The placement of the bars in a beam will depend on the shear force and the bending moment diagram.
The diagram of bending and shearing forces shows the moments and shearing forces acting on any part of the beam. The layout of reinforcement is dependent on the size and sign (or direction) of diagrams in order to withstand these active forces. For instance, shear reinforcement or stirrups are tightly spaced on the region of the beam where shear force is greater. Additionally, maximal longitudinal reinforcement is supplied in the area with the highest bending moment. Similar to this, some longitudinal reinforcement is reduced when the influence of the bending moment decreases, when the direction of the bending stresses changes, or when the stresses in the web reinforcement increase.
Learn more about stresses here-
brainly.com/question/13261407
#SPJ4
Answer:
I do i do it everyday
Explanation:
Press windows and prt sc at the same time
Answer:
a) The one that is loosing electron is oxidation process and the one that is gaining electron is reduction.
b) Reduction reaction occur at cathode while oxidation reaction occur at Anode.
Explanation:
The main difference between the reduction and oxidation process is based on gaining and loosing of electron.
Reduction and oxidation both occur simultaneously. one is loosing electron and other is gaining. The one that is loosing electron is oxidation process and the one that is gaining electron is reduction.
B) Reduction reaction occur at cathode while oxidation reaction occur at Anode.
Explanation:
here you go. the answer is 303x+4
Answer:
critical stress required for the propagation is 27.396615 ×
N/m²
Explanation:
given data
specific surface energy = 0.90 J/m²
modulus of elasticity E = 393 GPa = 393 ×
N/m²
internal crack length = 0.6 mm
to find out
critical stress required for the propagation
solution
we will apply here critical stress formula for propagation of internal crack
( σc ) =
.....................1
here E is modulus of elasticity and γs is specific surface energy and a is half length of crack i.e 0.3 mm = 0.3 ×
m
so now put value in equation 1 we get
( σc ) =
( σc ) =
( σc ) = 27.396615 ×
N/m²
so critical stress required for the propagation is 27.396615 ×
N/m²