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1

The graph below shows a fixed-end frame subjected to a concentrated load in the mid-span of the horizontal member and the resulting deflected shape. Let's draw tangent lines on the deflected shapes at joints B, we note that both the curved segments of BF & BE are deflected away from the respective tangent lines with the angle of rotation $\theta = 0$, ...


0

Let's start from the horizontal beam on top. if the joint at the ends of this beam where a pin joint you would get a maximum moment at the center $$M_{c}= \frac{PL}{4} $$ But because they are fixed after the beam deflects under the load the corner joint rotates a little and eventually stops at equilibrium. Both beam and the column bend a bit with the joint ...


0

The energy required to break an object will be independent of the speed of the thing hitting it. This is known as the toughness of a material. That energy is what the striking object will need to impart and this will have the same impact energy imparted to itself. The way this is used in practice on the roadways is by having wooden posts at the end of ...


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These scenarios are too simplified to be able to form an opinion. But generally "unyielding barrier" will cause more damage. Let's say the yielding barrier never breaks, only bends and moves one meter. assume these factors just for the sake of a perspective. car mass M kg Car speed at crash V d, distance car moves to a full stop 2 meters for 1st ...


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For simplicity, let's hold the speed constant, the rigidity of the walls variable. The sketch below conceptually depicts two scenarios - the car runs into a non-yielding (rigid) wall and then runs into a defective yielding wall. Car A suffers damage due to the full impact force. The impact of car B will be absorbed partially by the crushed wall, which ...


0

Your graph tells a different story. From the property of the Mohr's Circle, $\tau_{max} = \dfrac {\sigma_1 - \sigma_2}{2}$ For $\sigma_2 = 0$, (uniaxial loading) $\sigma_1 = 2\tau_{max}$, or as indicated in the graph, $\sigma_y = 2\tau_y$ In which, $\sigma_1$ is the tensile stress. Does this answer your question?


1

Actually it is. When a material is loaded the stress state in the material is the same - it does not depend on the orientation that the stresses are analysed. Mohr's circle is a way of visualising a stress state at different orientations/planes. At yielding of the material, the stress state is that the maximum shear strength at yielding is equal to the ...


2

One does not want the rupture in the grips; that makes elongation and reduction of area measurements very difficult. Also, notch sensitive materials will rupture in the grip area unless the the grips are significantly larger cross-section than the gage length. For notch sensitive materials even the shoulders must have a long gradual taper to avoid shoulder ...


3

The stress is not uniform near the grip because there is a biaxial state of stress. There is the axial force the compression force from the grips. That is the reason that if the Cross-section was uniform the failure would start from the grip. the mental image is the following: If you have a ballon that you apply force from one end, and at the same time ...


2

As far as I know, the enlarged part is used to grip the specimen. The fillet and smaller cross-section area is where the necking happens and that smaller cross-section is designed to avoid necking on the grip. I think the failure does not happen at the end. If its completely axial load, the whole specimen will elongate and become thin, then the failure would ...


1

Assuming the force is placed at a distance L from the support, then the moment of the Force P with respect to the foundation will be $ M = P\cdot L$. If the horizontal square tubing tilts by an angle $ theta$ then the will be equal to $M = O \cdot L\cdot \cos\theta$ For angle up to 10 degrees, the $\cos \theta$ has a minimal change ($\cos 10^o =0.984$), so ...


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Even if the connection between the cantilever tube and the short tubing is loose and the cantilever tube will rotate a bit before it engages the bracket, it still works and imparts the same moment as long as the rotation doesn't cause it to slide off. You'd be surprised that many lifting jacks count on the biting force of a miss-aligned connection to work. ...


1

Q: Can I assume the bending moment is simply the load times the distance from the load to the center of the column? A: Yes. You have to make the assumption that the cantilever and the short tubing are fitted tightly, so the resultant forces remain in the same locations throughout the process. That is, there are no excessive geometry changes at the beam-...


1

If the load is applied in flexure (i.e. bending), then a small amount of material a long way from the neutral plane can contribute more to the load-bearing capacity than a large amount of material near the neutral plane (that's what the second moment of area is all about). Similarly, if the load is applied in torsion then a small amount of material a long ...


5

I just found an answear, the keyword is "stress concentration". Sometimes, the material is removed to make a stress relief in an element. Some examples: The origin source I found it: https://www.youtube.com/watch?v=QtSki5nfO2g


2

First courses on engineering statics tend to start from the forces where the structure is loaded and constrained, and at the interface between components, because in simple situations (for example statically determinate structures) the forces can be calculated independently of the internal stresses in the structure. The stresses can then by found (...


1

In the mechanics of material, the most important strength parameters are measured and expressed in terms of "stress" rather than force. For instance, the "yield" strength of a Grade 60 steel is 60 ksi, which is a well-known standard in the engineering community. However, how much force a piece of grade 60 steel can sustain at yield is ...


3

A good reading about that is the: History of strength of materials by Timoshenko. It describes the history behind Galileo Galilei and Hooke until the era of Cauchy, and the intermediate steps Galileo (sometime between 1592 and 1610) figured out that for tension, when you double the cross-section the load to break also doubles. But run into problem with ...


0

one example would be a very long tower under self weight. if you remove mass from the columns on the higher levels where it is not needed, the structure Will be stronger, because you have reduced the dead load. a cantilever heavy beam will be stronger if you taper the section and remove some of the self weight.


2

If a compressive force is applied, the bar will only experience compressive forces. If a tensile force is applied, the bar will only experience tensile force. They can only occur simultaneously on the same bar if the bar is subjected to bending. The picture you have shared, it can be one of these cases. 1) Either you are pulling (tensile) or pushing (...


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In engineering we have external forces and internal forces or reactions. extrnal forces don't necessarily have to comtogether. eg, you can apply tension force tangentially to a bicycle wheel and it will act as a torque and wheel will turn.


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This answer is to point out the possible reason for the "wrong" result getting from the superposition method. The method of superposition is valid and is desirable for this problem, but your assumptions on the support rigidity could have been wrong, thus causing the mistake. The end of the beam on the right (point C) is sitting on another beam ...


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The brackets don't play a role unless they have some Moment or external load acting on them, or their self-weight is significant, which is not shown here. This is a typical problem and the solution can readily be found in many design aids. by searching for a two-span beam with a concentrated load on the mid-span. An easy way of calculating the reactions is ...


2

I think that if you apply tensile loading, and a uniform cross-section is used, then if you remove material you will end up with a lower strength. The only class of examples I can think of (although not for static loading) is removing mass from a rotating element like a gear or a pulley. The reason is that the added mass increases the inertial forces (when ...


0

This answer addresses design concerns as well as structural reactions. As the pictures are shown below, $T, C, R$ are the global reactions required for the anchorage design, and $V, M$ are internal reactions for checking the connection rod/pin and the tube steel. Note the local reactions $H$ & $m$ may not be significant comparing to $V$ and $M$ if $d$ is ...


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This was intended more a comment regarding the third image, (however I opted to put a The confusing part for me is, the arch like section in the middle, and the differences between the style of the walls, and the ceiling (which appears to be wood). So it seems to me to be some type of renovation. If you blow up the image, you will notice that the final row ...


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All structure openings require either lintel/edge beams or lintel plates to support the load above. For brick and reinforced concrete structures, the lintel/edge beam can be made of reinforced concrete or CMU lintel blocks with steel reinforcement for large openings. For structural steel construction, the lintel/edge beam can either be made of reinforced ...


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These are veneered brickwork used over either concrete or steel structure with proper dowels or fasteners supporting the bricks. These fasteners are regulated by code to support both vertical loads and lateral loads of a potential earthquake and also depending on the type of the building and its function meet fire codes as well. . Here is an example of false ...


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Your first photo looks like it is a concrete structure and the bricks are only glued onto it.


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