Problem 12-12

Determine the deflection at B of the bar in Prob. 12-11.

Problem 12-13

The fence board weaves between the three smooth fixed posts. If the posts remain along the same line,determine the maximum bending stress in the board. The board has a width of 150 mm. and athickness of 12 mm. E = 12 GPa. Assume the displacement of each end of the board relative to itscenter is 75 mm.

Given: b 150mm:= t 12mm:= L 2.4m:=

E 12GPa:= δ 75mm:=

Solution:

Support Reactions : By symmetry, RA=RC=R

+ ΣFy=0; 2R P− 0= R 0.5P=Moment Function :

M x( ) R x⋅= M x( ) 0.5 P⋅ x⋅=Mmax R 0.5L( )⋅= Mmax 0.25P L⋅=

Section Property :c

t2

:= Ib t3⋅12

:=

Slope and Elastic Curve :

E I⋅d2 v⋅

dx2⋅ M x( )= E I⋅

d2v

dx2⋅

P x⋅2

=

E I⋅dvdx⋅

P x2⋅4

C1+= (1)

E I⋅ v⋅P x3⋅12

C1 x⋅+ C2+= (2)

Boundary Conditions : Due to symmetry, dv/dx=0 at x=L/2. Also, v=0 at x=0.

From Eq. (1): 0P4

L2

⎛⎜⎝

⎞⎠

2⋅ C1+= C1

P L2⋅16

−=

From Eq. (2): 0 0 0+ C2+= C2 0:=

The Elastic Curve : Substitute the values of C1 and C2 into Eq. (2),

E I⋅ v⋅P x3⋅12

P L2⋅16

x⋅−= vP x⋅

48E I⋅4x2 3L2−( )⋅= (3)

Require at x=L/2, v=-δ. From Eq. (3),

δ−P

48E I⋅L2

⎛⎜⎝

⎞⎠

⋅ 4L2

⎛⎜⎝

⎞⎠

23L2−

⎡⎢⎣

⎤⎥⎦

⋅= δ−P− L3⋅

48 E⋅ I⋅= P

48 E⋅ I⋅

L3δ⋅:=

Bending Stress: Mmax 0.25P L⋅:=

cmaxt2

:= σmaxMmax cmax⋅

I:= σmax 11.25 MPa= Ans

Problem 12-14

Determine the equation of the elastic curve for the beam using the x coordinate. Specify the slope at Aand the maximum deflection. EI is constant.

Problem 12-15

Determine the deflection at the center of the beam and the slope at B. EI is constant.

Problem 12-16

A torque wrench is used to tighten the nut on a bolt. If the dial indicates that a torque of 90 N·m isapplied when the bolt is fully tightened, determine the force P acting at the handle and the distance s theneedle moves along the scale. Assume only the portion AB of the beam distorts. The cross section issquare having dimensions of 12 mm by 12 mm. E = 200 GPa.

Given: b 12mm:= h 12mm:= L 0.45m:=

E 200GPa:= δ 75mm:= R 0.3m:=

Tz 90N m⋅:=

Solution:

Equations of Equilibrium : +

ΣFy=0; Ay P− 0= (1)

ΣΜB=0; Tz P L⋅− 0= (2)

Solving Eqs. (1) and (2): PTzL

:= Ay P:= Ay 200 N=

P 200 N= Ans

Moment Function : M x( ) Ay x⋅ Tz−=

Section Property : Ib h3⋅12

:=

Slope and Elastic Curve :

E I⋅d2 v⋅

dx2⋅ M x( )=

E I⋅d2v

dx2⋅ Ay x⋅ Tz−=

E I⋅dvdx⋅

Ay x2⋅

2Tz x⋅− C1+= (1)

E I⋅ v⋅Ay x3⋅

6

Tz x2⋅

2− C1 x⋅+ C2+= (2)

Boundary Conditions : Due to symmetry, dv/dx=0 at x=0, and v=0 at x=0.

From Eq. (1): 0 0 0− C1+= C1 0:=

From Eq. (2): 0 0 0− 0+ C2+= C2 0:=

The Elastic Curve : Substitute the values of C1 and C2 into Eq. (2),

E I⋅ v⋅Ay x3⋅

6

Tz x2⋅

2−= v

x2

6E I⋅Ay x 3Tz−( )⋅= (3)

At x=R, v=-s. From Eq. (3),

sR2−

6E I⋅Ay R 3Tz−( )⋅:= s 9.11 mm= Ans

Problem 12-17

The shaft is supported at A by a journal bearing that exerts only vertical reactions on the shaft and at Bby a thrust bearing that exerts horizontal and vertical reactions on the shaft. Draw the bending-momentdiagram for the shaft and then, from this diagram, sketch the deflection or elastic curve for the shaft'scenterline. Determine the equations of the elastic curve using the coordinates x1 and x2 . EI is constant.

Given: h 60mm:= L1 150mm:=

P 5kN:= L2 400mm:=

Solution: L L1 L2+:=

Support Reactions:+

ΣFy=0; A B+ 0= (1)

ΣΜB=0; P h⋅ A L2⋅+ 0= (2)

Solving Eqs. (1) and (2): AP− h⋅L2

:= A 0.75− kN=

B A−:= B 0.75 kN=

Moment Function : M1 x1( ) P h⋅:=

M2 x2( ) B x2⋅:=

Section Property : EI kN m2⋅:=

Slope and Elastic Curve :

EId2 v1⋅

dx12

⋅ M1 x1( )= EId2 v2⋅

dx22

⋅ M2 x2( )=

EId2v1

dx12

⋅ P h⋅= EId2v2

dx22

⋅ B x2⋅=

EIdv1dx1⋅ P h⋅( ) x1⋅ C1+= (1) EI

dv2dx2⋅

B2

x22⋅ C3+= (3)

EI v1⋅P h⋅ x1

2⋅

2C1 x1⋅+ C2+= (2) EI v2⋅

B x23⋅

6C3 x2⋅+ C4+= (4)

Boundary Conditions :

v1=0 at x1=0.15m, From Eq. (2): 0P h⋅ 0.15m( )2⋅

2C1 0.15m( )⋅+ C2+= (5)

v2=0 at x2=0, From Eq. (4): 0 0 0+ C4+= C4 0:= Ans

v2=0 at x2=0.4m, From Eq. (4): 0B 0.4m( )3⋅

6C3 0.4m( )⋅+ C4+=

C3B6

− 0.4m( )2⋅:= C3 0.02− kN m2⋅= Ans

Continuity Condition:dv1/dx1= - dv2/dx2 at A (x1=0.15m and x2=0.4m)

From Eqs. (1) and (3), P h⋅ 0.15m( )⋅ C1+B2

0.4m( )2⋅ C3+=

C1B2

0.4m( )2⋅ C3+ P h⋅ 0.15m( )⋅−:= C1 0.005− kN m2⋅= Ans

From Eq. (5): C2P h⋅ 0.15m( )2⋅

2− C1 0.15m( )⋅−:= C2 0.00263− kN m3⋅= Ans

The Elastic Curve : Substitute the values of C1 and C2 into Eq. (2), and C3 and C4 into Eq. (4),

v11EI

P h⋅2

x12⋅ C1 x1⋅+ C2+⎛⎜

⎝⎞⎠

= Ans

v21EI

B6

x23⋅ C3 x2⋅+ C4+⎛⎜

⎝⎞⎠

= Ans

BMD :

x'1 0 0.01 L1⋅, L1..:= x'2 L1 1.01 L1⋅, L..:=

M'1 x'1( ) P h⋅kN m⋅

:= M'2 x'2( ) P h⋅ A x'2 L1−( )⋅+⎡⎣ ⎤⎦1

kN m⋅⋅:=

0 0.2 0.40

0.2

0.4

Distane (m)

Mom

ent (

kN-m

)

M'1 x'1( )M'2 x'2( )

x'1 x'2,

Problem 12-18

Determine the equations of the elastic curve using the coordinates x1 and x2 , and specify the slope anddeflection at C. EI is constant.

Problem 12-19

Determine the equations of the elastic curve using the coordinates x1 and x2 , and specify the slope atA. EI is constant.

Problem 12-20

Determine the equations of the elastic curve using the coordinates x1 and x2 , and specify the slope anddeflection at B. EI is constant.

Problem 12-21

Determine the equations of the elastic curve using the coordinates x1 and x3 , and specify the slope anddeflection at B. EI is constant.

Problem 12-22

Determine the maximum slope and maximum deflection of the simply-supported beam which issubjected to the couple moment M0 . EI is constant.

Problem 12-23

The two wooden meter sticks are separated at their centers by a smooth rigid cylinder having adiameter of 50 mm. Determine the force F that must be applied at each end in order to just make theirends touch. Each stick has a width of 20 mm and a thickness of 5 mm. Ew = 11 GPa.Given: do 50mm:= L 0.5m:=

b 20mm:= t 5mm:= E 11GPa:=

Solution:

Section Property: Ib t3⋅12

:=

Moment Function : M x( ) F− x⋅=Slope and Elastic Curve :

E I⋅d2 v⋅

dx2⋅ M x( )=

E I⋅d2v

dx2⋅ F− x⋅=

E I⋅dvdx⋅

F x2⋅2

− C1+= (1)

E I⋅ v⋅F x3⋅

6− C1 x⋅+ C2+= (2)

Boundary Conditions :

dv/dx=0 at x=L. From Eq. (1): 0F L2⋅

2− C1+= C1

F L2⋅2

:=

v=0 at x=L. From Eq. (2): 0F L3⋅

6− C1 L⋅+ C2+=

C2F L3⋅

6C1 L⋅−= C2

F L3⋅3

−:=

Require : v 0.5− do:= at x=0.

From Eq. (2): 0.5− do E⋅ I⋅ 0− 0F L3⋅

3−+=

F1.5do E⋅ I⋅

L3:=

F 1.375 N= Ans

Problem 12-24

The pipe can be assumed roller supported at its ends and by a rigid saddle C at its center. The saddlerests on a cable that is connected to the supports. Determine the force that should be developed in thecable if the saddle keeps the pipe from sagging or deflecting at its center. The pipe and fluid within ithave a combined weight of 2 kN/m. EI is constant.

Given:e 0.3m:= L 3.75m:= w

2kNm

:=

Solution:

Moment Function : M x( ) P x⋅12

w⋅ x2⋅−=Slope and Elastic Curve :

E I⋅d2 v⋅

dx2⋅ M x( )=

E I⋅d2v

dx2⋅ P x⋅

w2

x2⋅−=

E I⋅dvdx⋅

P x2⋅2

w6

x3⋅− C1+= (1)

E I⋅ v⋅P x3⋅

6w24

x4⋅− C1 x⋅+ C2+= (2)

Boundary Conditions : v=0 at x=0 and at x=L.

From Eq. (2): 0 0 0− 0+ C2+= C2 0:=

0P L3⋅

6w24

L4⋅− C1 L⋅+= (3)

Also, dv/dx=0 at x=L.

From Eq. (1): 0P L2⋅

2w6

L3⋅− C1+= (4)

Solving Eqs. (3) and (4) for P,

P L2⋅6

w24

L3⋅− C1+P L2⋅

2w6

L3⋅− C1+=

P L2⋅3

w8

L3⋅= P3w L⋅

8:= P 2.813 kN=

Equations of Equilibrium :

+ ΣFy=0; 2P F+ w 2L( )⋅− 0=F 2w L⋅ 2P−:= F 9375 N=

+At C : ΣFy=0; 2Tcablee

e2 L2+⋅ F− 0= Tcable

e2 L2+2e

F⋅:=

Tcable 58.78 kN= Ans

Problem 12-25

Determine the equations of the elastic curve using the coordinates x1 and x2 , and specify the slope at Cand displacement at B. EI is constant.

Problem 12-26

Determine the equations of the elastic curve using the coordinates x1 and x3 , and specify the slope at Band deflection at C. EI is constant.

Problem 12-27

Determine the elastic curve for the simply supported beam using the x coordinate . Also,determine the slope at A and the maximum deflection of the beam. EI is constant.

2/0 Lx ≤≤

Problem 12-28

Determine the elastic curve for the cantilevered beam using the x coordinate. Also determine themaximum slope and maximum deflection. EI is constant.