Please, Where the coefficient: 0.45; -0.69; -0.42; -0.35, for Load case A (East-West)????. I found on the code at page 313. Figure 28.3-1, the coefficient should be: 0.4;-0.69; -0.37; -0.29, for roof angle 0-5 degree
The inclination angle for our roof is about 11.6 degrees. We need to interpolate the values given in the table in Figure 28.3-1 to arrive at the correct values for the roof.
Thank you Dr. structure for the comprehensive video explaining the application of the wind pressures calculated by ASCE 7 procedure which I could not find anywhere on the web or elsewhere. I have to say finding those required pressures is a piece of cake but how you applied those to the building is none existence. A lot of videos out there are elementary knowledge and have no value at all. I was stuck for months on that and now I can proceed!. Thanks.
@@DrStructure Follow up questions: Referring M&S diagram for Member AB, Case A-1 , is the 85KN vertical load , the gravity axial wind load for use to design the foundation and 45KN, the base shear due to wind load?
All the shown loads are due to wind. Needless to say, for the actual design of the structure, other loads and load combinations must be considered. Hypothetically speaking, if wind was the sole cause of member forces, then we would need to design AB as a beam (since it is subjected to bending and shear) and as a tension member (since it is subjected to an axial tensile force). All three forces at the base of AB transfer to the foundation, which means we would need to design the foundation to resist the vertical force, the horizontal force, and the bending moment.
In ASCE Wind load calculation for Enclosed and Partially Enclosed building requires to design a shed building using: 1) both Positive internal pressure coefficient +GCpi and Negative internal pressure coefficient -GCpi 2) both External pressure coefficient, Cp (e.g. -0.9, -0.18) for ROOF. If I consider the wind direction is Normal to the Ridge: +Wx (Roof angle< 10 degrees & h/L
Yes, we do need to consider all (often seemingly too many) load combinations for each element that is being designed. The consequence of not doing so, potentially overlooking a critical load combination case, could result in structural failure and possible lose of life.
We are treating the hangar as a closed structure, with the hangar door kept closed during high winds. Consequently, the wind impact on the hangar is identical for both the North-South and South-North directions.
At around 6:50 onward the B-1 and B-2 load cases utilise pressure coefficients Cpe and Cpi for what looks like a transverse (wind across the ridge) loading case. I thought that those transverse loadings were calculated as load cases A-1 and A-2. I was expecting the second set of load cases at B-1 and B-2 to utilise longitudinal loading coefficients as the wind is applied N-S on gable-ended walls (including the hangar door which is closed in adverse wind events and making it a Type C normal enclosed building). Can you please explain why the transverse coefficients were utilised for the Case B load cases? If this is confirmed to be an error, what are the longitudinal coefficients to use? Overall, I can readily follow the general calculation processes and find these videos a fantastic resource for underlying analysis principles. Thanks.
The external pressure coefficients for the East-West (transversal) wind direction are obtained, via interpolation, from Figure 28.3-1 of AISC 7-16 (under the heading Load Case A on top of page 313). These coefficients (shown @5:22) are: (-0.35,-0.42,-0.69,0.45) The coefficients for North-South (longitudinal) wind direction (Figure 28.3-1 under heading Load Case B) are: (-0.45,-0.37,-0.69,-0.45). These are the coefficients used for loading case B @7:00. The two sets of coefficients are not the same, although they are some similarities between them. Figure 28.3-1 of AISC 7-16: lab101.space/qa/windpressurecoefficients.pdf
At 9:50, it's stated that: We then transform the member loads to nodal forces. This is done member by member by first calculating the member fixed end forces then transforming the forces from the local coordinate system to the global coordinate system. I can't follow how the forces are transformed from the local coordinate system to the global coordinate system, e.g., member AB.
F = Qf where f is the force vector in local system, F is the force vector in the global system, and Q is the transformation matrix. The details are provided in Lectures SA49 and SA50.
@@DrStructure At 5:58 of SA49, the forces F1, F2, F3, F4 of member AB correspond to the degrees of the freedom D1, D2, D3, D4. At 10:04 of SA52, the forces 18.24, 0, -29.18, 18.24, 0, 29.18 doesn't correspond to the degrees of freedom of member AB since member AB has 3 degrees of freedom, not 6 DoF.
@@hoshangmustafa9672 Each beam/frame member in isolation (without any boundary conditions imposed) has 6 degrees of freedom. If we label the ends of the member i and j, we get three degrees of freedom at i and 3 degrees of freedom at j. That means the local and system force vectors for such a member are vectors of dimension 6x1. @10:10 of SA52, we are given the 6x1 member force vectors in the system coordinate system. Each vector is obtained by multiplying the 6x6 transformation vector by the 6x1 force vector in the local coordinate system. All of this is done independently of the degrees of freedom of the structure. While for members BC and CD all six force vector elements are used to determine the system force vector, for members AB and ED, since they are fixed at the base (there are no system degrees of freedom associated with the support joints) only 3 of the vector elements are used in determining the global force vector. For member AB, elements 4, 5, and 6 of the force vector correspond to system DOFs 1, 2, and 3. For member ED, force vector elements 4, 5, and 6 correspond to the system DOFs 7, 8, 9.
@@hoshangmustafa9672 @5:58 SA49, we have the transformation of the 6x1 force vector from the local to global/system coordinate system. That is consistent with what we have discussed above. Each member in isolation has a 6x1 force vector, 6x6 stiffness matrix, and 6x1 displacement vector in both local and global coordinate systems. Here, since there are no member loads applied to the frame, the system force vector is composed directly using the joint loads.
We use several software tools including: SketchUp, Adobe Illustrator, and Camtasia Studio. The 3D model was created, and animated using SketchUp. Text and basic graphics were created and animated using Illustrator, additional animation were added and the entire video produced in Camtasia Studio.
Excellent and informative video. However in the code it mentions a minimum design load of 0.77 kN/m2 to be used why didn’t we use it here as some of the values (at minute 06:40 ) are bellow it 323 ,63 kN/m2 ... Again Thank you so much for this excellent video.
8:38s , To calculate the concentrated load to the frame from the roof panel (at each joint), why you don't get the reaction at each support of beam (=concentrated load of each joint)? I saw you multiple the distributed load with 6m (e.g., 2.64KN/m x 6m=15.84 KN)
Yes, we can calculate the vertical support reactions for those long transversal beams and then apply them to the frames. But that requires analyzing those indeterminate beams first. What we did here is an approximation technique that works in practice.
Design of structural systems is often governed by local/state/regional codes. For example, in the US, the ASCE code offers the following basic load combinations for designing members: 1.4D 1.2D + 1.6L + 0.5(Lr or S or R) 1.2D + 1.6(Lr or S or R) + (L or 0.5W) 1.2D + 1.0W + L + 0.5(Lr or S or R) 0.9D + 1.0W where D is dead load, L is live load, Lr is roof live load, W is wind load, S is snow load, R is rain load. There are another set of equations for dealing with seismic loads. The equation that results in the largest value for the given member controls the design of the member. The method described in the lecture treats wind load for the roof and the walls. We cannot separate them. So the calculations presented encompass both. But, for design purposes, we do need to analyze the frame under dead load, live load and possibly snow load, then use the combination equations to determining the governing set of values when designing the individual frame members.
Hi ! i have a question at 10:40 where we get the Member-end Forces for each member, shouldn't F1 and F2 for each force switch places because when drawing the Shear and Moment diagram later thay seem to be drawn in the oposite way for example, F1 wich is the force to X direction is 42.99 kn but in our results at 10:40 it is -84.94 kN. looking forward to your reply, Thank you :)
Member forces are always given/computed in the local (member) coordinate system. So, for member AB at 10:40, -84.94 is an axial force (in local x which corresponds to global y). And 42.99 is the force in local y (or global x), which means the shear force at the left/bottom end of the member is 42.99 kN.
@@DrStructure Maybe i am missing something here, at SA51: Analysis of a Tea House Frame at 7:15 when you draw the forces on the member you keep the same rule ? The forces drawn are ok, but they are not in the order you state above...
The stiffness matrix for a member can be defined in two related ways: in local coordinate system and in global coordinate system. We can go back and forth between the two via a coordinate transformation matrix. This allows us to express member forces and displacements in both coordinate systems. When formulating the system matrix, for determining joints displacements/rotations, we operate in the global coordinate system. Once the displacements/rotations are computed, and we want to examine each member individually, we often work in the local coordinate system. This means, after calculating the displacements, and member forces, in the global coordinate system, we multiply them by the transformation matrix to get them into the local coordinate system. In SA51, given the simplicity of the problem, that was not done. I will have SA52 checked to make sure it is error free.
Computing F for member AB we have F=KAB*D F={-43,-85,228.66,79.47,83,359.15} which corresponds to the diagram of member AB case 1. In the exercise, to get the correct sense, you just have to multiply the transformation matrix by F of member AB, because F is in local coordinates.
For load case A, there are two data. The data for A-1 (positive internal pressure) & the data for A-2 (negative internal pressure). A-1 & A-2 (@6.45 minutes) shows different values for a single frame. which load to consider in between two cases for the final design? Though you have considered only case A-1 in your video. But, In practical design, which values to consider? The maximum value between two values? Please help me with your answer.
Note that @11:00 the shear and moment diagrams for both cases (A-1 and A-2) are drawn, meaning we need to consider both (or all load cases) when designing a member. More specifically, we need to draw the shear and moment diagrams for the beam under all load cases, then compare and contrast them in order to understand how shear and moment vary in the beam, and where the maximum values occur. For example, while maximum positive moment in the member may develop under one load case, maximum negative moment could occur under a different load case, and maximum shear under a third load case. We need to know them all if we are going to properly design the member.
For the life of me, I find it so unbelievable that the leeward side can ever experience anything close to even half the windward side. I work on projects both here in the US and in Canada, and I'm constantly calculating Wind Loads (Windward WW) of Leeward W=.625WW, Parallel W=.71WW, Roof W=.585WW. That just seems insane to me. Because I've been on the Leeward and parallel sides of buildings during high winds. I've never seen anything close to those kinds of loads.
@@DrStructure most of my engineering is to ASCE 7-10 and -16. I have some current projects that I have to produce the loading according to Canada's NBC. I just never have seen the kinds of loads for parallel and leeward sides that the codes develop.
As you know, most modern design standards are probability based, taking into account the risk (mainly to human life) associated with severe loading scenarios, even though the structure may not encounter such a scenario during its life cycle.
@@DrStructure yes, I am aware. Not that you would know, but I am a PE and own an engineering firm, Quanta Engineering LLC. Still, I've just never seen the wind pressures on the parallel and leeward sides ever experience what the codes show...
An explanation is more than wonderful, but if I want to use an artificial intelligence network to predict the response of this structure to wind loads at different speeds, is there a Matlab code or an analysis program that can find the response of the structure by simply changing the parameters affecting the wind load and its speed
Dynamic or time-varying analysis of structures is well beyond the scope of our video lectures. For that, simulation software can be used to model and analyze the behavior of the system under various loading scenarios. You may want to look into the commercially available structural analysis (finite element analysis) tools and see which ones best fit your computational needs. Except for a few well-defined problems, "Artificial Intelligence" technologies, beyond the perception being projected, is still in the research stage when it comes to reasoning and problem solving that require deep and non-trivial human expertise.
Dr. Structure I would like to say a big thanks for the video. But I'm wondering if we add crain console to the columns how can we re-write the system stiffness matrix?
Any additional (structural) element added to the system should be treated as a separate element with its own stiffness matrix. For example, a crane console attached to a column could be treated as a (cantilever) beam with its own stiffness. This would not alter the stiffness of the column. Having its own stiffness, the cantilever beam simply contributes to the stiffness of the frame, like any other beam or column.
The internal coefficient ci = 0.18 seems to be too low for hanger type structure. Should it be 0.55 instead (partially enclosed building, or, open front structure)? The difference is not insignificant. Please confirm. Thanks.
Yes, if the structure is truly only partially enclosed, a higher value for the internal pressure coefficient can be used. But if the hanger is equipped with doors that can be closed, and there is no significant opening in any of the walls, the lower internal pressure coefficient would be applicable.
Could you kindly explain why the bending moment diagram resulting from lateral uniformly distributed load on the frame (column) is a linear pattern? isn't it supposed to be a second degree curve?
Yes, indeed! The shape of the moment diagram for AB, and ED, is not correct, it should be a second degree polynomial since shear is linear in the member. Also, the moment diagrams for BC and CD should be of linear type since shear is piece-wise constant in those members. Thanks for point this point.
Thanks for the clarification. I have had a discussion with friends about the bending and shear moment diagram of a statically indeterminate (two-story two bay, fix supports) frames subjected to lateral uniformly distributed load along its length from one side, but we all were not so sure about the accurate solution. It would be very helpful if you could explain (or make a video) how to calculate and draw the diagrams for this particular case. I have tried to use SAP2000 vs. Robot to draw them, at the column subjected to UDL, SAP2000 gave me linear pattern while Robot showed me second degree curve.
We will work on a video for that, but that will take some time. The analysis of the frame can be done using a standard method such as the displacement method, or one can use software tools such as SAP2000 for that purpose. Regardless of the method in use, the analysis gives us the member-end forces. In the case of a two-story two bay frame, since we have 6 columns and 4 beams, we end up with 10 sets of member-end forces. In each set, we have three forces for each end of the member (an axial force, a shear force and a bending moment). Shear and moment diagrams need to be done member by member. For each member, if it is subjected to a uniformly distributed load, like the two columns that are subjected to the UDL, shear diagram is going to be linear and moment diagram would be nonlinear (second degree polynomial). For all the other members, since they are not loaded directly, each is going to have a constant shear and a linear moment diagram. Given the member-end forces from the analysis step, you should be able to construct these diagrams. In doing so, you need to pay particular attention to the members subjected to the UDL as the maximum moment in the member needs to be computed as well. This moment is located at the point along the column where shear is zero (where the shear diagram cross the axis.) Some software applications approximate the shape of the diagrams using straight lines. This may not be a critically important factor if the beam is finely discretized into small enough segments.
We use several software tools to create the lectures including Adobe Illustrator, SketchUp, Camtasia Studio, and a few more as the need arises. The choice of software is dictated by the conceived organization of the presentation and how best to realize it. We spend a significant amount of time conceptually constructing the presentation on paper. We then employ the fitting (software) tools to implement what we have on paper.
The lateral bracings at the top of the frame seem unnecessary given that, as you stated, the roof itself could serve that purpose. However, without looking at the details of the analysis and design of the building, and the design code in use, it would be difficult to assess the rationale for such decisions.
@@rawjor I too am not certain if those bracings are actually needed. The 3D model was done based on an actual hanger in a remote location in Eastern Europe, but we don't have the design details of the hanger available to us...
Thanks for your feedback. Yes, you may contribute to our efforts through PayPal. The link is at the bottom of the following web page: lab101.space/Sponsor.asp
Dr. Structure, I have seen in practice engineers for quick sizing and cost estimation of rafters would (skip step) not transfer the load to the purlins but would use the tributary area of the rafter times the area load to compute the line-load for rafter. Here you would not have point loads on rafters, Is this bad practice or even wrong?
How the load is being transferred from the roof panels to the supporting beams depends on the configuration of the assembly. If the roof panels rest directly on the rafters, as opposed to the purlins, or if the rafters and purlins are on the same plane, then the method of estimation that you described is correct; we would end up having line loads on the rafters. However, when the roof panels are resting on the purlins only, and the purlins sit on top of the rafters, as is the case here, then the purlins are subjected to line loads only. The rafters take the roof load at the points of contact with the purlins. Here, distributing the roof load on a rafter uniformly is not a good idea as it would result in underestimating maximum bending moment in the member.
The wall panels are in contact with the columns directly. That causes wind pressure to act on the entire length of the column, as a distributed load. However, the roof panels are not in direct connect with the frame beams. The panels rest on the cross beams which then transfer their loads to the frame at eleven (11) points.
Do you mean how the lateral bracings are attached to concrete beams or columns? There are various techniques for creating attachments to concrete. For example, anchor bolts with sufficient lengths can be placed inside concrete for creating such attachments.
Sorry. I was asking about how lateral loads due to wind or earthquake acting at joints of columns and beams are computed or obtained. In other words, how do we compute for the applied loads due to wind or earthquake acting on concrete frames in buildings?
The forces at a joint of the structure are the same as internal forces at the ends of the members connected to the joint. These forces are computed by analyzing the structure an determining the member-end forces. If lateral forces are significant and present in the structure, we often resort to three-dimensional structural analysis where each joint would has 6 degrees of freedom, three displacements (in x, y and z directions) and three rotations. In such a case, a beam or column would bend along two planes, and it may go under torsion as well. Then, a joint/connection needs to be designed to accommodate in-plane and out-of-pane bending as well as torsion.
Is there a chance that you can be able to assist me with the codes on for the coefficients of the wind loads that you used in the video like ASCE 7-16, or just the important relevant pages or tables
The relevant provisions are covered only in a few pages of ASCE 7-16 which are referenced in the video. The selection of the relevant coefficients depends on the location, purpose and shape of the building. If you have ASCE 7-16 in your possession and have studies it, yet have specific questions, feel free to post them here. I'll do my best to answer them.
Dr. Structure : Thank you for the Response, don't actually have the code because i am based in South Africa for the moment. So i was hoping that you might be having the relevant reference pages because the codes that i have for South Africa are a little bit confusing, i thought getting yours would help me compare which of the procedures is better and easier to get the coefficients.
Unfortunately, due to the copyright issue, I am unable to send you the relevant pages of ASCE 7-16. You might be able to find some information on this online. There might be some powerpoint presentations or pdf files out there ....
The procedure described here is for low-rise buildings only. Mid-rise or high-rise buildings require the use of more comprehensive techniques, such as the use of wind tunnels to accurately estimate the governing wind forces.
@@DrStructure Tqvm. I am still learning on the wind loads. Can you compare your analysis with BS standard CP3 on the walls/cladding? It is like q=0.613*Vs*Vs Vs=1.06*1*1*Vb=1.06*1*1*58m/s =61.48m/s Thus, Vs= 0.613*61.48*62.48 =2317 N/m2 Is this correct? A lot higher than th ASCE standard?
@yikee7592 British standards differ from the American Society of Civil Engineers (ASCE) guidelines in determining the applied loads on a structure, leading to variations in load values. Since we don't have access to CP3, we won't be able to do a direct comparison at this time.
The factor is determined based on several numbers including: Velocity pressure exposure coefficient, Topographic factor, Wind velocity factor, and Ground elevation factor. These are explained in ASCE 7-16.
Hello, you need to have access to the code to know where i’m getting the numbers from but it’s as follows: q = 0.613 kz kzt kd V2 Where, q is the velocity pressure in kN / m2 Kz is the wind speed coefficient due to height Kzt is the topographic factor ( if the building is up by a cliff...etc) otherwise = 1 Kd is the wind direction factor = 0.85 for building, but varies if you design tanks (see the code) v = 58 m/s given ( also in the code) To get kz we have this equation: Kz = 2.01 ( z/zg ) ^ (2 / alpha ) Where z is the height to the mid roof so 9.6 + (13.3-9.6)/2 = 11.45 m = (x3.28 = 37.54 feet) Zg & alpha depend on exposure categories there is type B or C or D Since the hanger is in the open by the sea its type C , B is for city and D for very isolated areas like Siberia xD Form the code type C gives Zg = 900 feet & alpha = 9.5 So kz = 2.01 (37.54/900)^(2/9.5) = 1.0297 So q = 0.613 x 1.0297 x 1x 0.85 x V2 = 0.5365 V2 ( i got some error due to rounding she got 0.534) Good bye!
A canopy can be treated as an open building (not enclosed). There are provisions for open buildings in the code. Similarly, the code provides provisions for roof overhangs. Similar to the enclosed or partially enclosed cases, such provisions specify the required wind pressure for the components of the structure under consideration.
ASCE 7-16 is the most recent version of the document. Certainly it contains updated provisions as compared to the older versions. With regard to the specific provisions used in this lecture, since we don't have a copy of ASCE 7-05 available, we cannot tell how the two sets of provisions differ.
Over all, we need to consider dead load, live load, snow load, and probably a few more depending on the location of the building. There are various load combination equations given in ASCE 7-16 for determining critical shear and moment values for design purposes.
A standard W30x124 steel member was used in the example. The nominal dimensions (in inches) are: Area: 36.5 in^2 Depth: 30.17 in Web thickness: 0.585 in Flange width: 10.515 in
Thanks so much@@DrStructure for your prompt reply. May I ask you a favor? I check the area of W30x124 (A=36.5 in^2 ~ 0.023716 m^2)---> thus, it acceptable. But the moment inertia of W30x124 (Ix=5,360 in^4 , and Iy=181 in^4). The moment inertia is not equal as shown in the video (I=0.00567136 m^4)
I stand corrected. For this example, the standard section that I mentioned was not used. Instead a rectangular section with a height of 1694 mm and width of 14 mm was used. For design purposes, this is not a good choice. But since here the focus was on analysis as opposed to design, we should make a note of it but consider it okay.
I tried to find out the value of "0.534" in formulas: q=0.534V^2. I understand "0.534" is the product factor of : 0.613, Kz, Kzt, Kd, Ke. I tried to use Kz=1.04 (Exposure C); Kzt=1; Kd=0.85; Ke=1 (Page 268-code). But I can't achieve the number "0.534". Plz help me?
To determine Kz, we use the mean height of the roof. In this case, it is about 11.45 meters. Again, we need to interpolate to get the right value for Kz. It comes out to about 1.025, not 1.04. 0.613(1.025)(1)(0.85)(1) = 0.534
@@DrStructure I always do calculation on paper. All your video make me more comprehensive. So, I want more video wind and quake (Static & Dynamic Lecture Video). Dynamic Mechanic for Lateral Load. Quake (Equivalent Lateral Force Procedure,Responce Spectrum,. . .) I'm wating to study from you
You can get the publication directly from ASCE. Below is the link for the latest version of ASCE 7. Scroll to the bottom of the page for info for getting the printed and online versions of the standard. www.asce.org/publications-and-news/civil-engineering-source/article/2021/12/02/updated-asce-7-22-standard-now-available
WHY USE METRIC UNITS ? AMERICA DOESN'T USE METRIC UNITS. QUIT TALKING METRIC WHILE THINKING IMPERIAL. NOBODY HAS THE FEEL OF KN/METER SQUARE.....EVERYBODY KNOWS LBS/SQUARE FEET. I DON'T HAVE TIME TO CONVERT AND THEN GET THE FEEL. KILO-NEWTON TO ME = 1000 ISSAC NEWTONS.
A gentle reminder - the world does not revolve around our preferences and biases. Learning to communicate in multiple languages (be they natural, technical, or measurement related) is a strength, not an inconvenience. Part of the mission of engineering education is to help students across the globe develop a growth mindset, one that potentially could produce more creative and culturally fluid engineers who can then address emerging complex technical problems, both locally and globally. One’s insistence in using a specific way of solving problems (in this case, measuring), although an acceptable personal choice, invariably hinders one’s professional and personal growth in the long run.
For Load Case B, the code (Page 313) offers a single entry table with values for the surfaces of the building. Columns 1 thru 4 give us the coefficients for the side walls and the roof panels. For example, column 3 gives the coefficient for one of the roof panels as -0.37.
In this lecture, we are considering the analysis of a lateral frame only. So, we are interested in the loads that are acting on the plane of the frame. Since the wind pressure on Side 6 does not cause an in-plane load on our frame, it was neglected in the analysis.
Please, Where the coefficient: 0.45; -0.69; -0.42; -0.35, for Load case A (East-West)????. I found on the code at page 313. Figure 28.3-1, the coefficient should be: 0.4;-0.69; -0.37; -0.29, for roof angle 0-5 degree
The inclination angle for our roof is about 11.6 degrees. We need to interpolate the values given in the table in Figure 28.3-1 to arrive at the correct values for the roof.
thanks
Example: In the case, the roof is flat (angle =0), what is the coefficient for Load Case A & B?
We use the values given in the table for 0-5 angles. We interpolate only when the angle falls in between the table entries.
@@samiabhuiyan4607 Enclosed does not mean air-tight; air can still get inside the enclosed building through windows and other small openings.
I love poeple doing quality works, it shows how much they respect themselves and the others, thanks!
you can't imagine how much i respect your work ;)
I am so impressed with this video, the quality of work, and how concisely it it conveyed.
Released what couldn't Learnt at college learnt here. Thank you very much
You are the best online professor Dr.Structure
Thank you Dr. structure for the comprehensive video explaining the application of the wind pressures calculated by ASCE 7 procedure which I could not find anywhere on the web or elsewhere. I have to say finding those required pressures is a piece of cake but how you applied those to the building is none existence. A lot of videos out there are elementary knowledge and have no value at all. I was stuck for months on that and now I can proceed!. Thanks.
Thank you for your feedback.
@@DrStructure Follow up questions: Referring M&S diagram for Member AB, Case A-1 , is the 85KN vertical load , the gravity axial wind load for use to design the foundation and 45KN, the base shear due to wind load?
All the shown loads are due to wind. Needless to say, for the actual design of the structure, other loads and load combinations must be considered.
Hypothetically speaking, if wind was the sole cause of member forces, then we would need to design AB as a beam (since it is subjected to bending and shear) and as a tension member (since it is subjected to an axial tensile force).
All three forces at the base of AB transfer to the foundation, which means we would need to design the foundation to resist the vertical force, the horizontal force, and the bending moment.
This video was so good and clear! Amazing!
This is the best explaining video Iv ever seen. Subscribed☑️
So detailed and very clear, thank you so much!
In ASCE Wind load calculation for Enclosed and Partially Enclosed building requires to design a shed building using:
1) both Positive internal pressure coefficient +GCpi and Negative internal pressure coefficient -GCpi
2) both External pressure coefficient, Cp (e.g. -0.9, -0.18) for ROOF.
If I consider the wind direction is Normal to the Ridge: +Wx
(Roof angle< 10 degrees & h/L
Yes, we do need to consider all (often seemingly too many) load combinations for each element that is being designed. The consequence of not doing so, potentially overlooking a critical load combination case, could result in structural failure and possible lose of life.
@@DrStructure thank you. Sir may i know your designation? if you don't mind
@@engrkaribul There are several individuals answering questions on this channel located across the globe.
Excellent lecture.
Excellent presentation
Why did not you consider the wind direction from the south to the north that might be more severe than the opposite wind direction?
We are treating the hangar as a closed structure, with the hangar door kept closed during high winds. Consequently, the wind impact on the hangar is identical for both the North-South and South-North directions.
Fantastic high quality video as is expected, alrought I must say that I liked the cartoonish avatars a lot better.
Excellent. Thank you.
At around 6:50 onward the B-1 and B-2 load cases utilise pressure coefficients Cpe and Cpi for what looks like a transverse (wind across the ridge) loading case. I thought that those transverse loadings were calculated as load cases A-1 and A-2. I was expecting the second set of load cases at B-1 and B-2 to utilise longitudinal loading coefficients as the wind is applied N-S on gable-ended walls (including the hangar door which is closed in adverse wind events and making it a Type C normal enclosed building). Can you please explain why the transverse coefficients were utilised for the Case B load cases? If this is confirmed to be an error, what are the longitudinal coefficients to use? Overall, I can readily follow the general calculation processes and find these videos a fantastic resource for underlying analysis principles. Thanks.
The external pressure coefficients for the East-West (transversal) wind direction are obtained, via interpolation, from Figure 28.3-1 of AISC 7-16 (under the heading Load Case A on top of page 313). These coefficients (shown @5:22) are: (-0.35,-0.42,-0.69,0.45)
The coefficients for North-South (longitudinal) wind direction (Figure 28.3-1 under heading Load Case B) are:
(-0.45,-0.37,-0.69,-0.45). These are the coefficients used for loading case B @7:00.
The two sets of coefficients are not the same, although they are some similarities between them.
Figure 28.3-1 of AISC 7-16: lab101.space/qa/windpressurecoefficients.pdf
thanks Dr.stracture :)
very understandable explanation! thank you educating the new ones.
Thanks for sharing
You have good knowledge
Great video!!!!!
At 9:50, it's stated that:
We then transform the member loads to nodal forces. This is done member by member by first calculating the member fixed end forces then transforming the forces from the local coordinate system to the global coordinate system.
I can't follow how the forces are transformed from the local coordinate system to the global coordinate system, e.g., member AB.
F = Qf where f is the force vector in local system, F is the force vector in the global system, and Q is the transformation matrix. The details are provided in Lectures SA49 and SA50.
@@DrStructure
At 5:58 of SA49, the forces F1, F2, F3, F4 of member AB correspond to the degrees of the freedom D1, D2, D3, D4.
At 10:04 of SA52, the forces 18.24, 0, -29.18, 18.24, 0, 29.18 doesn't correspond to the degrees of freedom of member AB since member AB has 3 degrees of freedom, not 6 DoF.
@@hoshangmustafa9672 Each beam/frame member in isolation (without any boundary conditions imposed) has 6 degrees of freedom. If we label the ends of the member i and j, we get three degrees of freedom at i and 3 degrees of freedom at j. That means the local and system force vectors for such a member are vectors of dimension 6x1.
@10:10 of SA52, we are given the 6x1 member force vectors in the system coordinate system. Each vector is obtained by multiplying the 6x6 transformation vector by the 6x1 force vector in the local coordinate system. All of this is done independently of the degrees of freedom of the structure.
While for members BC and CD all six force vector elements are used to determine the system force vector, for members AB and ED, since they are fixed at the base (there are no system degrees of freedom associated with the support joints) only 3 of the vector elements are used in determining the global force vector. For member AB, elements 4, 5, and 6 of the force vector correspond to system DOFs 1, 2, and 3. For member ED, force vector elements 4, 5, and 6 correspond to the system DOFs 7, 8, 9.
@@DrStructure
What about 5:58 of SA49?
@@hoshangmustafa9672 @5:58 SA49, we have the transformation of the 6x1 force vector from the local to global/system coordinate system. That is consistent with what we have discussed above. Each member in isolation has a 6x1 force vector, 6x6 stiffness matrix, and 6x1 displacement vector in both local and global coordinate systems. Here, since there are no member loads applied to the frame, the system force vector is composed directly using the joint loads.
Very nice animations which software do you use to prepare a video like this ?
We use several software tools including: SketchUp, Adobe Illustrator, and Camtasia Studio. The 3D model was created, and animated using SketchUp. Text and basic graphics were created and animated using Illustrator, additional animation were added and the entire video produced in Camtasia Studio.
@@DrStructure You give it so much efforts, i like them thank you so much.
Thanks for the feedback!
Excellent and informative video.
However in the code it mentions a minimum design load of 0.77 kN/m2 to be used why didn’t we use it here as some of the values (at minute 06:40 ) are bellow it 323 ,63 kN/m2 ...
Again Thank you so much for this excellent video.
8:38s , To calculate the concentrated load to the frame from the roof panel (at each joint), why you don't get the reaction at each support of beam (=concentrated load of each joint)? I saw you multiple the distributed load with 6m (e.g., 2.64KN/m x 6m=15.84 KN)
Yes, we can calculate the vertical support reactions for those long transversal beams and then apply them to the frames. But that requires analyzing those indeterminate beams first. What we did here is an approximation technique that works in practice.
Thanks so much. The video is excellent.
Thank you for a great lecture can you please tell how to take load combinations if we calculate the wind loads of roof and building seperately?
Design of structural systems is often governed by local/state/regional codes. For example, in the US, the ASCE code offers the following basic load combinations for designing members:
1.4D
1.2D + 1.6L + 0.5(Lr or S or R)
1.2D + 1.6(Lr or S or R) + (L or 0.5W)
1.2D + 1.0W + L + 0.5(Lr or S or R)
0.9D + 1.0W
where D is dead load, L is live load, Lr is roof live load, W is wind load, S is snow load, R is rain load. There are another set of equations for dealing with seismic loads.
The equation that results in the largest value for the given member controls the design of the member.
The method described in the lecture treats wind load for the roof and the walls. We cannot separate them. So the calculations presented encompass both. But, for design purposes, we do need to analyze the frame under dead load, live load and possibly snow load, then use the combination equations to determining the governing set of values when designing the individual frame members.
Hi ! i have a question at 10:40 where we get the Member-end Forces for each member, shouldn't F1 and F2 for each force switch places because when drawing the Shear and Moment diagram later thay seem to be drawn in the oposite way for example, F1 wich is the force to X direction is 42.99 kn but in our results at 10:40 it is -84.94 kN. looking forward to your reply, Thank you :)
Member forces are always given/computed in the local (member) coordinate system. So, for member AB at 10:40, -84.94 is an axial force (in local x which corresponds to global y). And 42.99 is the force in local y (or global x), which means the shear force at the left/bottom end of the member is 42.99 kN.
@@DrStructure Maybe i am missing something here, at SA51: Analysis of a Tea House Frame at 7:15 when you draw the forces on the member you keep the same rule ? The forces drawn are ok, but they are not in the order you state above...
The stiffness matrix for a member can be defined in two related ways: in local coordinate system and in global coordinate system. We can go back and forth between the two via a coordinate transformation matrix. This allows us to express member forces and displacements in both coordinate systems.
When formulating the system matrix, for determining joints displacements/rotations, we operate in the global coordinate system. Once the displacements/rotations are computed, and we want to examine each member individually, we often work in the local coordinate system. This means, after calculating the displacements, and member forces, in the global coordinate system, we multiply them by the transformation matrix to get them into the local coordinate system.
In SA51, given the simplicity of the problem, that was not done. I will have SA52 checked to make sure it is error free.
Computing F for member AB we have F=KAB*D F={-43,-85,228.66,79.47,83,359.15} which corresponds to the diagram of member AB case 1.
In the exercise, to get the correct sense, you just have to multiply the transformation matrix by F of member AB, because F is in local coordinates.
For load case A, there are two data. The data for A-1 (positive internal pressure) & the data for A-2 (negative internal pressure). A-1 & A-2 (@6.45 minutes) shows different values for a single frame. which load to consider in between two cases for the final design? Though you have considered only case A-1 in your video. But, In practical design, which values to consider? The maximum value between two values? Please help me with your answer.
Note that @11:00 the shear and moment diagrams for both cases (A-1 and A-2) are drawn, meaning we need to consider both (or all load cases) when designing a member. More specifically, we need to draw the shear and moment diagrams for the beam under all load cases, then compare and contrast them in order to understand how shear and moment vary in the beam, and where the maximum values occur. For example, while maximum positive moment in the member may develop under one load case, maximum negative moment could occur under a different load case, and maximum shear under a third load case. We need to know them all if we are going to properly design the member.
For the life of me, I find it so unbelievable that the leeward side can ever experience anything close to even half the windward side.
I work on projects both here in the US and in Canada, and I'm constantly calculating Wind Loads (Windward WW) of Leeward W=.625WW, Parallel W=.71WW, Roof W=.585WW.
That just seems insane to me. Because I've been on the Leeward and parallel sides of buildings during high winds. I've never seen anything close to those kinds of loads.
What design standard are you using? Perhaps it’s worth time and effort to compare and contrast with ASCE 7.
@@DrStructure most of my engineering is to ASCE 7-10 and -16. I have some current projects that I have to produce the loading according to Canada's NBC.
I just never have seen the kinds of loads for parallel and leeward sides that the codes develop.
As you know, most modern design standards are probability based, taking into account the risk (mainly to human life) associated with severe loading scenarios, even though the structure may not encounter such a scenario during its life cycle.
@@DrStructure yes, I am aware.
Not that you would know, but I am a PE and own an engineering firm, Quanta Engineering LLC.
Still, I've just never seen the wind pressures on the parallel and leeward sides ever experience what the codes show...
An explanation is more than wonderful, but if I want to use an artificial intelligence network to predict the response of this structure to wind loads at different speeds, is there a Matlab code or an analysis program that can find the response of the structure by simply changing the parameters affecting the wind load and its speed
Dynamic or time-varying analysis of structures is well beyond the scope of our video lectures. For that, simulation software can be used to model and analyze the behavior of the system under various loading scenarios. You may want to look into the commercially available structural analysis (finite element analysis) tools and see which ones best fit your computational needs.
Except for a few well-defined problems, "Artificial Intelligence" technologies, beyond the perception being projected, is still in the research stage when it comes to reasoning and problem solving that require deep and non-trivial human expertise.
Dr. Structure I would like to say a big thanks for the video.
But I'm wondering if we add crain console to the columns how can we re-write the system stiffness matrix?
Any additional (structural) element added to the system should be treated as a separate element with its own stiffness matrix. For example, a crane console attached to a column could be treated as a (cantilever) beam with its own stiffness. This would not alter the stiffness of the column. Having its own stiffness, the cantilever beam simply contributes to the stiffness of the frame, like any other beam or column.
The internal coefficient ci = 0.18 seems to be too low for hanger type structure. Should it be 0.55 instead (partially enclosed building, or, open front structure)? The difference is not insignificant. Please confirm. Thanks.
Yes, if the structure is truly only partially enclosed, a higher value for the internal pressure coefficient can be used. But if the hanger is equipped with doors that can be closed, and there is no significant opening in any of the walls, the lower internal pressure coefficient would be applicable.
Could you kindly explain why the bending moment diagram resulting from lateral uniformly distributed load on the frame (column) is a linear pattern? isn't it supposed to be a second degree curve?
Yes, indeed! The shape of the moment diagram for AB, and ED, is not correct, it should be a second degree polynomial since shear is linear in the member. Also, the moment diagrams for BC and CD should be of linear type since shear is piece-wise constant in those members. Thanks for point this point.
Thanks for the clarification.
I have had a discussion with friends about the bending and shear moment diagram of a statically indeterminate (two-story two bay, fix supports) frames subjected to lateral uniformly distributed load along its length from one side, but we all were not so sure about the accurate solution. It would be very helpful if you could explain (or make a video) how to calculate and draw the diagrams for this particular case. I have tried to use SAP2000 vs. Robot to draw them, at the column subjected to UDL, SAP2000 gave me linear pattern while Robot showed me second degree curve.
We will work on a video for that, but that will take some time.
The analysis of the frame can be done using a standard method such as the displacement method, or one can use software tools such as SAP2000 for that purpose. Regardless of the method in use, the analysis gives us the member-end forces. In the case of a two-story two bay frame, since we have 6 columns and 4 beams, we end up with 10 sets of member-end forces. In each set, we have three forces for each end of the member (an axial force, a shear force and a bending moment).
Shear and moment diagrams need to be done member by member. For each member, if it is subjected to a uniformly distributed load, like the two columns that are subjected to the UDL, shear diagram is going to be linear and moment diagram would be nonlinear (second degree polynomial). For all the other members, since they are not loaded directly, each is going to have a constant shear and a linear moment diagram. Given the member-end forces from the analysis step, you should be able to construct these diagrams. In doing so, you need to pay particular attention to the members subjected to the UDL as the maximum moment in the member needs to be computed as well. This moment is located at the point along the column where shear is zero (where the shear diagram cross the axis.)
Some software applications approximate the shape of the diagrams using straight lines. This may not be a critically important factor if the beam is finely discretized into small enough segments.
Thank you for the answer. You have cleared out my puzzle.
Good
How to make these type of videos. and whcih software you use. please guide. Thanks
We use several software tools to create the lectures including Adobe Illustrator, SketchUp, Camtasia Studio, and a few more as the need arises.
The choice of software is dictated by the conceived organization of the presentation and how best to realize it. We spend a significant amount of time conceptually constructing the presentation on paper. We then employ the fitting (software) tools to implement what we have on paper.
@@DrStructure Appreciated work...
why is the braces required at the roof level (where the steel deck is) is that required? does the deck not provide enough lateral stability?
The lateral bracings at the top of the frame seem unnecessary given that, as you stated, the roof itself could serve that purpose. However, without looking at the details of the analysis and design of the building, and the design code in use, it would be difficult to assess the rationale for such decisions.
@@DrStructure thanks - I was just trying to figure out why that is required. maybe for more stiffness.
@@rawjor I too am not certain if those bracings are actually needed. The 3D model was done based on an actual hanger in a remote location in Eastern Europe, but we don't have the design details of the hanger available to us...
Your presentation is so awesome! Do you have a page where one can donate to show some gratitude?
Thanks for your feedback. Yes, you may contribute to our efforts through PayPal. The link is at the bottom of the following web page: lab101.space/Sponsor.asp
Dr. Structure, I have seen in practice engineers for quick sizing and cost estimation of rafters would (skip step) not transfer the load to the purlins but would use the tributary area of the rafter times the area load to compute the line-load for rafter. Here you would not have point loads on rafters, Is this bad practice or even wrong?
How the load is being transferred from the roof panels to the supporting beams depends on the configuration of the assembly. If the roof panels rest directly on the rafters, as opposed to the purlins, or if the rafters and purlins are on the same plane, then the method of estimation that you described is correct; we would end up having line loads on the rafters.
However, when the roof panels are resting on the purlins only, and the purlins sit on top of the rafters, as is the case here, then the purlins are subjected to line loads only. The rafters take the roof load at the points of contact with the purlins. Here, distributing the roof load on a rafter uniformly is not a good idea as it would result in underestimating maximum bending moment in the member.
why is the load on the column is not concentrated as well as the beam
The wall panels are in contact with the columns directly. That causes wind pressure to act on the entire length of the column, as a distributed load. However, the roof panels are not in direct connect with the frame beams. The panels rest on the cross beams which then transfer their loads to the frame at eleven (11) points.
great thanks !
In concrete frame structures, lateral joints are acting on joints of columns and beams. How are these obtained?
Do you mean how the lateral bracings are attached to concrete beams or columns? There are various techniques for creating attachments to concrete. For example, anchor bolts with sufficient lengths can be placed inside concrete for creating such attachments.
Sorry. I was asking about how lateral loads due to wind or earthquake acting at joints of columns and beams are computed or obtained. In other words, how do we compute for the applied loads due to wind or earthquake acting on concrete frames in buildings?
The forces at a joint of the structure are the same as internal forces at the ends of the members connected to the joint. These forces are computed by analyzing the structure an determining the member-end forces.
If lateral forces are significant and present in the structure, we often resort to three-dimensional structural analysis where each joint would has 6 degrees of freedom, three displacements (in x, y and z directions) and three rotations. In such a case, a beam or column would bend along two planes, and it may go under torsion as well. Then, a joint/connection needs to be designed to accommodate in-plane and out-of-pane bending as well as torsion.
Is there a chance that you can be able to assist me with the codes on for the coefficients of the wind loads that you used in the video like ASCE 7-16, or just the important relevant pages or tables
The relevant provisions are covered only in a few pages of ASCE 7-16 which are referenced in the video. The selection of the relevant coefficients depends on the location, purpose and shape of the building. If you have ASCE 7-16 in your possession and have studies it, yet have specific questions, feel free to post them here. I'll do my best to answer them.
Dr. Structure : Thank you for the Response, don't actually have the code because i am based in South Africa for the moment. So i was hoping that you might be having the relevant reference pages because the codes that i have for South Africa are a little bit confusing, i thought getting yours would help me compare which of the procedures is better and easier to get the coefficients.
Unfortunately, due to the copyright issue, I am unable to send you the relevant pages of ASCE 7-16. You might be able to find some information on this online. There might be some powerpoint presentations or pdf files out there ....
Dr. Structure ok thanks, i will find some means. Thanks a lot
If a building has 5 floors ( five storey) building, do we do the coefficients of the wind loads in each floor or we do them for the whole building?
The procedure described here is for low-rise buildings only. Mid-rise or high-rise buildings require the use of more comprehensive techniques, such as the use of wind tunnels to accurately estimate the governing wind forces.
Dr. Structure : Ok thank you.
Pls clarify why
q= 0.534*sq(V)= 0.534*58*58
= 1796 N/m2 , whilst your calc is 1196.4 N/m2
We have made the correction in a new version of the lecture which you can find in the free online course referenced in the video description field.
@@DrStructure Tqvm. I am still learning on the wind loads.
Can you compare your analysis with BS standard CP3 on the walls/cladding? It is like
q=0.613*Vs*Vs
Vs=1.06*1*1*Vb=1.06*1*1*58m/s
=61.48m/s
Thus,
Vs= 0.613*61.48*62.48
=2317 N/m2
Is this correct? A lot higher than th ASCE standard?
@yikee7592 British standards differ from the American Society of Civil Engineers (ASCE) guidelines in determining the applied loads on a structure, leading to variations in load values.
Since we don't have access to CP3, we won't be able to do a direct comparison at this time.
How factor 0.534 factor is calculated in Wind Velocity Pressure Equation?
The factor is determined based on several numbers including:
Velocity pressure exposure coefficient,
Topographic factor,
Wind velocity factor, and
Ground elevation factor.
These are explained in ASCE 7-16.
Hello, you need to have access to the code to know where i’m getting the numbers from but it’s as follows:
q = 0.613 kz kzt kd V2
Where,
q is the velocity pressure in kN / m2
Kz is the wind speed coefficient due to height
Kzt is the topographic factor ( if the building is up by a cliff...etc) otherwise = 1
Kd is the wind direction factor = 0.85 for building, but varies if you design tanks (see the code)
v = 58 m/s given ( also in the code)
To get kz we have this equation:
Kz = 2.01 ( z/zg ) ^ (2 / alpha )
Where z is the height to the mid roof so 9.6 + (13.3-9.6)/2 = 11.45 m = (x3.28 = 37.54 feet)
Zg & alpha depend on exposure categories there is type B or C or D
Since the hanger is in the open by the sea its type C , B is for city and D for very isolated areas like Siberia xD
Form the code type C gives Zg = 900 feet & alpha = 9.5
So kz = 2.01 (37.54/900)^(2/9.5) = 1.0297
So q = 0.613 x 1.0297 x 1x 0.85 x V2 = 0.5365 V2 ( i got some error due to rounding she got 0.534)
Good bye!
Thanks you sir , if we have a canopy or extension member how to design winload ?
A canopy can be treated as an open building (not enclosed). There are provisions for open buildings in the code. Similarly, the code provides provisions for roof overhangs. Similar to the enclosed or partially enclosed cases, such provisions specify the required wind pressure for the components of the structure under consideration.
@@DrStructure acording to slope of roof 11.62 , your coeficient has a problem with Linear Interpolation .
Thanks for the comment, we will look into it.
What about the torsional load case sir ? I do not understand about torsional load case clearly . Could you give me any comment . Thank you
@@DrStructure what about wind load on sidewall ?
Can you explain the design of space frame in strctutal steel
Thanks for your note. Eventually we will cover design related topics, after we have exhausted analysis discussions.
Is this a steel structure?
Yes, the example frame used in this lecture is assumed to be made of steel. The procedure however is independent of the structural material.
I think it's the same as ASCE7-05 ,Is there any difference ?
ASCE 7-16 is the most recent version of the document. Certainly it contains updated provisions as compared to the older versions. With regard to the specific provisions used in this lecture, since we don't have a copy of ASCE 7-05 available, we cannot tell how the two sets of provisions differ.
how many loading cases need to be considered?
Over all, we need to consider dead load, live load, snow load, and probably a few more depending on the location of the building. There are various load combination equations given in ASCE 7-16 for determining critical shear and moment values for design purposes.
Hi Dr. Structure, May I know the dimension of the cross-section of member structure that u used to obtain: A=0.023716 (m2), I=0.00567136 (m4) . Thanks
A standard W30x124 steel member was used in the example. The nominal dimensions (in inches) are:
Area: 36.5 in^2
Depth: 30.17 in
Web thickness: 0.585 in
Flange width: 10.515 in
Thanks so much@@DrStructure for your prompt reply. May I ask you a favor? I check the area of W30x124 (A=36.5 in^2 ~ 0.023716 m^2)---> thus, it acceptable. But the moment inertia of W30x124 (Ix=5,360 in^4 , and Iy=181 in^4). The moment inertia is not equal as shown in the video (I=0.00567136 m^4)
Let me check with the person who did the calculations and get back to you on this.
I stand corrected. For this example, the standard section that I mentioned was not used. Instead a rectangular section with a height of 1694 mm and width of 14 mm was used. For design purposes, this is not a good choice. But since here the focus was on analysis as opposed to design, we should make a note of it but consider it okay.
@@DrStructure Thanks so much
I tried to find out the value of "0.534" in formulas: q=0.534V^2. I understand "0.534" is the product factor of : 0.613, Kz, Kzt, Kd, Ke. I tried to use Kz=1.04 (Exposure C); Kzt=1; Kd=0.85; Ke=1 (Page 268-code). But I can't achieve the number "0.534". Plz help me?
To determine Kz, we use the mean height of the roof. In this case, it is about 11.45 meters. Again, we need to interpolate to get the right value for Kz. It comes out to about 1.025, not 1.04.
0.613(1.025)(1)(0.85)(1) = 0.534
Thanks for giving the reply,as soon as possible can u explain .
how do we design the members?
That would fall under structural design, in this case steel design. A topic to be covered at a later time.
sorry....why : q=0.534(58)^2=1196.4?
A typographical error
0.534 * 58^2 is not the result of 1196.4 N/m^2.
It is 1796.4.
Thanks for pointing out the calculation error.
@@DrStructure I always do calculation on paper.
All your video make me more comprehensive.
So, I want more video wind and quake (Static & Dynamic Lecture Video).
Dynamic Mechanic for
Lateral Load.
Quake (Equivalent Lateral Force Procedure,Responce Spectrum,. . .)
I'm wating to study from you
thanks a lot for your kind information... can you give link asce7-10, 16 standard code,,,,,
You can get the publication directly from ASCE. Below is the link for the latest version of ASCE 7. Scroll to the bottom of the page for info for getting the printed and online versions of the standard.
www.asce.org/publications-and-news/civil-engineering-source/article/2021/12/02/updated-asce-7-22-standard-now-available
We need example in pdf
Yes, we need to document all the analysis details.
+Dr. Structure o.k
sir your videos excelant practical explanation.
thanking you sir.
can you explain the connection design in steel structures
WHY USE METRIC UNITS ? AMERICA DOESN'T USE METRIC UNITS. QUIT TALKING METRIC WHILE THINKING IMPERIAL. NOBODY HAS THE FEEL OF KN/METER SQUARE.....EVERYBODY KNOWS LBS/SQUARE FEET. I DON'T HAVE TIME TO CONVERT AND THEN GET THE FEEL. KILO-NEWTON TO ME = 1000 ISSAC NEWTONS.
A gentle reminder - the world does not revolve around our preferences and biases. Learning to communicate in multiple languages (be they natural, technical, or measurement related) is a strength, not an inconvenience. Part of the mission of engineering education is to help students across the globe develop a growth mindset, one that potentially could produce more creative and culturally fluid engineers who can then address emerging complex technical problems, both locally and globally. One’s insistence in using a specific way of solving problems (in this case, measuring), although an acceptable personal choice, invariably hinders one’s professional and personal growth in the long run.
Can you tell me how you get the external pressure coefficient in Load Case B? From the code, it only has a value for each member.
For Load Case B, the code (Page 313) offers a single entry table with values for the surfaces of the building. Columns 1 thru 4 give us the coefficients for the side walls and the roof panels. For example, column 3 gives the coefficient for one of the roof panels as -0.37.
@@DrStructure For load case B, why the pressure coefficient of windward wall "6" is not considered?
In this lecture, we are considering the analysis of a lateral frame only. So, we are interested in the loads that are acting on the plane of the frame. Since the wind pressure on Side 6 does not cause an in-plane load on our frame, it was neglected in the analysis.