For finding reactions for simply supported beam with uniformly distributed load, first we have to convert the u.d.l into a single point load. And then we have to consider it to be a simply supported beam with a point load and solve it. I think you know how to calculate the reactions for beam with point load.
You can solve this question with help of your brain.
If you dont have brain you cant solve it.
Thank you for faithfully mallo
only one support in y-axis direction
A uniformly distributed load is one which the load is spread evenly across the full length of the beam (i.e. there is equal loading per unit length of the beam).
these are the load varrying uniformly from zero to a particular value and spread over a certain length of the beam.Such load is also called triangular load.The total load can be obtained by calculating the total area of triangle & multiplied if by the intensity or rate of loading.The total load will act through the centroid of the triangle.
SO that the oil cup gets uniformly heated from all the sides
I'm not sure if you're asking for just the equation for the reactions, or how to calculate the FEM of a cantilever via the Moment Distribution method (aka Hardy Cross method), so I'll anser both.If you assume clockwise moments are negative...If your cantilevered span has a length "L", and a uniform load "W" acting downward along the entirelength of the span, then the fixed end moment is (W*L2)/8If the fixed end is to the left, the equation is positive (the load is acting clockwise, so the reacting FEM is counter-clockwise, therefore positive.) If the fixed end is on the right, the equation becomes -(W*L2)/8If your uniform load W is acting downward, and is acting along length "d" (which is NOT the entire span), its easiest to convert the uniform load W to a point load "P", where P = W*dThen your FEM = [P*a*b*(2*L-a)]/(2*L2)where:L = entire length of spana = distance P is acting from the fixed end,b = distance P is acting from the cantilevered end (also, b = L - a)OR, if W is centered along the span (therefore Pis also centered), then the equation becomes (3*P*L)/16When going through the distributive process, the cantilevered end will ALWAYS have a final moment of zero. At the fixed point, the cantilevered side has a distribution factor of zero, while the other side has a D.F. of one. So, if you have length ABCD, where A is a cantelever, and B, C and D are assumed fixed, FEMAB = 0, D.F.BA = 0, D.F.BC = 1, and D.F.CB = D.F.CD = D.F.DC = 0.5So what ends up happening is that the final FEM for the cantilever ("AB") is zero, and the opposite end of the span ("BA") remains as its initial value. The reason for this is when you sum up the reactions at the point (in this case B) supporting the cantilever and flip the sign, the distribution factors tell you to distribute everything to the side of the support opposite the cantilever ( -(FEMBA + FEMBC) is added to FEMBC). So you can go through the Cross Method as normal, but you don't touch AB or BA, and BC is adjusted only the once that I just mentioned (you don't carry over the 1/2 reactions from CB).
The modulus of elasticity is a property specific to a given material and in practice is derived through laboratory testing. The modulus of elasticity is defined as stress/strain. One would have to apply a force uniformly over a known cross section of a material and monitor the strain utilising strain gauges. When the results are plotted you will notice that you get elastic behaviour up to a point of yield (this is known as the yield stress in normal carbon steels, however in stainless steel where the yield point is not as defined, we normally accept it to be the 0.2% strain) and the material should behave linearly in this area. If you take the gradient of the stress/strain, this will be your Elastic modulus. Please note that the plotted curve will begin to flatten off roughly at the 0.2% strain line and this is due to the fact that the material has yielded. even after this point the material will not fail but will act 'plastically' up to a point where the material fractures which we call the ultimate stress.
Homogeneous mixture -uniformly distributed throughout the composition heterogeneous mixture -not uniformly distributed throughout the composition
equilibrium
no. the greener the part is the more it has, as a rule of thumb
A heterogeneous mixture is a mixture of two or more substances that are not uniformly distributed. The iodine and water in the question are not homogeneous because they are not uniformly distributed.
A uniformly distributed load is one which the load is spread evenly across the full length of the beam (i.e. there is equal loading per unit length of the beam).
w(l^2)/8 w = 38N l = 5m
Solution
loads are carried out as point load uniformly distributed load and uniformly varying load
The two categories of mixtures are heterogeneous and homogeneous. In a homogeneous mixture the components are uniformly distributed throughout the mixture. Homogeneous mixtures are solutions, such as salt water. In a heterogeneous mixture, the components are not uniformly distributed, such as granite, or Pizza.
Actually, its' all of them because anything can be distributed into different types of matter.
Uniformly distributed loads, also known as uniformly distributed loads (UDL), refer to loads that are evenly distributed over a given length or area of a structural element. They exert a constant magnitude per unit length or unit area along the specified region. In the case of one-dimensional structural elements like beams or slabs, a uniformly distributed load applies a constant force or weight per unit length. For example, a beam with a UDL of 10 kN/m means that there is a load of 10 kilonewtons acting on every meter of the beam's length. In two-dimensional elements like plates or surfaces, uniformly distributed loads apply a constant pressure or weight per unit area. For instance, a floor slab with a UDL of 5 kN/m² means that there is a load of 5 kilonewtons per square meter acting on the entire surface area of the slab. Uniformly distributed loads are commonly encountered in various structural applications, such as floor loads in buildings, self-weight of structural elements, dead loads, or evenly distributed loads from equipment or storage. They allow for simplified analysis and design calculations since the load intensity remains constant over the specified area or length. When analyzing or designing structures subjected to uniformly distributed loads, engineers consider the load magnitude, the span or length of the element, and the support conditions. By applying principles of structural mechanics and equilibrium, they can determine the internal forces, moments, deflections, and overall behavior of the structure under the UDL. It's important to note that UDLs are an idealization of real-life loading conditions. In practice, actual loads may vary or have different distributions, requiring engineers to consider more complex load patterns and combinations to accurately analyze and design structures.
A uniformly distributed load (UDL) is a load which is spread over a beam in such a way that each unit length is loaded to the same extent.