Answer: ε₁+ε₂+ε₃ = 0
Explanation: Considering the initial and final volume to be constant which gives rise to the relation:-
l₀l₀l₀=l₁l₂l₃
taking natural log on both sides
Considering the logarithmic Laws of division and multiplication :
ln(AB) = ln(A)+ln(B)
ln(A/B) = ln(A)-ln(B)
Use the image attached to see the definition of true strain defined as
ln(l1/1o)= ε₁
which then proves that ε₁+ε₂+ε₃ = 0
Answer:
mechanical engineer is the best answer
Answer:
Los aditivos que deben incorporarse a la masa de concreto para aumentar su resistencia a los ciclos alternos de congelación y descongelación son;
1. Agentes de arrastre de aire (AEA) o
2. Materiales poliméricos súper absorbentes
Explanation:
La resistencia alterna de los ciclos de congelación y descongelación en el concreto puede aumentarse mediante la adición de agentes de arrastre de aire.(AEA) que es un surfactante, crea burbujas de aire muy pequeñas en el concreto resultante para mejorar la durabilidad y resistencia del cemento al ciclo repetido de congelación y descongelación o materiales poliméricos súper absorbentes
Ejemplos de agentes de arrastre de aire son;
Sulfonatos alcalinos
Acidos de resinas sulfonadas
Sales de ácidos grasos
Ejemplos de materiales poliméricos superabsorbentes son;
SAP0.26CT
SAP0.39PT.
Batter boards (or battre boards, Sometimes mispronounced as "battle boads") are temporary frames, set beyond the corners of a planned foundation at precise elevations. These batter boards are then used to hold layout lines (construction twine) to indicate the limits (edges and corners) of the foundation.
Answer:
The inventor's claim is false in the sense that no thermal machine can violate the first thermodynamic law.
Explanation:
The inventor's claim could not be possible as no thermal machine can transfer more heat than the input work consumed. If we expose the thermal efficiency:
Where Q and W both must be in the same power unit, so we will convert the remove heat from BTU/hr to hp:
Therefore by comparing, we notice that the removing heat of 4.75 hp is large than the delivered work of 1.11 hp. By evaluating the efficiency:
[tex]n=4.75 hp / 1.1 hp = 4.3 > 1[/tex]
