Answer: b) Area under the elastic portion of the stress-strain curve
Explanation:
By definition, resilience is the strain-energy density stored by the material when it is stressed to the proportional limit defined by Hooke's Law. Resilience is given by the following expression:
μ(r) = / 2E
μ(r) is the modulus of resilience
σ(pl) is the stress to the proportional limit
E is the elastic modulus
When you look at the stress-strain curve, the area under the elastic portion (up to the proportional limit) can be obtained by the area of a triangle with base equal to the strain (σ) and height equal to the stress (ε):
Ω = (b . h) / 2 = (σ(pl) . ε) / 2
Using Hooke's Law: σ = E . ε → ε = σ/E
Replacing the expression in the area equation:
Ω = (σ(pl) . ε) / 2 = / 2E = μ(r)
Answer:
The table is attached as a picture.
a)
Select VENDOR_CONTACT_LAST_NAME || ', ' || VENDOR_CONTACT_FIRST_NAME "full_name" from VENDORS where VENDOR_CONTACT_LAST_NAME like 'A%' or VENDOR_CONTACT_LAST_NAME like 'E%' order by VENDOR_CONTACT_LAST_NAME,VENDOR_CONTACT_FIRST_NAME;
concatenation operator || is used . Also LIKE is used for pattern matching. full_name is alias for the concatenated column
b) As sample data is not given ,Please test the query for the data given in table
Explanation:
Answer:
the angle of twist of B with respect to D is -1.15°
the angle of twist of C with respect to D is 1.15°
Explanation:
The missing diagram that is supposed to be added to this image is attached in the file below.
From the given information:
The shaft is made of A992 steel. It has a diameter of 1 in and is supported by bearing at A and D.
For the Modulus of Rigidity G = 11 × 10³ Ksi = 11 × 10⁶ lb/in²
The objective are :
1) To determine the angle of twist of B with respect to D
Considering the Polar moment of Inertia at the shaft
shaft =
where ;
r = 1 in /2
r = 0.5 in
shaft =
shaft = 0.098218
Now; the angle of twist at B with respect to D is calculated by using the expression
where;
are the torques at segments CD and length at segments CD
are the torques at segments BC and length at segments BC
Also ; from the diagram; the following values where obtained:
= 2.5 in
= 0.098218
G = 11 × 10⁶ lb/in²
= -60 lb.ft
= 0 lb.ft
= 5.5 in
− 0.02 rad
To degree; we have
Since we have a negative sign; that typically illustrates that the angle of twist is in an anti- clockwise direction
Thus; the angle of twist of B with respect to D is 1.15°
(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places
For the angle of twist of C with respect to D; we have:
0.02 rad
To degree; we have
Answer:
E = 2940 J
Explanation:
It is given that,
Mass, m = 12 kg
Position at which the object is placed, h = 25 m
We need to find the potential energy of the mass. It is given by the formula as follows :
E = mgh
g is acceleration due to gravity
So, the potential energy of the mass is 2940 J.