From
6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886)
NOTES TO THE INSTRUCTOR
The Solutions Manual to accompany the text AdvancedMechanicsofMaterialsand AppliedElasticity supplements the study of stress and deformation analyses developed in the book. The main objective of the manual is to provide efficient solutions for problems dealing with variously loaded members. This manual can also serve to guide the instructor in the assignments of problems, in grading these problems, and in preparing lecture materials as well as examination questions. Every effort has been made to have a solutions manual that can cut through the clutter and is self - explanatory as possible thus reducing the work on the instructor. It is written and class tested by the author, Ansel Ugural.
As indicated in its preface, the text is designed for the senior and/or first year graduate level courses in stress analysis. In order to accommodate courses of varying emphasis, considerably more material has been presented in the book than can be covered effectively in a single three-credit course. The instructor has the choice of assigning a variety of problems in each chapter. Answers to selected problems are given at the end of the text. A description of the topics covered is given in the introduction of each chapter throughout the text. It is hoped that the foregoing materials will help instructor in organizing his or her course to best fit the needs of his or her students.
Holmdel, NJ
Ansel C. Ugural
CHAPTER 1
SOLUTION (1.1)
We have 50753.75(10)m32 A =×= , o 50 = θ , and AP x = σ
Equations (1.11), with o 50 = θ : P x o xx 18.110 413.050 cos )10(700 2 3 ' == = = σ σ σ or 6.35kN P = and 3 '' 560(10)sin50cos500.492131.2 oo xyxx P τσσ == ==
Solving 4.27kN P = allP =
SOLUTION (1.2)
Normal stress is 3 125(10) 0.050.05 50MPa P xAσ × ===
( a ) Equations (1.11), with o 20 = θ :
2 ' 50cos2044.15MPa o x σ == '' 50sin20cos2016.08MPa oo xy τ = −= 2 ' 50cos(2090)5.849MPa oo y σ =+=
MPa
( b ) Equations (1.11), with o45 = θ : 2 ' 50cos4525MPa o x σ == '' 50sin45cos4525MPa oo xy τ = −= 2 ' 50cos(4590)25MPa oo y σ =+=
SOLUTION (1.3)
From Eq. (1.11a), ' 22 75 coscos30 100MPa x xo σ θ σ ===−
For o 50 = θ , Eqs. (1.11) give then 2 ' 100cos5041.32MPa o x σ =−=− oo yx 50 cos 50 sin )100( '' −−= τ 49.24MPa =
Similarly, for o 140 = θ : 2 ' 100cos14058.68MPa o x σ = −= '' 49.24MPa xy τ =−
SOLUTION (1.4)
Refer to Fig. 1.6c. Equations (1.11) by substituting the double angle-trigonometric relations, or Eqs. (1.18) with 0 = y σ and 0 = xy τ , become θσσσ 2 cos 2 1 2 1 ' xxx += and θστ 2 sin 2 1 '' yxx = or cos)2 1(20 2 θ += A P and θ2 sin 10 2 A P =
The foregoing lead to 12 cos 2 sin 2 θθ=− (a)
By introducing trigonometric identities, Eq. (a) becomes 0 cos 2 cossin 4 2 θθθ=− or 21 tan = θ . Hence o 56.26 = θ
Thus, 2(1300) 20(10.6) P =+ gives 32.5kN P = It can be shown that use of Mohr’s circle yields readily the same result.
SOLUTION (1.5)
Equations (1.12):
SOLUTION (1.6)
Shaded transverse area: 22(10)(75)1.5(10)mm32Aat===
Metal is capable of supporting the load 63 90(10)(1.510)135kN PA σ ==×=
Apply Eqs. (1.11): 62 ' 3 25(10)(cos55) 1.5(10)
Thus, 38.3kN allP =
SOLUTION (1.7)
Use Eqs. (1.11): 62 ' 3 20(10)(cos40) 1.5(10) o x P σ == , 51.1kN P = 6 '' 3 8(10) sin40cos40 1.5(10)
Thus, 24.4kN allP =
SOLUTION (1.8) 2 1530450mm A =×=
Apply Eqs. (1.11): 3 2 ' 6 120(10) (cos40)156MPa 45010 o x σ = = × 3 '' 6 120(10) sin40cos40131MPa 45010 oo xy τ = −= ×
SOLUTION (1.9)
We have 450(10)m62 A = . Use Eqs. (1.11): 3 2 ' 6 100(10) (cos60)55.6MPa 45010 o x σ = = × 3
'' 6 100(10) sin60cos6096.2MPa 45010 oo xy τ = −= ×
SOLUTION (1.10)
ooo1309040=+= θ 3
Equations (1.11): 2 ' 31.83cos13013.15MPa o x σ = −= '' 31.83sin130cos13015.67MPa oo xy τ = =−
SOLUTION (1.11)
Use Eqs. (1.14), 0)()2()2(=++−+ x xxyFx
0)0()2()( 2 =+++−+− y Fxyyz
0)2()0()4(=+−++− z zxyFz
Solving, we have (in 3 MNm ): xFxy x 23+−= 2 2 y Fxyyz =−++ Fxyz z += 4 (a)
Substituting x=-0.01 m, y=0.03 m, and z=0.06 m, Eqs. (a) yield the following values 333 29.4kNm14.5kNm58.8kNm xyz FFF = = =
Resultant body force is thus 222 3 67.32kNm xyz FFFF=++=
SOLUTION (1.12)
Equations (1.14):
No. Eqs. (1.14) are not satisfied.
SOLUTION (1.13)
( a ) No. Eqs. (1.14) are not satisfied.
( b ) Yes. Eqs. (1.14) are satisfied.
SOLUTION (1.14)
Eqs. (1.14) for the given stress field yield:
SOLUTION (1.15)
1.16 (CONT.)
' '' 0:50sin25cos25 oo yxy FAA τ =∆−∆
2 90sin25cos2515cos25 oooAA−∆ −∆ 2 15sin250 Ao +∆= '' 19.1534.4712.322.6863.3MPa xy τ =++−=
SOLUTION (1.17)
' 11 22(4060)(4060)cos4050sin40 oo x σ =−++−−+ 1038.332.13.8MPa =−+=
'' 1 (4060)sin4050cos40 2 oo xy τ =−−−+
32.1438.370.4MPa=+=
SOLUTION (1.18)
' 11 22(9050)(9050)cos23015sin230 oo x σ =−++− 204511.513.5MPa=−+=−
'' 1 (9050)sin23015cos230 2 oo xy τ =−+− 53.629.6463.3MPa=+=
SOLUTION (1.19)
Transform from 40 o θ = to 0 θ = . For convenience in computations, Let 160MPa,80MPa,40MPaxyxy σστ = −= −= and 40 o θ =−
Then ' 11 22()()cos2sin2xxyxyxy σσσσσθτθ =++−+ 11
22(16080)(16080)cos(80)40sin(80) oo=−−+−+−+−
166.3MPa=− '' 1 ()sin2cos2 2 xyxyxy τσσθτθ =−−+ 1 (16080)sin(80)40cos(80) 2 oo=−−+−+− 32.4MPa=−
So '' 16080166.373.7MPa yxyx σσσσ=+−=−−+=−
For 0 o θ = :
SOLUTION (1.21)
070 o xy τθ==
(a) '' 60 30sin140 2 o xy σ τ =−=− 153.3MPa σ =
(b) ' 6060 80 cos140 22 o x σσ σ +− ==+ 231MPa σ =
SOLUTION (1.22)
Equations(1.18) with 60 o θ = , 110MPa x σ = , 0 y σ = , 50MPa xy τ = give 11 ' 22(110)(110)cos12050sin12070.8MPa oo x σ =++=
1 '' 2 (110)sin12050cos12072.6MPa oo xy τ =−+=− 11 ' 22(110)(110)cos12050sin12039.2MPa oo y σ =−−=
SOLUTION (1.23)
Equations(1.18) with 30 o θ = , 110MPa x σ = , 0 y σ = , 50MPa xy τ = result in
1 ' 2 (110)55cos6050sin60125.8MPa oo x σ =++= 1 '' 2 (110)sin6050cos6022.6MPa oo xy τ =−+=−
1 ' 2 (110)55cos6050sin6015.8MPa oo y σ =−−=−
SOLUTION (1.24)
We have 2590115o θ =+= 10MPa x σ =− 30MPa y σ = 0 xy τ = (a) ' 11 22()()cos2xxyxy σσσσσθ =++− 11 22(1030)(1030)cos23022.86MPa o =−++−−=
Thus, ' 22.86MPa wx σσ==
(CONT.) θ x’ 10 MPa ' x σ ''xy τ 25o 30 MPa x y’
1.24 (CONT.)
(b) '' 1 ()sin2 2 xyxy τσσθ =−− 1 (1030)sin23015.32MPa 2 o =−−−=− So '' 15.32MPa wxy ττ==−
SOLUTION (1.25)
(a) 22 1 050050 80 ( ) 22 στ +− ==++ 49MPa τ =
SOLUTION (1.26) (CONT.) w τ
(b) 22 max 50 ()4955MPa 2 τ =+= 50 '25MPa 2 σ == 1 050 2tan[]27 2(49) o s θ =−= '' 50 sin2749cos2755MPa 2 oo xy τ =+= Thus, '13.5o s θ =
1.26 (CONT.)
40804080 cos(60)601050MPa 22 o x σ +− =+−=−= 601070MPa y σ =+= 4080 sin(60)17.32MPa 2 o xy τ =−−=−
' 5070 60cos(15)2.68sin(15) 2 oo x σ =+−+− 609.660.69449.65MPa =−−= Thus, "7.5o p θ =−
1.27 (CONT.)
1 50 tan45 50 o α ==
1 2 22 (5050)70.7 R =+=
'' sin85(70.7)70.4MPa o xy τ = = ' 10cos85(70.7)3.84MPa o x σ =−=
SOLUTION (1.28)
1 15 tan12.1 70 o α ==
1 2 22 (1570)71.6 R =+=
'' 71.6sin62.163.3MPa o xy τ = = ' 71.6cos62.120 o x σ = −+ 13.5MPa=−
1.29 (CONT.)
'' 22.5sin73.821.6MPa o xy τ = = ' 67.522.5cos73.873.8MPa o x σ =+=
Sketch of results is as shown in solution of Prob. 1.20.
SOLUTION (1.30)
(a) '' 60 30sin(40);153.3MPa 2 o xy σ τσ =−=−=
(b) ' 60 8060[1cos(40)] 2 o x σ σ ==+−− 231MPa σ =
SOLUTION (1.31)
( a ) From Mohr’s circle, Fig. (a): 1 2 max 121MPa71MPa96MPaσστ == −= o s o p 7.25'3.19' = −= θ θ By applying Eq. (1.20):
Eq. (1.19):
(CONT.)
1.31 (CONT.)
( b ) From Mohr’s circle, Fig. (b): 1 2 max 200MPa50MPa125MPaσστ == −= o s o p 55.71'55.26' = = θ θ
Figure (b)
Through the use of Eq. (1.20), [ ] 12575000,10 75 2 1 4 500,22 2,1 ±=+±= σ or 12200MPa50MPa σσ = =−
Using Eq. (1.19),342 tan = p θ : '26.57'71.57 oo ps θθ = =
SOLUTION (1.32)
Referring to Mohr’s circle, Fig. 1.15:
From Eqs. (a), 21'' σσσσ+=+ yx
By using sin12 2 cos 2 2 θθ=+ , and Eqs. (a) and (b), we have const yxyx =⋅=−⋅ 21 2 ''''σστσσ . SOLUTION (1.33)
We have
(a)
(b)
1.33 (CONT.)
Equations (1.18):
SOLUTION (1.34)
Substituting
SOLUTION (1.35)
Transform from o 60 = θ to o 0 = θ with ''20MPa,60MPa xy σσ = −= , '' 22MPa xy τ =− , and o 60 θ−= . Use Eqs. (1.18): 20602060 22 cos2(60)22sin2(60)59MPa oo x
y x 19 MPa
SOLUTION (1.36)
Figure (a)
( a ) Figure (a): 14sin6012.12MPa o y σ == 14cos607MPa o xy τ ==
Figure (b): 060 sin 60 cos 12.12 = =
o xy o y F τ or 7MPa xy τ = (as before)
sin oo xx F σ or 38.68MPa x σ =
( b ) Equation (1.20) is therefore:
2 12.1268.38 2,1 7)( + ±= + σ or 1240.41MPa,10.39MPa σσ = =
Also, o p 9.13 tan12.1268.38 )7(2 1 2 1 = = θ
Note: Eq. (1.18a) gives, ' 40.41MPa x σ = Thus, o p 9.13' = θ
xsin60o
Figure (b)
SOLUTION (1.37)
Figure (a):
100cos4570.7MPa o x σ ==
100sin4570.7MPa o y σ ==
100cos4570.7MPa o xy τ ==
Now, Eqs. (1.18) give (Fig. b):
' 70.7070.7sin2409.47MPa o x σ =++=
'' 070.7cos24035.35MPa o xy τ =−+ =−
' 70.7070.7sin240131.9MPa o y σ =−−=
SOLUTION (1.38)
70sin3035MPa o y σ =−=−
70cos3060.6MPa o xy τ ==
( a ) Figure (a): 0)866.0(6.605.0150 = ++−= ∑ xx F σ or 195MPa x σ =
150 MPa
(a)
xsin30o
Area=1
Figure (a)
Figure (b)
Figure