S20A Pelengkung 3 sendi part 1
Summary
TLDRIn this lecture on statics, the focus is on analyzing a three-jointed arch (pelengkung tig sendi) structure under a uniform load. The lecturer explains how to calculate reactions at the supports (Va, Vb, Ha, Hb) using equilibrium equations. By applying symmetry and moments, the reactions are determined to be 30 kN for vertical forces and 15 kN for horizontal forces at both supports. The lecture emphasizes the importance of understanding static equilibrium in indeterminate structures and provides a step-by-step approach for solving such problems in structural analysis.
Takeaways
- 😀 The topic of the lecture is statics, specifically focusing on a three-joint arch structure (pelengkung tig sendi).
- 😀 The three-joint arch structure is statically indeterminate due to more unknown forces than equations of equilibrium.
- 😀 The structure features three key joints: two at the ends and one in the middle, which is considered a hinge.
- 😀 The total number of unknown forces is greater than three, necessitating the use of additional methods for solving static problems.
- 😀 The structure experiences a uniform load (Q = 10 kN/m) across its span.
- 😀 Equilibrium equations are used to calculate the reactions at the supports (Va, VB, Ha, HB).
- 😀 The symmetry of the structure simplifies calculations, as reactions at the supports are equal (Va = VB).
- 😀 To find reactions, the lecture applies the horizontal force equilibrium equation (ΣFx = 0), where Ha = HB.
- 😀 The vertical force equilibrium equation (ΣFy = 0) is used to determine Va and VB, considering the symmetrical load distribution.
- 😀 Moment equilibrium (ΣM = 0) is used to calculate internal forces and reactions at various points of the structure, including point S.
- 😀 After calculations, the reactions are found to be Va = VB = 30 kN and Ha = HB = 15 kN, confirming the structure's balance under the given load.
Q & A
What is a three-pin arch and why is it called so?
-A three-pin arch is a type of structure where three points are fixed to support it, with one pin at each end and one in the middle. It is called a 'three-pin' arch because of these three pin support points which make the structure statically indeterminate.
Why is the three-pin arch considered statically indeterminate?
-The three-pin arch is considered statically indeterminate because the number of unknown forces (reactions) is greater than the number of available equilibrium equations, making it impossible to solve using only the static equations of equilibrium.
What is the purpose of using equilibrium equations in statics problems?
-Equilibrium equations are used to calculate unknown forces in a structure by ensuring that the sum of forces in both the horizontal and vertical directions, as well as the sum of moments, all equal zero. This ensures the structure is in a state of balance and is not moving.
How does symmetry affect the calculation of reactions in a three-pin arch?
-Symmetry simplifies the calculations of reactions because it allows the assumption that the horizontal reaction forces at both ends of the arch (HA and HB) are equal. This reduces the complexity of solving for the unknown forces.
What is the first step in analyzing the forces in a three-pin arch?
-The first step is to apply the force balance equations, starting with the horizontal forces (ΣFX = 0) and then the vertical forces (ΣFY = 0). This helps establish relationships between the reactions at the supports.
What is the significance of calculating the moment in structural analysis?
-Calculating the moment (ΣM = 0) is crucial because it allows for the determination of unknown forces by considering the rotational equilibrium of the structure. Moment calculations help balance the forces causing rotation about a specific point.
What happens when a structure is under a uniform load, as in this case?
-When a structure is under a uniform load, the load is distributed evenly along the structure. In the case of the three-pin arch, the reactions at the supports are calculated based on the symmetry of the load and the structure’s geometry.
How were the reactions (Va and VB) calculated in this case?
-The reactions Va and VB were calculated by using the equilibrium equation for vertical forces (ΣFY = 0) and by recognizing that the load is symmetrically distributed, leading to equal vertical reactions at both ends of the arch. Thus, Va = VB = 30 kN.
What are the values of the horizontal reactions (HA and HB) in this case, and why are they equal?
-The horizontal reactions HA and HB were calculated to be 15 kN each. These are equal because the load is symmetrically applied, and the structure is symmetric, leading to equal horizontal forces at both support points.
What is the role of moment balance around point S in the analysis?
-Moment balance around point S is used to calculate the unknown horizontal reactions (HA and HB). By considering the moments caused by forces at different points of the arch, it is possible to determine the magnitudes of these reactions.
Outlines
Dieser Bereich ist nur für Premium-Benutzer verfügbar. Bitte führen Sie ein Upgrade durch, um auf diesen Abschnitt zuzugreifen.
Upgrade durchführenMindmap
Dieser Bereich ist nur für Premium-Benutzer verfügbar. Bitte führen Sie ein Upgrade durch, um auf diesen Abschnitt zuzugreifen.
Upgrade durchführenKeywords
Dieser Bereich ist nur für Premium-Benutzer verfügbar. Bitte führen Sie ein Upgrade durch, um auf diesen Abschnitt zuzugreifen.
Upgrade durchführenHighlights
Dieser Bereich ist nur für Premium-Benutzer verfügbar. Bitte führen Sie ein Upgrade durch, um auf diesen Abschnitt zuzugreifen.
Upgrade durchführenTranscripts
Dieser Bereich ist nur für Premium-Benutzer verfügbar. Bitte führen Sie ein Upgrade durch, um auf diesen Abschnitt zuzugreifen.
Upgrade durchführenWeitere ähnliche Videos ansehen
S21A- STATIKA PORTAL 3 SENDI- REAKSI PERLETAKAN
Equilibrium of Rigid Bodies 3D force Systems | Mechanics Statics | (solved examples)
Equilibrium of Rigid Bodies (2D - Coplanar Forces) | Mechanics Statics | (Solved examples)
VIDEO PEMBELAJARAN ANALISIS STRUKTUR 1 MATERI GAYA DALAM PADA KONSTRUKSI BALOK SEDERHANA
SFD and BMD - Problem 1 - Part 1 - Shear Force and Bending Moment Diagram - Strength of Materials
Mechanics of Materials: F1-1 (Hibbeler)
5.0 / 5 (0 votes)