Resultant internal loadings acting on cross section | 1-23 | Stress | Mechanics of materials rc hibb
Summary
TLDRThis video demonstrates the solution to Problem 1-23 from the 'Mechanics of Materials' by R.C. Hibbeler. It involves calculating internal loadings on cross-sections at points E and BC using equilibrium equations. The instructor applies methods like moment calculation, resolving forces, and equilibrium analysis to determine normal forces, shear forces, and moments. Using the previous problem's data, they find key results, such as the reaction forces and internal loadings at various points. Viewers are encouraged to engage by subscribing and asking questions in the comments section.
Takeaways
- 🔧 The video explains how to solve problem 1-23 from chapter 1 (Stress) of 'Mechanics of Materials' by RC Hibbeler, using data from problem 1-22.
- 📏 The problem involves finding the internal loading at cross-section E and in short link BC, given a 120N load applied to the handle.
- 🛠 The first step is removing the link at point B and replacing it with a reaction force (FBC), then analyzing the left-hand side of the system.
- 🔄 To find FBC, the moment equilibrium equation (ΣM = 0) is used, resulting in FBC = 1385.6 N (1.386 kN).
- 🔗 After cutting member BC, it is determined that the normal force in BC is 1385.6 N, with zero shear force (V = 0).
- 🔄 Moments about point 1 also result in zero moment in link BC.
- 🔍 The focus shifts to cross-section E, where a normal force (NE), shear force (V), and moment (ME) are calculated.
- 📐 NE at cross-section E is found to be zero, while the shear force (V) is 120 N.
- ⚖️ The moment at cross-section E (ME) is calculated as 48 N·m using the moment equilibrium equation (ΣM = 0).
- 🛑 The video concludes by summarizing the internal loadings at cross-section E (NE = 0, V = 120 N, ME = 48 N·m) and in short link BC (normal force = 1385.6 N, shear force = 0, moment = 0).
Q & A
What is the main problem being solved in this video?
-The main problem being solved in this video is Problem 1-23 from Chapter 1 of 'Mechanics of Materials' by RC Hibbeler. The problem involves finding the internal loading at cross-section E and in link BC for a structure subjected to a 120 Newton load on the handle.
How is the reaction force at point B represented?
-The reaction force at point B is represented as FBC. When the link at point B is removed, FBC acts as the reaction force applied by the link at point B.
What forces and moments are involved in calculating FBC?
-To calculate FBC, the video applies the equilibrium equation for moments around point A, considering the 120 Newton load and its perpendicular distance (500 mm), along with the components of FBC at a 30° angle and its perpendicular distance (50 mm).
What is the value of FBC, and how is it calculated?
-FBC is calculated as 1385.6 Newtons (or 1.386 kN). This is determined using the moment equation: FBC = (120 N * 500 mm) / (50 mm * cos(30°)).
What are the internal forces acting in member BC after cutting it?
-The internal forces acting in member BC are the normal force (N), shear force (V), and moment (M). In this case, the normal force is equal to FBC (1385.6 Newtons), while the shear force and moment are found to be zero.
What equilibrium equation is used to find the normal force in member BC?
-The equilibrium equation used to find the normal force in member BC is the sum of all forces in the Y direction (∑Fy = 0), which gives the normal force as equal to FBC (1385.6 Newtons).
What is the purpose of cutting the handle at point E in the second part of the problem?
-The handle is cut at point E to find the internal loadings at that cross-section, including the normal force (NE), shear force (V), and moment (ME).
What is the result of applying equilibrium equations at point E?
-At point E, the normal force (NE) is found to be zero, the shear force (V) is 120 Newtons, and the moment (ME) is 48 Newton-meters.
How is the moment at point E (ME) calculated?
-The moment at point E (ME) is calculated by using the equilibrium equation for moments: ∑M about E = 0. The moment produced by the 120 Newton load at a distance of 400 mm (0.4 meters) is subtracted from ME, resulting in ME = 48 Newton-meters.
What are the final internal loadings at cross-section E?
-The final internal loadings at cross-section E are: normal force (NE) = 0, shear force (V) = 120 Newtons, and moment (ME) = 48 Newton-meters.
Outlines
This section is available to paid users only. Please upgrade to access this part.
Upgrade NowMindmap
This section is available to paid users only. Please upgrade to access this part.
Upgrade NowKeywords
This section is available to paid users only. Please upgrade to access this part.
Upgrade NowHighlights
This section is available to paid users only. Please upgrade to access this part.
Upgrade NowTranscripts
This section is available to paid users only. Please upgrade to access this part.
Upgrade NowBrowse More Related Video
Determine internal resultant loading | 1-22 | stress | shear force | Mechanics of materials rc hibb
Mechanics of Materials: F1-1 (Hibbeler)
Understanding Shear Force and Bending Moment Diagrams
Equilibrium of a particle - Example : ExamSolutions
Statika/Mekanika Teknik #4: Gaya Dalam
Equilibrium of Rigid Bodies (2D - Coplanar Forces) | Mechanics Statics | (Solved examples)
5.0 / 5 (0 votes)