If the shear stress in steel exceeds about 4.30*10^8 N/m2, the steel ruptures.?
954946
force required would be area under shear stress multiplied by the shear stress value(tau) of a particular material, in this case steel.\n\nthe first question is easy. the required area is the cross sectional area of that bolt. pi(r)². please substitute r in meters only.\n\nfor the second question, the shear area is not the area of the circle plate but the area of the ring around it. this area is effectively a rectangle, and can be calculated by multiplying the circle's circumference (2pi x r) by the thickness of the circle (0.0058m).
1187223
Well, a clear question would help. \n\nAre you trying to calculate the power that a certain material can withstand?   Or are you trying to calculate strength?
609246
*When an external force is acting along the section of the body then the force is termed as shear force (Q) and the correspondingstress (q) is called shear stress.\n                    \nshear stress =  SHEAR FORCE / AREA UNDER SHEAR\n                  \n  q = Q / A   \n\n   \n A rivetted joint is a typical example for shear,where the rivet resists shear force across its cross-sectional area A.
4239
it has to do with estimating the steel weight (of a pipe)
438539
You should find that under the ASTM Manual for steel. Your engineering department should have a copy, otherwise try finding it on line.
715379
Kips thousand instructions/second data processing.\n\nTwelve grades of steel, having yield strengths varying between 42 kips per square inch (ksi) and 100 ksi, were used to fabricate the perimeter column and spandrel plates as dictated by the computed gravity and wind demands.
1093237
you could have a stress fracture
128277
Typically not. In engineering calculations we employ a strength allowable knockdown in the Heat-effected-zone near the weld.
117533
Hardened steel cannot bend like iron, due to it's crystallic structure, if it is forced to bend, it brakes.
311499
Work = force * distance\n= 1000 N * 10 m\n= 10000 Nm\n\nYou wouldn't take into consideration the mass of the car because the force you are applying is horizontal (assuming that you are pushing straight on the car and not down at an angle).  The only time you would use the 3000 N is if you took friction into consideration.  In that case, you would have a friction force opposing your 1000 N, thus reducing it.  (Force of friction = coefficient of friction * normal force)\n\nPower = work / time\n= 10000 Nm / 100 s\n= 100 Nm/s
456275
9.81 m/s² or 32.2 ft/s²
93591
The rather loose grammar in use on Yahooanswers makes your question ambiguous.  Do you want to know what of what value stiff steel is? If so, the answer is obvious; the uses of steel in construction, consumer products, automobiles, etc. are myriad.\n\nIf you mean instead, what numerical value of "stiffness" does steel posses, there are several possible answers.  One would be the shear modulus of steel, which is a measure of how much a substance deforms when a shearing force is applied.  It is define as S= Shear stress/shear strain= F/A)/ (DEL X/h) where F is the force applied parallel to an object's fixed end, A is the area over which the force is applied, DEL x is the distance the object deforms, and h is the height of the object.  For steel, this is 8.4 x 10^10 Pascals.  Compare this to 2.5 x 10^10 Pascals for aluminum, and 14 x 10^10 Pa for tungsten.
763604
a) The stress in the wire is given by force per unit area = m*g/A\nIf Y is the Young's modulus of the material of the wire, \nY = stress / strain. Therefore \nstrain = stress / Y.\nElongation = strain*original length.\nb) When the object is accelerated upward with an acceleration of a, the force acting in the wire = m*(g+a)\nThe rest of the equations above are valid.\nc) If the elastic limit of the wire is E, the greatest mass that can be accelerated up with an acceleration of a is got from \nE = Mmax*(g+a). Hence\nMmax = E/(g+a)
896228
Let's see.. if the steel is thick, probably a lot of it
687946
In fatigue analysis where you apply a load range to your FEA model, i.e. if you have a load of +/- 10kN then you would apply a single load of 20kN to represent the whole load amplitude.\n\nWhen you come to look at the stresses you look at maximum principal and minimum principal stresses to determine the location and level of peak stress range.  The rationale behind this is that you could just as easily apply your load in the opposite direction, in which case all the stresses would be reversed.\n\nDon't forget that the magnitude of the minimum principal stress isn't necessarily smaller than the maximum, it can be large and negative.
172509
Only partly in jest... beat it with a hammer and shatter the ceramic, or heat the steel til it melts and flows away from the ceramic -- both assume some sort of bond between the ceramic and steel.  The actual quickest answer depends on how the ceramic and steel are attached.  It could be as simple as lifting which ever is on top.  \n\nSee what happens when you as an ill-defined question?
278543
that depends on which ceramic you're talking about.  There are hundreds of different ceramics which will interact with steel in different ways.\n\nAlso, the coefficients of static and kinetic friction are different