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Solutions Manual For Advanced Mechanics of Materials and Applied Elasticity 6th Edition By Ansel Ugural, Saul Fenster (All Chapters, 100% Original Verified, A+ Grade) $28.49
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Advanced Mechanics Of Materials And Applied Elasti
Advanced Mechanics of Materials and Applied Elasti
Exam (elaborations)
Solutions Manual For Advanced Mechanics of Materials and Applied Elasticity 6th Edition By Ansel Ugural, Saul Fenster (All Chapters, 100% Original Verified, A+ Grade)
Advanced Mechanics of Materials and Applied Elasti
Institution
Advanced Mechanics Of Materials And Applied Elasti
This Is The Original 6th Edition Of The Solution Manual From The Original Author All Other Files In The Market Are Fake/Old Editions. Other Sellers Have Changed The Old Edition Number To The New But The Solution Manual Is An Old Edition.
Solutions Manual For Advanced Mechanics of Materials and ...
We have
A = 50 × 75 = 3.75(10−3 ) m 2 , θ = 50o , and σ x = P A .
Equations (1.11), with θ = 50o :
σ x ' = 700(10 3 ) = σ x cos 2 50o = 0.413σ x = 110.18P
or P = 6.35 kN
and
τ x ' y ' 560(10
= = 3
) σ x sin 50o =
cos 50o 0.492
= σ x 131.2 P
Solving
P = 4.27 kN = Pall
______________________________________________________________________________________
SOLUTION (1.2)
Normal stress is
125(103 )
σ x= P
A= = 50 MPa
0.05×0.05
( a ) Equations (1.11), with θ = 20o :
=σ x ' 50
= cos 2 20o 44.15 MPa
τ x' y' =
−50sin 20o cos 20o =
−16.08 MPa
σ y ' 50 cos 2 (20o +=
= 90o ) 5.849 MPa
5.849 MPa
y’
44.15 MPa
16.08 MPa x’
20 o
x
θ = 45o :
( b ) Equations (1.11), with
=σ x ' 50
= cos 2 45o 25 MPa
τ x' y' =
−50sin 45o cos 45o =
−25 MPa
σ y ' 50 cos 2 (45o +=
= 90o ) 25 MPa
25 MPa
25 MPa
x’
y’
25 MPa
45 o
x
______________________________________________________________________________________
Refer to Fig. 1.6c. Equations (1.11) by substituting the double angle-trigonometric relations,
or Eqs. (1.18) with σ y = 0 and τ xy = 0 , become
σ x ' = 12 σ x + 12 σ x cos 2θ and τ x ' y ' = 12 σ x sin 2θ
or
20 = P
2A (1 + cos 2θ ) and 10 = P
2A sin 2θ
The foregoing lead to
2 sin 2θ − cos 2θ = 1 (a)
By introducing trigonometric identities, Eq. (a) becomes
4 sin θ cos θ − 2 cos 2 θ = 0 or tan θ = 1 2 . Hence
θ = 26.56o
Thus,
=20 P
2(1300) (1 + 0.6)
gives
P = 32.5 kN
It can be shown that use of Mohr’s circle yields readily the same result.
______________________________________________________________________________________
SOLUTION (1.5)
Equations (1.12):
P −150(103 )
σ1 = = = −76.4 MPa
A π 2
(50)
4
P
τ max
= = 38.2 MPa
2A
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