Concept explainers
(a)
Find the maximum shearing stress
(a)
Answer to Problem 138P
The maximum shearing stress
Explanation of Solution
Given information:
The length of the steel member (L) is
The provided section of the member is
The torque in the member (T) is
The modulus rigidity of the steel (G) is
Assume the angle of twist in flange and web is equal.
Calculation:
Consider flange:
Refer Appendix C, “Properties of Rolled-Steel shapes”.
The width of the flange (a) is
The thickness of the flange (b) is
Calculate the ratio of width to thickness of the steel
Substitute
Hence, the ratio of
Calculate the ratio of thickness to width of the steel
Substitute
Calculate the coefficient for rectangular bar
Substitute 0.0544 for
Calculate the angle of twist in flange
Here,
Substitute 0.32191 for
Consider web:
Refer Appendix C, “Properties of Rolled-Steel shapes”.
The thickness of the web (b) is
The depth of the member (D) is
Calculate the width of the web (a) using the formula:
Here,
Substitute
Calculate the ratio of width to thickness of the steel
Substitute
Hence, the ratio of
Calculate the ratio of thickness to width of the steel
Substitute
Calculate the coefficient for rectangular bar
Substitute 0.039972 for
Calculate the angle of twist in web
Substitute 0.32494 for
Since the angle of twist in flange and web is equal, therefore,
Substitute
By taking the sum of torque exerted on two flanges and web in the member is equal to the total torque T applied to member. Therefore,
Substitute
Substitute
Calculate the maximum shearing stress
Substitute
Therefore, maximum shearing stress
(b)
Find the maximum shearing stress
(b)
Answer to Problem 138P
The maximum shearing stress
Explanation of Solution
Given information:
The length of the steel member (L) is
The provided section of the member is
The torque in the member (T) is
The modulus rigidity of the steel (G) is
Assume the angle of twist in flange and web is equal.
Calculation:
Calculate the torque in the web
Substitute
The maximum shearing stress
(c)
Find the angle
(c)
Answer to Problem 138P
The angle
Explanation of Solution
Given information:
The length of the steel member (L) is
The provided section of the member is
The torque in the member (T) is
The modulus rigidity of the steel (G) is
Assume the angle of twist in flange and web is equal.
Calculation:
From the above calculation of angle of twist, take the critical angle to compute the angle of twist.
Calculate the angle of twist
Consider the torque equation,
Substitute
Assume
Calculate the value of
Substitute 0.32191 for
Calculate the value of
Substitute 0.32494 for
Find the angle of twist:
Substitute
Therefore, the angle of twist of the section is
Want to see more full solutions like this?
Chapter 3 Solutions
Mechanics of Materials, 7th Edition
- 5. Two steel springs are arranged in series as shown. The upper spring has 24 turns of 50-mm- diameter wire on a mean radius of 186 mm. The lower spring consists of 16 turns of 28-mm diameter wire on a mean radius of 109 mm . If the ultimate shearing stress in either spring must not exceed 596 MPa, (a) compute the maximum value of P and total elongation using factor of safety equal to 3.0. (b) What is the equivalent spring constant? Use G = 83 GPa.arrow_forwardFor the bronze segment AB, the maximum shearing stress is limited to 8000 psi and for the steel segment BC, it is limited to 12 ksi. Determine the diameters of each segment so that each material will be simultaneously stressed to its permissible limit when a torque T = 12 kip-ft is applied. For bronze, G = 6 × 106 psi and for steel, G = 12 × 106 psi. Bronze Steel B 4 ft 6 ftarrow_forwardE' 3. For the 3-in.-diameter solid cylinder and loading shown, T = 40 kip-in., determine: (a) the polar moment of inertia, (b) the maximum shearing stress, (c) the maximum shearing strain, and (d) the angle of twist d. Use G=77 GPa for the modulus of rigidity and the length of the rod as 10 ft. 3 in.arrow_forward
- A composite shaft is made by slipping a bronze tube of 3-in. outer diameteran -in. inner diameter over a solid steel shaft of the same length and 2-in.-diameter. The two components are then fastened rigidly together at their ends. What is the largest torque that can be carried by the composite shaft if the working shear stresses are 8 ksi for bronze and 12 ksi for the steel? For bronze, G = 6 x 106 psi, and for steel, G = 12 x 106 psiarrow_forward1. A steel shaft is subjected to the torques shown. The shaft is solid with a diameter of 1 in and G = 12,000 ksi. Determine: (a) internal torque (lb ft) in each segment. Include the necessary diagram to show these internal forces (b) maximum shear stress (psi) in each segment (c) rotation angle of pulley D w/r to the support of A. Follow the rule of signs given in the lecture. Include the necessary FBDs of sections.arrow_forwardPROBLEM 3.52 A 4 kNm torque T is applied at end A of the composite shaft shown. Knowing that the shear modulus is 77 GPa for the steel and 27 GPa for the aluminium, determine (a) the maximum shear stress in the steel core, (b) the maximum shear stress in the aluminium jacket, and (c) the angle of twist at A. [Ans. (a) 73.6 MPa (b) 34.4 MPa (c) 5.07°] 72 mm 54 mm 2.5 m Steel core Aluminium jackeť Fig. P3.52 and P3.53 22:37 e dx D 14/04/2022 BANG & OLUFSEN 40 delete home end pg up pg dn num backspace lock W ERT U %23 home og up 4.arrow_forward
- PROBLEM 3.52 A 4 kNm torque T is applied at end A of the composite shaft shown. Knowing that the shear modulus is 77 GPa for the steel and 27 GPa for the aluminium, determine (a) the maximum shear stress in the steel core, (b) the maximum shear stress in the aluminium jacket, and (c) the angle of twist at A. [Ans. (a) 73.6 MPa (b) 34.4 MPa (c) 5.07°] Hint: angle of twist at 72 mm end A is same for core and jacket 54 mm A 2.5 m Steel core Aluminium jacket Fig. P3.52 and P3.53 22:35 BANG & OLUFSEN delete home end og up pg dn num backspace 4 lock Q WE T U 080 home pg uparrow_forward3. Two solid 1.75 in.-diameter steel shafts are connected by the gears shown. The shaft lengths are L₁= 6 ft and L2 = 10 ft. Assume that the shear modulus of both shafts is G= 12,000 ksi and that the bearings shown allow free rotation of the shafts. If the torque applied at gear D is TD= 225 lb-ft, determine (a) the internal torques T₁ and 72 in the two shafts. (b) the angles of twist 1 and 2. (c) the rotation angles 3 and þc of gears B and G. (d) the rotation angle of gear D. 4₁ (1) 30 teeth 54 teeth B (2)arrow_forwardProblem 3.3, I’m not asking for problem 3.4 just 3.3 pleasearrow_forward
- The composite shaft shown consists of a 0.2-in.-thick brass jacket (Gbrass= 5.6 × 106 psi) bonded to a 1.2-in.-diameter steel core (Gsteel=11.2 × 106 psi). Knowing that the shaft is being subjected to the torques shown, determine the largest angle through which it can be twisted if the following allowable stresses are not to be exceeded: τsteel=15 ksi and τbrass= 8 ksi.arrow_forwardA 88-N·m torque is applied to a hollow shaft having the cross section shown. Neglecting the effect of stress concentrations, determine the shearing stress at points a and b. The shearing stress at point a is MPa. The shearing stress at point b is MPa.arrow_forwardPlease answer number 3.42arrow_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