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Mechanics Question : Maximum Load needed ?? Urgent Please Help
http://img694.imageshack.us/img694/5462/66814773.jpg
A new steel bracket was designed to support various loads being hung from the end. The designer would like
to know the largest load that can be hung on the bracket.
What is known:
-Allowable yield stress for the steel is 36 ksi.
-The bracket safety of factor must be at least 2.
-Each arm of the bracket is 6 in long.
2. The Question
What is the maximum load that can be placed on the bracket?
First of all i converted all the units into SI units ,, Ksi into N/m2 & in into m
The question gave me the procedure to solve this problem but i am stucking on each step ,, so help is really appreciated friends.
The procedures given :
-Calculate the maximum torsion in the circular bar.
-Calculate the maximum bending stress in the circular bar.
-Determine the stress state at the wall (location of the maximum bending stress and torsional stress)
-Find the principal stresses and using the given factor of safety and the yield stress of steel calculate the
allowable Load.
For the first procedure :
Calculate the maximum torsion in the circular bar.
Do they mean the shear stress or it is different .
I used This formula : σmax = T R / Ip
where , σmax is shear stress , T is the torque , R is bar of shaft radius , Ip is moment of inertia.
I found torque by using this formula : T = G * σall
G = Z = Ip/c
σall is given = 248.2e6 N/m2
I found these numbers please if there is any mistake please inform me ,,,
c = diameter / 2 = 0.009525 m
Ip = J = pi/2 * R4 = 1.2929e-8 m4
Then G = Z = J/c = 1.357e-6 m3
Then torque T = G * σall = 336.807 N.m
Finally i found shear stress to be = 248.13e6 N/m2
Which is as same as the σall that is given ,, Is this reasonable ??
Please correct the mistakes if found by you ,,, help me to finish the first step ??
Thx in advance for all.
Homework Statement
http://img694.imageshack.us/img694/5462/66814773.jpg
A new steel bracket was designed to support various loads being hung from the end. The designer would like
to know the largest load that can be hung on the bracket.
What is known:
-Allowable yield stress for the steel is 36 ksi.
-The bracket safety of factor must be at least 2.
-Each arm of the bracket is 6 in long.
2. The Question
What is the maximum load that can be placed on the bracket?
The Attempt at a Solution
First of all i converted all the units into SI units ,, Ksi into N/m2 & in into m
The question gave me the procedure to solve this problem but i am stucking on each step ,, so help is really appreciated friends.
The procedures given :
-Calculate the maximum torsion in the circular bar.
-Calculate the maximum bending stress in the circular bar.
-Determine the stress state at the wall (location of the maximum bending stress and torsional stress)
-Find the principal stresses and using the given factor of safety and the yield stress of steel calculate the
allowable Load.
For the first procedure :
Calculate the maximum torsion in the circular bar.
Do they mean the shear stress or it is different .
I used This formula : σmax = T R / Ip
where , σmax is shear stress , T is the torque , R is bar of shaft radius , Ip is moment of inertia.
I found torque by using this formula : T = G * σall
G = Z = Ip/c
σall is given = 248.2e6 N/m2
I found these numbers please if there is any mistake please inform me ,,,
c = diameter / 2 = 0.009525 m
Ip = J = pi/2 * R4 = 1.2929e-8 m4
Then G = Z = J/c = 1.357e-6 m3
Then torque T = G * σall = 336.807 N.m
Finally i found shear stress to be = 248.13e6 N/m2
Which is as same as the σall that is given ,, Is this reasonable ??
Please correct the mistakes if found by you ,,, help me to finish the first step ??
Thx in advance for all.
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