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The solution for the exercise problem is provided in the free online course referenced in the video description field.

Correct, we need to calculate the deflection when the a vertical reaction is the redundant forces.

Yes, if you don’t have access to your notes or a textbook, to apply the method, you need to remember the two slope-deflection equations as well as the equations for calculating the fixed-end moments.

@@DrStructure is it practical? I mean anyone could do it if he knows the formula unlike the force method

Yes, it is practical. When you use the method several times, the equations can be recalled from memory with ease.

You are correct. We have updated the solution file. Thank you for bringing this error to our attention.

Please elaborate and justify your assertion. The pressure coefficients (GCp) have been determined through wind tunnel experiments as a function of the area of the model canopy used in the testing. The wind provisions in the ASCE manual provide wind pressure values for surfaces subjected to wind, whether they are roofs, walls, or canopies. These pressure values represent the force per unit area on the system. It's important to note that the method by which this pressure is transmitted to the supporting structure is not a factor in determining the wind pressure on the system itself.

To solve for the unknowns, including Ey, we need to apply linear algebra techniques. Specifically, we can use established methods such as Gaussian Elimination to solve the system of linear equations.

The solutions for the exercise problems are available in the free online course referenced in the video description field.

We can determine shear, moment, and deflection by integrating the load function while considering the beam’s boundary conditions. This integration works when the load function is continuous. However, when point loads or support reactions are present (which generally cannot be expressed as continuous functions), the beam must be divided into segments where the load is continuous. This results in piecewise equations for shear, moment, and deflection.

@@DrStructure Thank you!

Correct, the classical methods of structural analysis are time-consuming, especially when applied by hand without the use of computers. Although they are not practical for solving problems with more than a few degrees of indeterminacy, these classical techniques offer useful insights into the behavior of structural systems and a conceptual framework for thinking about their analysis.

​@@DrStructure Yea I do know that,, in work we use computers to analyze beam and frame, truss structures however, I just wanna understand this in order to pass structural analysis 2 tho. Btw I have a question, can we use conjugate beam method for slope and deflection for the first case instead of virtual work? And then apply virtual work on the virtual beam

In principle, yes. We can use any of the available techniques to determine slope and/or deflection in beams.

Do you mean bridge design or lecture design?

@@DrStructure bridge design. I have to design a bridge which is exposed to hazards, for my Final thesis in civil engineering.

Here are a few commercial bridge software options available for professional use. Depending on your location and university, you may be eligible to obtain one of the following tools at a discounted (educational) price: CSIBridge: Bridge Analysis, Design, and Rating www.csiamerica.com/products/csibridge AutoDesk Structural Bridge Design www.autodesk.com/products/structural-bridge-design/overview?term=1-YEAR&tab=subscription Larsa 4D Bridge Series www.larsa4d.com/products/features/4dbridge-overview.aspx There are others, which you should be able to find online via search.

@@DrStructure thank you for your support. I'll chek on them.

Conceptually, the virtual load is different from the real load. The principle of virtual work cannot be properly explained without having a virtual load. Computationally, if there is only one real load applied to the truss, and the virtual load is placed at the same location and in the same direction as the real load, then we do not need to analyze the structure twice. We can adjust the analysis result from the real load to obtain the effect of the virtual load.

@@DrStructure i see..

We have determined the end moments for segment BD. This makes the segment statically determinate, with the shear forces at the ends of the segment being unknown. We can determine these forces using static equilibrium equations. Writing the moment equation about point B, we get: (2 kN/m) (6 m)(3 m)+ 5.17 - 3.19 + 6Vd= 0 By solving the above equation for the shear force at D, we get Vd= 6.33 kN. Next, summing the forces in the y-direction, we get: Vb + 6.33 = (2 kN/m)(6 m) which yields Vb = 5.67 kN.

We are using the sign convention adopted for the slope-deflection method: an anticlockwise moment at the ends of the member is considered positive. Note that an anticlockwise moment at the end of the member (connected to a joint) translates into a clockwise moment at the joint since the sum of the two moments must be zero. So, at 3:28, the moment shown is at the joint (not at the end of the member). That anticlockwise moment at the joint results in a clockwise moment at the end of the member; therefore, we treat it as a negative moment. At 8:38, however, the referenced moment is at the end of the member. The counterclockwise moment at the left end of the member is positive, and the clockwise moment at the right end is negative.

Please post your questions about the lecture topics here. Our volunteers are available to help in this forum, but their availability to respond to email messages is limited.

The wind directionality factor is a probability-based coefficient related to the structure's dimensions and geometric shape. ASCE Table 26.6-1 provides this coefficient for various types of structures. To determine this coefficient, we don’t need to decide which wind direction produces the maximum pressure on a structural member; that is considered elsewhere in the analysis. Wind analysis for an attached canopy, not part of the main wind force resisting system (MWFRS), falls under the Components and Cladding (C&C) provisions of the ASCE. For canopies, we don’t need to consider different wind directions to determine the maximum wind pressure. The net pressure coefficients (GCp and GCpn), based on experimental data, are given for the most severe cases. In a previous lecture (see link below), we accounted for wind directions in designing a frame structure (MWFRS) in an airplane hangar. You may want to review it if you haven't already. ruclips.net/video/dDNaw87nyN4/видео.html

Dal punto di vista ingegneristico, il Ponte sullo Stretto di Messina è destinato a diventare un punto di riferimento ingegneristico e il ponte sospeso più lungo del mondo. La propensione della regione all'attività sismica richiede un design che prioritizzi la sicurezza, sfidando indubbiamente gli ingegneri a innovare e risolvere problemi complessi. Progetti grandi come questo devono essere analizzati da tutte le angolazioni, non solo tecniche. Sviluppare un consenso tra le parti interessate sugli impatti economici, culturali e ambientali del sistema è fondamentale. L'idea di collegare la Sicilia all'Italia continentale esiste dai tempi dei Romani, quindi suppongo che la gente in Italia e in Sicilia abbia dibattuto i pro e i contro di un tale ponte per decenni. Qual è il consenso tra la gente in Italia?

@@DrStructure Grazie per la tua disponibilità e risposta. Il problema in esterna sintesi è il seguente: il Ponte ipotizzato sullo stretto di Messina è antisismico a livello progettuale, ma l'intorno immediato è caratterizzato da sistemi edilizi che non rispondono o rispecchiano le norme antisismiche. Quindi si dovrebbero prima eseguire degli interventi di consolidamento antisismico e poi nel caso attivare i processi e metodi di realizzazione del ponte di Messina. Forze posso sembrare "improprio" ma io sarei a favore di un Ponte in legno. Un caro saluto da molto lontano. Pisa Centrale - Italia. Grazie.

​@@DrStructure your team is literally the best in terms of teaching. Hats off to you.

The solutions for the expertise problems are available in the free online course referenced in the video description field.

Please feel free to post your questions or comments here.

Yes, a negative value indicates that the actual displacement is in the opposite direction of the assumed displacement. Therefore, if we initially assume the frame displaces to the right and the displacement value is negative, it means the frame is actually moving to the left.

@@DrStructure in the lecture did you assume it was going right? But we got negative values of delta so youre telling me the frame actually went left?

No, the frame displaces as shown. The delta term in the slope-deflection equation is considered positive when it is produced by a counterclockwise rotation of the member. Since the columns are rotating clockwise (causing the frame to sidesway to the right), the delta term associated with each column is negative.

There are 2 and half (2.5) holes along the shear plane on each side of the connection. That is where 2.5 comes from. We are subtracting the area due to the holes from the area of the flange in the shear plane, since the holes don’t contribute to the shear strength of the member. The vertical spacing between the holes is at least 7 inches, considering the section dimensions (WT7x66) with a flange width of 14.7 inches. However, this spacing is not relevant to our analysis.

for problem 1*

You are correct. Contrary to the problem definition (7/8” bolt diameter), the solution uses 3/4” bolt diameter. Thanks for pointing out the error.

The same discrepancy in the problem definition. So, for the solution to be correct, we can change the specified bolt diameter to 3/4 in the problem definitions.

@@DrStructure I appreciate you sincerely! I have my steel midterm tomorrow will update with results (:

No, we don’t have any lectures on concrete design.

The equivalent concentrated load is placed at the centroid of the right triangle. The centroid of a right triangle with base L is located L/3 from the right-angled vertex. Therefore, since L = 2.5, we get 2.5/3 = 0.83 m.

In exercise problem 2, the notation 4@2” means the spacing between two consecutive vertical lines (shown on the drawing) is 2 inches, and we have 4 of them, which makes the overall length of the connection 8 inches. That is why we use 8 in the denominator when determining U, since the length in the denominator for U is the overall length of the connection.

Consider the beam segment between the two internal hinges. A load of 50 kN is applied at the left hinge. There are reaction forces at the left roller (R1) and right roller (R2). The total length between the left hinge and the right roller is 12 m, with the distance between R1 and R2 being 8 m. The moment equilibrium equation about the right roller can be expressed as: (50 kN)(9 m) - R1(8 m) = 0. By solving this equation for R1, we find R1 = 56.25 kN.

This is a stylistic choice to reduce the size of the matrix coefficients, making the matrix more compact. This is analogous to factoring in algebra. Such manipulations do not affect the solution of the system.