagnetic brake that applies a force FEB to the block, opposing its motion. The magnitude of the electromagnetic force varies quadratically with the distance moved along the brake (x): 10 FEB = kx², with k = 5 N m² V₁ = 8 m/s m = 10 kg FEB Frictionless surface Electromagnetic brake ⇒x Determine how far the block slides along the electromagnetic brake before stopping, in m.arrow_forwardQ1: Determine the length, angle of contact, and width of a 9.75 mm thick leather belt required to transmit 15 kW from a motor running at 900 r.p.m. The diameter of the driving pulley of the motor is 300 mm. The driven pulley runs at 300 r.p.m. and the distance between the centers of two pulleys is 3 meters. The density of the leather is 1000 kg/m³. The maximum allowable stress in the leather is 2.5 MPa. The coefficient of friction between the leather and pulley is 0.3. Assume open belt drive.arrow_forward5. A 15 kW and 1200 r.p.m. motor drives a compressor at 300 r.p.m. through a pair of spur gears having 20° stub teeth. The centre to centre distance between the shafts is 400 mm. The motor pinion is made of forged steel having an allowable static stress as 210 MPa, while the gear is made of cast steel having allowable static stress as 140 MPa. Assuming that the drive operates 8 to 10 hours per day under light shock conditions, find from the standpoint of strength, 1. Module; 2. Face width and 3. Number of teeth and pitch circle diameter of each gear. Check the gears thus designed from the consideration of wear. The surface endurance limit may be taken as 700 MPa. [Ans. m = 6 mm; b= 60 mm; Tp=24; T=96; Dp = 144mm; DG = 576 mm]arrow_forward
- Elements Of ElectromagneticsMechanical EngineeringISBN:9780190698614Author:Sadiku, Matthew N. O.Publisher:Oxford University PressMechanics of Materials (10th Edition)Mechanical EngineeringISBN:9780134319650Author:Russell C. HibbelerPublisher:PEARSONThermodynamics: An Engineering ApproachMechanical EngineeringISBN:9781259822674Author:Yunus A. Cengel Dr., Michael A. BolesPublisher:McGraw-Hill Education
- Control Systems EngineeringMechanical EngineeringISBN:9781118170519Author:Norman S. NisePublisher:WILEYMechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage LearningEngineering Mechanics: StaticsMechanical EngineeringISBN:9781118807330Author:James L. Meriam, L. G. Kraige, J. N. BoltonPublisher:WILEY