Is My Beam Hinge Pin Calculation Correct?

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The discussion focuses on the calculation of a hinge pin size for a beam subjected to specific loading conditions. The original poster provided their calculations for reaction forces, maximum working shear stress, and pin diameter, which were confirmed as correct by another participant. It was noted that the pin is in double shear, allowing for an effective allowable shear stress of 8 MN/m² instead of the initial 4 MN/m² after applying the safety factor. The method for calculating shear stress involved dividing the shear force by the pin's cross-sectional area. The conversation concluded with the original poster expressing understanding of the concepts discussed.
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Appreciate if someone can check my answers to the following question:

A beam hinged at one end and simply supported at the other is subjected to the loading shown. A design calculation is required to determine the suitable size of hinge pin. If wrong, could you point me in the right direction.

Data Shear strength of pin material = 20 MN/m^2
Factor of safety = 5

Drawing of system can be found here:

http://img206.imageshack.us/img206/6549/oc2q3gc5.jpg a) Use these conditions to determine the reaction forces RP and R1b) State the maximum working shear stress in the pin.c) Calculate a suitable diameter for the pin to the nearest millimeter.a) My answers are RP = 1700 N, and R1 = 800 N.

b) My answer 4 MN/m^2

c) My answer = 16 mm
 
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BIGEYE said:
Appreciate if someone can check my answers to the following question:

A beam hinged at one end and simply supported at the other is subjected to the loading shown. A design calculation is required to determine the suitable size of hinge pin. If wrong, could you point me in the right direction.

Data Shear strength of pin material = 20 MN/m^2
Factor of safety = 5

Drawing of system can be found here:

http://img206.imageshack.us/img206/6549/oc2q3gc5.jpg


a) Use these conditions to determine the reaction forces RP and R1


b) State the maximum working shear stress in the pin.


c) Calculate a suitable diameter for the pin to the nearest millimeter.


a) My answers are RP = 1700 N, and R1 = 800 N.

b) My answer 4 MN/m^2

c) My answer = 16 mm
Your answers look correct to me. The pin is in a classic case of double shear.
 
Last edited by a moderator:
Thanks for taking the time.
 
Determine Pin Size?

PhanthomJay said:
Your answers look correct to me. The pin is in a classic case of double shear.

hey guys .. I am just wondering how you went about calculating the maximum shearing stress in the pin, and from that determining pin size ...

your help would be of most value ..
Thanks in advance
 
komy said:
hey guys .. I am just wondering how you went about calculating the maximum shearing stress in the pin, and from that determining pin size ...

your help would be of most value ..
Thanks in advance
The allowable shear stress in the pin with a 5.0 safety factor is "given" as 20/5 = 4MN/m^2. However, the pin is in double shear, so effectively the allowable shear stress can be taken as twice that value (8MN/m^2). The 1700N pin force was calculated from statics. Pin shear stress is shear force divided by cross sectional area of pin (1700N/A = 8MN/m^2). Solve for A, then diameter; watch units and decimal point.

Please ask again if you don't understand the concept of double shear.
 
PhanthomJay said:
The allowable shear stress in the pin with a 5.0 safety factor is "given" as 20/5 = 4MN/m^2. However, the pin is in double shear, so effectively the allowable shear stress can be taken as twice that value (8MN/m^2). The 1700N pin force was calculated from statics. Pin shear stress is shear force divided by cross sectional area of pin (1700N/A = 8MN/m^2). Solve for A, then diameter; watch units and decimal point.

Please ask again if you don't understand the concept of double shear.


mate i totally understand it now ..
thanks for your help !
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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