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MECH-321 Mechanics of deformable solids

Solution for Problem set-1

PROBLEM 2.101 Rod ABC consists of two cylindrical portions AB and BC; it is made of a mild steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 250 MPa. A force P is applied to the rod and then removed to give it a permanent set δ p = 2 mm. Determine the maximum value of the force P and the maximum amount δ m by which the rod should be stretched to give it the desired permanent set.

SOLUTION AAB = ABC =

π 4

π

(30)2 = 706.86 mm 2 = 706.86 × 10−6 m 2 (40) 2 = 1.25664 × 103 mm 2 = 1.25664 × 10−3 m 2

4 Pmax = Aminσ Y = (706.86 × 10−6 )(250 × 106 ) = 176.715 × 103 N Pmax = 176.7 kN 

δ′ =

P′LAB P ′LBC (176.715 × 103 )(0.8) (176.715 × 103 )(1.2) + = + EAAB EABC (200 × 109 )(706.86 × 10−6 ) (200 × 109 )(1.25664 × 10−3 )

= 1.84375 × 10−3 m = 1.84375 mm

δ p = δ m − δ ′ or δ m = δ p + δ ′ = 2 + 1.84375

δ m = 3.84 mm 

MECH-321 Mechanics of deformable solids

Solution for Problem set-1

PROBLEM 2.102 Rod ABC consists of two cylindrical portions AB and BC; it is made of a mild steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 250 MPa. A force P is applied to the rod until its end A has moved down by an amount δ m = 5 mm. Determine the maximum value of the force P and the permanent set of the rod after the force has been removed.

SOLUTION AAB =

4

(30)2 = 706.86 mm 2 = 706.86 × 10−6 m 2

π

(40) 2 = 1.25664 × 103 mm 2 = 1.25644 × 10−3 m 2 4 = Aminσ Y = (706.86 × 10−6 )(250 × 106 ) = 176.715 × 103 N

ABC = Pmax

π

Pmax = 176.7 kN 

δ′ =

P′LAB P ′LBC (176.715 × 103 )(0.8) (176.715 × 103 )(1.2) + = + EAAB EABC (200 × 109 )(706.68 × 10−6 ) (200 × 109 )(1.25664 × 10−3 )

= 1.84375 × 10−3 m = 1.84375 mm

δ p = δ m − δ ′ = 5 − 1.84375 = 3.16 mm

δ p = 3.16 mm 

MECH-321 Mechanics of deformable solids

Solution for Problem set-1

PROBLEM 2.103 The 30-mm square bar AB has a length L = 2.2 m; it is made of a mild steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 345 MPa. A force P is applied to the bar until end A has moved down by an amount δ m . Determine the maximum value of the force P and the permanent set of the bar after the force has been removed, knowing that (a) δ m = 4.5 mm, (b) δ m = 8 mm.

SOLUTION A = (30)(30) = 900 mm2 = 900 × 10−6 m 2

δY = Lε Y =

Lσ Y (2.2)(345 × 106 ) = = 3.795 × 10−3 = 3.795 mm E 200 × 109

If δ m ≥ δ Y , Pm = Aσ Y = (900 × 10−6 )(345 × 106 ) = 310.5 × 103 N Unloading: δ ′ =

Pm L σ Y L = = δY = 3.795 mm AE E

δ P = δm − δ ′ (a)

δ m = 4.5 mm > δY Pm = 310.5 × 103 N δ perm = 4.5 mm − 3.795 mm

(b)

δ m = 8 mm > δY Pm = 310.5 × 103 N



δ perm = 8.0 mm − 3.795 mm

δ m = 310.5 kN  δ perm = 0.705 mm  δ m = 310.5 kN  δ perm = 4.205 mm 

MECH-321 Mechanics of deformable solids

Solution for Problem set-1

PROBLEM 2.104 The 30-mm square bar AB has a length L = 2.5 m; it is made of mild steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 345 MPa. A force P is applied to the bar and then removed to give it a permanent set δ p . Determine the maximum value of the force P and the maximum amount δ m by which the bar should be stretched if the desired value of δ p is (a) 3.5 mm, (b) 6.5 mm.

SOLUTION A = (30)(30) = 900 mm 2 = 900 × 10−6 m 2

δY = Lε Y =

Lσ Y (2.5)(345 × 106 ) = = 4.3125 × 103 m = 4.3125 mm 9 E 200 × 10

When δ m exceeds δ Y , thus producing a permanent stretch of δ p , the maximum force is Pm = Aσ Y = (900 × 10−6 )(345 × 106 ) = 310.5 × 103 N

= 310.5 kN 

δ p = δ m − δ ′ = δ m − δY ∴ δ m = δ p + δY (a)

δ p = 3.5 mm δ m = 3.5 mm + 4.3125 mm

= 7.81 mm 

(b)

δ p = 6.5 mm δ m = 6.5 mm + 4.3125 mm

= 10.81 mm 

MECH-321 Mechanics of deformable solids

Solution for Problem set-1

PROBLEM 2.105 Rod AB is made of a mild steel that is assumed to be elastoplastic with E = 29 × 106 psi and σ Y = 36 ksi. After the rod has been attached to the rigid lever CD, it is found that end C is 83 in. too high. A vertical force Q is then applied at C until this point has moved to position C ′. Determine the required magnitude of Q and the deflection δ1 if the lever is to snap back to a horizontal position after Q is removed.

SOLUTION Since the rod AB is to be stretched permanently, the peak force in the rod is P = PY , where PY = Aσ Y =

π 3

2

(36) = 3.976 kips 4  8 

Referring to the free body diagram of lever CD,

ΣM D = 0: 33 Q − 22 P = 0 22 (22)(3.976) = 2.65 kips Q= P= 33 33

Q = 2.65 kips 

During unloading, the spring back at B is

δ B = LABε Y =

LABσ Y (60)(36 × 103 ) = = 0.0745 in. E 29 × 106

From the deformation diagram, Slope:

θ=

δB 22

=

δC 33

∴ δC =

33 δ B = 0.1117 in. −22

δ C = 0.1117 in. 

MECH-321 Mechanics of deformable solids

Solution for Problem set-1

PROBLEM 2.106 Solve Prob. 2.105, assuming that the yield point of the mild steel is 50 ksi. PROBLEM 2.105 Rod AB is made of a mild steel that is assumed to be elastoplastic with E = 29 × 106 psi and σ Y = 36 ksi. After the rod has been attached to the rigid lever CD, it is found that end C is 83 in. too high. A vertical force Q is then applied at C until this point has moved to position C′. Determine the required magnitude of Q and the deflection δ1 if the lever is to snap back to a horizontal position after Q is removed.

SOLUTION Since the rod AB is to be stretched permanently, the peak force in the rod is P = PY , where PY = Aσ Y =

π 3

2

(50) = 5.522 kips 4  8 

Referring to the free body diagram of lever CD,

ΣM D = 0: 33Q − 22 P = 0 Q=

22 (22)(5.522) P= = 3.68 kips 33 33

Q = 3.68 kips 

During unloading, the spring back at B is

δ B = LAB ε Y =

LABσ Y (60)(50 × 103 ) = = 0.1034 in. E 29 × 106

From the deformation diagram, Slope:

θ=

δB 22

=

δC 33

∴

δC =

33 δB 22

δ C = 0.1552 in. 

MECH-321 Mechanics of deformable solids

Solution for Problem set-1

PROBLEM 2.107 Each cable has a cross-sectional area of 100 mm2 and is made of an elastoplastic material for which σ Y = 345 MPa and E = 200 GPa. A force Q is applied at C to the rigid bar ABC and is gradually increased from 0 to 50 kN and then reduced to zero. Knowing that the cables were initially taut, determine (a) the maximum stress that occurs in cable BD, (b) the maximum deflection of point C, (c) the final displacement of point C. (Hint: In Part c, cable CE is not taut.)

SOLUTION Elongation constraints for taut cables. Let θ = rotation angle of rigid bar ABC.

θ=

δ BD LAB

δ BD =

=

δ CE LAC

LAB 1 δ CE = δ CE LAC 2

(1)

Equilibrium of bar ABC.

M A = 0 : LAB FBD + LAC FCE − LAC Q = 0 Q = FCE +

LAB 1 FBD = FCE + FBD 2 LAC

Assume cable CE is yielded.

FCE = Aσ Y = (100 × 10−6 )(345 × 106 ) = 34.5 × 103 N

From (2),

FBD = 2(Q − FCE ) = (2)(50 × 103 − 34.5 × 103 ) = 31.0 × 103 N

Since FBD < Aσ Y = 34.5 × 103 N, cable BD is elastic when Q = 50 kN.

(2)

MECH-321 Mechanics of deformable solids

Solution for Problem set-1

PROBLEM 2.107 (Continued)

(a)

σ CE = σ Y = 345 MPa

Maximum stresses.

σ BD = (b)

FBD 31.0 × 103 = = 310 × 106 Pa −6 A 100 × 10

σ BD = 310 MPa 

Maximum of deflection of point C.

δ BD = From (1),

FBD LBD (31.0 × 103 )(2) = = 3.1 × 10−3 m EA (200 × 109 )(100 × 10−6 )

δ C = δ CE = 2δ BD = 6.2 × 10−3 m

Permanent elongation of cable CE: (δ CE ) p = (δ CE ) −

6.20 mm ↓ 

σ Y LCE E

FCE LCE σ L = (δ CE ) max − Y CE EA E 6 (345 × 10 )(2) = 6.20 × 10−3 − = 2.75 × 10−3 m 9 200 × 10

(δ CE ) P = (δ CE ) max −

(c)

Unloading. Cable CE is slack ( FCE = 0) at Q = 0. From (2),

FBD = 2(Q − FCE ) = 2(0 − 0) = 0

Since cable BD remained elastic, δ BD =

FBD LBD = 0. EA



MECH-321 Mechanics of deformable solids

Solution for Problem set-1

PROBLEM 2.108 Solve Prob. 2.107, assuming that the cables are replaced by rods of the same cross-sectional area and material. Further assume that the rods are braced so that they can carry compressive forces. PROBLEM 2.107 Each cable has a cross-sectional area of 100 mm2 and is made of an elastoplastic material for which σ Y = 345 MPa and E = 200 GPa. A force Q is applied at C to the rigid bar ABC and is gradually increased from 0 to 50 kN and then reduced to zero. Knowing that the cables were initially taut, determine (a) the maximum stress that occurs in cable BD, (b) the maximum deflection of point C, (c) the final displacement of point C. (Hint: In Part c, cable CE is not taut.)

SOLUTION Elongation constraints. Let θ = rotation angle of rigid bar ABC.

θ= δ BD =

δ BC LAB

=

δ CE LAC

LAB 1 δ CE = δ CE LAC 2

(1)

Equilibrium of bar ABC.

M A = 0: LAB FBD + LAC FCE − LAC Q = 0 Q = FCE +

LAB 1 FBD = FCE + FBD LAC 2

Assume cable CE is yielded. FCE = Aσ Y = (100 × 10−6 )(345 × 106 ) = 34.5 × 103 N From (2),

FBD = 2(Q − FCE ) = (2)(50 × 103 − 34.5 × 103 ) = 31.0 × 103 N

Since FBD < Aσ Y = 34.5 × 103 N, cable BD is elastic when Q = 50 kN.

(2)

MECH-321 Mechanics of deformable solids

Solution for Problem set-1

PROBLEM 2.108 (Continued)

(a)

σ CE = σ Y = 345 MPa

Maximum stresses.

σ BD = (b)

FBD 31.0 × 103 = = 310 × 106 Pa −6 A 100 × 10

σ BD = 310 MPa 

Maximum of deflection of point C.

δ BD =

FBD LBD (31.0 × 103 )(2) = = 3.1 × 10−3 m EA (200 × 109 )(100 × 10−6 )

δ C = δ CE = 2δ BD = 6.2 × 10−3 m

From (1),

6.20 mm ↓ 

′ = δ C′ Unloading. Q′ = 50 × 103 N, δ CE ′ = 12 δ C′ δ BD

From (1), ′′ = Elastic FBD ′ = FCE

(200 × 109 )(100 × 10−6 )( 12 δ C′ ) ′ EAδ BD = = 5 × 106 δ C′ LBD 2

′ EAδ CE (200 × 109 )(100 × 10−6 )(δ C′ ) = = 10 × 106 δ C′ LCE 2

From (2),

′ + 12 FBD ′ = 12.5 × 106 δ C′ Q′ = FCE

Equating expressions for Q′,

12.5 × 106 δ C′ = 50 × 103

δ C′ = 4 × 10−3 m (c)

Final displacement.

δ C = (δ C ) m − δ C′ = 6.2 × 10−3 − 4 × 10−3 = 2.2 × 10−3 m

2.20 mm ↓ 

MECH-321 Mechanics of deformable solids

Solution for Problem set-1

PROBLEM 2.109 Rod AB consists of two cylindrical portions AC and BC, each with a cross-sectional area of 1750 mm2. Portion AC is made of a mild steel with E = 200 GPa and σ Y = 250 MPa, and portion CB is made of a high-strength steel with E = 200 GPa and σ Y = 345 MPa. A load P is applied at C as shown. Assuming both steels to be elastoplastic, determine (a) the maximum deflection of C if P is gradually increased from zero to 975 kN and then reduced back to zero, (b) the maximum stress in each portion of the rod, (c) the permanent deflection of C.

SOLUTION Displacement at C to cause yielding of AC. L σ (0.190)(250 × 106 ) δ C ,Y = LAC ε Y , AC = AC Y , AC = = 0.2375 × 10−3 m 9 E 200 × 10 FAC = Aσ Y , AC = (1750 × 10−6 )(250 × 106 ) = 437.5 × 103 N

Corresponding force. FCB = −

EAδ C (200 × 109 )(1750 × 10−6 )(0.2375 × 10−3 ) =− = −437.5 × 103 N 0.190 LCB

For equilibrium of element at C, FAC − ( FCB + PY ) = 0

PY = FAC − FCB = 875 × 103 N

Since applied load P = 975 × 103 N > 875 × 103 N, portion AC yields. FCB = FAC − P = 437.5 × 103 − 975 × 103 N = −537.5 × 103 N

(a)

δC = −

FCB LCD (537.5 × 103 )(0.190) = = 0.29179 × 10−3 m EA (200 × 109 )(1750 × 10−6 ) 0.292 mm 

(b)

Maximum stresses: σ AC = σ Y , AC = 250 MPa

(c)

FBC 537.5 × 103 =− = −307.14 × 106 Pa = −307 MPa −6 A 1750 × 10 Deflection and forces for unloading. P′ L P′ L L ′ = − PAC ′ AC = − PAC ′ δ ′ = AC AC = − CB CB ∴ PCB EA EA LAB

σ BC =



250 MPa  −307 MPa 

′ − PCB ′ = 2 PAC ′ PAC ′ = 487.5 × 10−3 N P′ = 975 × 103 = PAC

δ′ =

(487.5 × 103 )(0.190) = 0.26464 × 103 m −6 9 (200 × 10 )(1750 × 10 )

δ p = δ m − δ ′ = 0.29179 × 10−3 − 0.26464 × 10−3 = 0.02715 × 10−3 m

0.0272 mm 

MECH-321 Mechanics of deformable solids

Solution for Problem set-1

PROBLEM 2.110 For the composite rod of Prob. 2.109, if P is gradually increased from zero until the deflection of point C reaches a maximum value of δ m = 0.3 mm and then decreased back to zero, determine (a) the maximum value of P, (b) the maximum stress in each portion of the rod, (c) the permanent deflection of C after the load is removed. PROBLEM 2.109 Rod AB consists of two cylindrical portions AC and BC, each with a cross-sectional area of 1750 mm2. Portion AC is made of a mild steel with E = 200 GPa and σ Y = 250 MPa, and portion CB is made of a high-strength steel with E = 200 GPa and σ Y = 345 MPa. A load P is applied at C as shown. Assuming both steels to be elastoplastic, determine (a) the maximum deflection of C if P is gradually increased from zero to 975 kN and then reduced back to zero, (b) the maximum stress in each portion of the rod, (c) the permanent deflection of C.

SOLUTION Displacement at C is δ m = 0.30 mm. The corresponding strains are

ε AC =

δm LAC

ε CB = −

=

δm LCB

0.30 mm = 1.5789 × 10−3 190 mm =−

0.30 mm = −1.5789 × 10−3 190 mm

Strains at initial yielding:

ε Y, AC = ε Y, CB = (a)

σ Y, AC E

σ Y, BC E

250 × 106 = 1.25 × 10−3 (yielding) 200 × 109 345 × 106 =− = −1.725 × 10−3 (elastic) 200 × 109 =

Forces: FAC = Aσ Y = (1750 × 10−6 )(250 × 106 ) = 437.5 × 10−3 N FCB = EAε CB = (200 × 109 )(1750 × 10−6 )(−1.5789 × 10−3 ) = −552.6 × 10−3 N

For equilibrium of element at C, FAC − FCB − P = 0 P = FAC − FCD = 437.5 × 103 + 552.6 × 103 = 990.1 × 103 N

(b)

Stresses: AC : σ AC = σ Y, AC CB : σ CB =

FCB 552.6 × 103 =− = −316 × 106 Pa −6 A 1750 × 10

= 990 kN  = 250 MPa  −316 MPa 

MECH-321 Mechanics of deformable solids

Solution for Problem set-1

PROBLEM 2.110 (Continued)

(c) Deflection and forces for unloading.

δ′ =

′ LAC PAC P′ L L ′ = − PAC ′ AC = − PAC = − CB CB ∴ PCB EA EA LAB

′ − PCB ′ = 2 PAC ′ = 990.1 × 103 N ∴ PAC ′ = 495.05 × 103 N P′ = PAC

δ′ =

(495.05 × 103 )(0.190) = 0.26874 × 10−3 m = 0.26874 mm (200 × 109 )(1750 × 10−6 )

δ p = δ m − δ ′ = 0.30 mm − 0.26874 mm

0.031mm 

MECH-321 Mechanics of deformable solids

Solution for Problem set-1

PROBLEM 2.111 Two tempered-steel bars, each 163 -in. thick, are bonded to a 12 -in. mild-steel bar. This composite bar is subjected as shown to a centric axial load of magnitude P. Both steels are elastoplastic with E = 29 × 106 and with yield strengths equal to 100 ksi and 50 ksi, respectively, for the tempered and mild steel. The load P is gradually increased from zero until the deformation of the bar reaches a maximum value δ m = 0.04 in. and then decreased back to zero. Determine (a) the maximum value of P, (b) the maximum stress in the tempered-steel bars, (c) the permanent set after the load is removed.

SOLUTION 1 For the mild steel, A1 =   (2) = 1.00 in 2 2

δY1 =

Lσ Y 1 (14)(50 × 103 ) = = 0.024138 in. E 29 × 106

 3 For the tempered steel, A2 = 2   (2) = 0.75 in 2  16 

δY 2 =

Lσ Y 2 (14)(100 × 103 ) = = 0.048276 in. E 29 × 103

Total area: A = A1 + A2 = 1.75 in 2

δ Y 1 < δ m < δY 2 . The mild steel yields. Tempered steel is elastic. (a)

Forces: P1 = A1σ Y 1 = (1.00)(50 × 103 ) = 50 × 103 lb P2 =

EA2δ m (29 × 103 )(0.75)(0.04) = = 62.14 × 103 lb L 14

P = P1 + P2 = 112.14 × 103 lb = 112.1kips

(b)

Stresses: σ1 =

σ2 = Unloading: (c)

P = 112.1 kips 

P1 = σ Y 1 = 50 × 103 psi = 50 ksi A1 P2 62.14 × 103 = = 82.86 × 103 psi = 82.86 ksi A2 0.75

δ′ =

82.86 ksi 

PL (112.14 × 103 )(14) = = 0.03094 in. EA (29 × 106 )(1.75)

Permanent set: δ p = δ m − δ ′ = 0.04 − 0.03094 = 0.00906 in.

0.00906in. 

MECH-321 Mechanics of deformable solids

Solution for Problem set-1

PROBLEM 2.112 For the composite bar of Prob. 2.111, if P is gradually increased from zero to 98 kips and then decreased back to zero, determine (a) the maximum deformation of the bar, (b) the maximum stress in the tempered-steel bars, (c) the permanent set after the load is removed. PROBLEM 2.111 Two tempered-steel bars, each 163 -in. thick, are bonded to a 12 -in. mild-steel bar. This composite bar is subjected as shown to a centric axial load of magnitude P. Both steels are elastoplastic with E = 29 × 106 psi and with yield strengths equal to 100 ksi and 50 ksi, respectively, for the tempered and mild steel.

SOLUTION Areas:

Mild steel:

1 A1=   (2) = 1.00 in 2 2

Tempered steel:

 3 A2 = 2   (2) = 0.75 in 2  16 

A = A1+ A2 = 1.75 in 2

Total: Total force to yield the mild steel:

σY1 =

PY ∴ PY = Aσ Y 1 = (1.75)(50 × 103 ) = 87.50 × 103 lb A

P > PY , therefore, mild steel yields.

Let P1 = force carried by mild steel. P2 = force carried by tempered steel. P1 = A1σ1 = (1.00)(50 × 103 ) = 50 × 103 lb P1 + P2 = P, P2 = P − P1 = 98 × 103 − 50 × 103 = 48 × 103 lb

(a)

δm =

P2 L (48 × 103 )(14) = EA2 (29 × 106 )(0.75)

(b)

σ2 =

P2 48 × 103 = = 64 × 103 psi A2 0.75

Unloading: δ ′ = (c)

= 0.03090in.  = 64 ksi 

PL (98 × 103 )(14) = = 0.02703 in. EA (29 × 106 )(1.75)

δ P = δ m − δ ′ = 0.03090 − 0.02703

= 0.00387 in. 

MECH-321 Mechanics of deformable solids

Solution for Problem set-1

PROBLEM 2.113 The rigid bar ABC is supported by two links, AD and BE, of uniform 37.5 × 6-mm rectangular cross section and made of a mild steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 250 MPa. The magnitude of the force Q applied at B is gradually increased from zero to 260 kN. Knowing that a = 0.640 m, determine (a) the value of the normal stress in each link, (b) the maximum deflection of point B.

SOLUTION ΣM C = 0 : 0.640(Q − PBE ) − 2.64 PAD = 0

Statics:

δ A = 2.64θ , δ B = aθ = 0.640θ

Deformation: Elastic analysis:

A = (37.5)(6) = 225 mm 2 = 225 × 10−6 m 2 PAD =

σ AD PBE

σ BE

EA (200 × 109 )(225 × 10−6 ) δA = δ A = 26.47 × 106 δ A LAD 1.7

= (26.47 × 106 )(2.64θ ) = 69.88 × 106 θ P = AD = 310.6 × 109 θ A (200 × 109 )(225 × 10−6 ) EA = δB = δ B = 45 × 106 δ B 1.0 LBE = (45 × 106 )(0.640θ ) = 28.80 × 106 θ P = BE = 128 × 109 θ A

From Statics, Q = PBE +

2.64 PAD = PBE + 4.125PAD 0.640

= [28.80 × 106 + (4.125)(69.88 × 106 ] θ = 317.06 × 106 θ

θY at yielding of link AD:

σ AD = σ Y = 250 × 106 = 310.6 × 109 θ θY = 804.89 × 10−6 QY = (317.06 × 106 )(804.89 × 10−6 ) = 255.2 × 103 N

MECH-321 Mechanics of deformable solids

Solution for Problem set-1

PROBLEM 2.113 (Continued)

(a)

Since Q = 260 × 103 > QY , link AD yields.

σ AD = 250 MPa 

PAD = Aσ Y = (225 × 10−6 )(250 × 10−6 ) = 56.25 × 103 N

From Statics, PBE = Q − 4.125PAD = 260 × 103 − (4.125)(56.25 × 103 ) PBE = 27.97 × 103 N

σ BE = (b)

δB =

PBE 27.97 × 103 = = 124.3 × 106 Pa −6 A 225 × 10

PBE LBE (27.97 × 103 )(1.0) = = 621.53 × 10−6 m EA (200 × 109 )(225 × 10−6 )

σ BE = 124.3 MPa  δ B = 0.622 mm ↓ 

MECH-321 Mechanics of deformable solids

Solution for Problem set-1

PROBLEM 2.114 Solve Prob. 2.113, knowing that a = 1.76 m and that the magnitude of the force Q applied at B is gradually increased from zero to 135 kN. PROBLEM 2.113 The rigid bar ABC is supported by two links, AD and BE, of uniform 37.5 × 6-mm rectangular cross section and made of a mild steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 250 MPa. The magnitude of the force Q applied at B is gradually increased from zero to 260 kN. Knowing that a = 0.640 m, determine (a) the value of the normal stress in each link, (b) the maximum deflection of point B.

SOLUTION ΣM C = 0 : 1.76(Q − PBE ) − 2.64 PAD = 0

Statics:

δ A = 2.64θ , δ B = 1.76θ

Deformation: Elastic Analysis:

A = (37.5)(6) = 225 mm 2 = 225 × 10−6 m 2 PAD =

σ AD PBE

σ BE

EA (200 × 109 )(225 × 10−6 ) δA = δ A = 26.47 × 106 δ A LAD 1.7

= (26.47 × 106 )(2.64θ ) = 69.88 × 106 θ P = AD = 310.6 × 109 θ A (200 × 109 )(225 × 10−6 ) EA = δB = δ B = 45 × 106 δ B 1.0 LBE = (45 × 106 )(1.76θ ) = 79.2 × 106 θ P = BE = 352 × 109 θ A

From Statics, Q = PBE +

2.64 PAD = PBE + 1.500 PAD 1.76

= [73.8 × 106 + (1.500)(69.88 × 106 ] θ = 178.62 × 106 θ

θY at yielding of link BE:

σ BE = σ Y = 250 × 106 = 352 × 109 θY θY = 710.23 × 10−6 QY = (178.62 × 106 )(710.23 × 10−6 ) = 126.86 × 103 N

Since Q = 135 × 103 N > QY , link BE yields. PBE = Aσ Y = (225 × 10−6 )(250 × 106 ) = 56.25 × 103 N



σ BE = σ Y = 250 MPa 

MECH-321 Mechanics of deformable solids

Solution for Problem set-1

PROBLEM 2.114 (Continued)

From Statics, PAD =

1 (Q − PBE ) = 52.5 × 103 N 1.500

(a) From elastic analysis of AD, (b)

σ AD =

PAD 52.5 × 103 = = 233.3 × 106 −6 A 225 × 10

θ=

PAD = 751.29 × 10−3 rad 69.88 × 106

δ B = 1.76θ = 1.322 × 10−3 m

σ AD = 233 MPa 

δ B = 1.322 mm ↓ 