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
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Chapter 3 Solutions
EBK MECHANICS OF MATERIALS
- I at nd. end 3.8 Attaching a hollow shaft to a solid shaft creates the steel shaft. Determine the maximum torque T that may be applied to the ends of the shaft in the 3.5-m length without exceeding a shear stress of 70 MPa or a twist angle of 2.5°. For steel, use G = 83 GPa. =12 T 120mm Alvern de ww08 1914 90 ****************** 2.5m 2007 099 wwg/ 1.8m Oarrow_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_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°] 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_forward
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