Riski states that any function written without exponents must be linear.Which function proves Riski 's statement is incorrect? A
)y=5x+3 B)y=x⁵+3 C)y=×/3+5 D)y=3/×+5
1 answer:
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Answer:
please see answers are as in the explanation.
Step-by-step explanation:
As from the data of complete question,
The question also has 3 parts given as
<em>Part a: Sketch the deformed shape for α=0.03, β=-0.01 .</em>
Solution
As w is 0 so the deflection is only in the x and y plane and thus can be sketched in xy plane.
the new points are calculated as follows
Point A(x=0,y=0)
Point A'(x+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>
Point A'(0+<em>(0.03)</em><em>(0),0+</em><em>(-0.01)</em><em>(0))</em>
Point A'(0<em>,0)</em>
Point B(x=1,y=0)
Point B'(x+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>
Point B'(1+<em>(0.03)</em><em>(1),0+</em><em>(-0.01)</em><em>(0))</em>
Point <em>B</em>'(1.03<em>,0)</em>
Point C(x=1,y=1)
Point C'(x+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>
Point C'(1+<em>(0.03)</em><em>(1),1+</em><em>(-0.01)</em><em>(1))</em>
Point <em>C</em>'(1.03<em>,0.99)</em>
Point D(x=0,y=1)
Point D'(x+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>
Point D'(0+<em>(0.03)</em><em>(0),1+</em><em>(-0.01)</em><em>(1))</em>
Point <em>D</em>'(0<em>,0.99)</em>
So the new points are A'(0,0), B'(1.03,0), C'(1.03,0.99) and D'(0,0.99)
The plot is attached with the solution.
<em>Part b: Calculate the six strain components.</em>
Solution
Normal Strain Components
Shear Strain Components
Part c: <em>Find the volume change</em>
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<em>Also the change in volume is 0.0197</em>
For the unit cube, the change in terms of strains is given as
As the strain values are small second and higher order values are ignored so
As the initial volume of cube is unitary so this result can be proved.
