# soluções resistência dos materiais hibbeler_ 7ª edição

Post on 12-Apr-2017

1.188 views

Embed Size (px)

TRANSCRIPT

Stamp

CD14/13 US ISM for

Problem 1-1

Determine the resultant internal normal force acting on the cross section through point A in eachcolumn. In (a), segment BC weighs 300 kg/m and segment CD weighs 400kg/m. In (b), the columnhas a mass of 200 kg/m.

a( ) Given: g 9.81m

s2:= wBC 300

kgm

:=

LBC 3m:= wCA 400kgm

:=

FB 5kN:= LCA 1.2m:=

FC 3kN:=

Solution:

+ Fy = 0; FA wBC g( ) LBC wCA g( ) LCA FB 2FC 0=

FA wBC g( ) LBC wCA g( ) LCA+ FB+ 2FC+:=

FA 24.5 kN= Ans

b( ) Given: g 9.81m

s2:= w 200

kgm

:=

L 3m:= F1 6kN:=

FB 8kN:= F2 4.5kN:=

Solution:

+ Fy = 0; FA w L( ) g FB 2F1 2F2 0=

FA w L( ) g FB+ 2F1+ 2F2+:=

FA 34.89 kN= Ans

Problem 1-2

Determine the resultant internal torque acting on the cross sections through points C and D of theshaft. The shaft is fixed at B.

Given: TA 250N m:=

TCD 400N m:=

TDB 300N m:=

Solution:

Equations of equilibrium:

+ TA TC 0=

TC TA:=

TC 250 N m= Ans

+ TA TCD TD+ 0=

TD TCD TA:=

TD 150 N m= Ans

Problem 1-3

Determine the resultant internal torque acting on the cross sections through points B and C.

Given: TD 500N m:=

TBC 350N m:=

TAB 600N m:=

Solution:

Equations of equilibrium:

Mx = 0; TB TBC TD+ 0=

TB TBC TD+:=

TB 150 N m= Ans

Mx = 0; TC TD 0=

TC TD:=

TC 500 N m= Ans

Problem 1-4

A force of 80 N is supported by the bracket as shown. Determine the resultant internal loadings actingon the section through point A.

Given: P 80N:=

30deg:= 45deg:=

a 0.3m:= b 0.1m:=

Solution:

Equations of equilibrium:

+ Fx'=0; NA P cos ( ) 0=

NA P cos ( ):=

NA 77.27 N= Ans

+ Fy'=0; VA P sin ( ) 0=

VA P sin ( ):=

VA 20.71 N= Ans

+ A=0; MA P cos ( ) a cos ( )+ P sin ( ) b a sin ( )+( ) 0=

MA P cos ( ) a cos ( ) P sin ( ) b a sin ( )+( )+:=

MA 0.555 N m= Ans

Note: Negative sign indicates that MA acts in the opposite direction to that shown on FBD.

Problem 1-5

Determine the resultant internal loadings acting on the cross section through point D of member AB.

Given: ME 70N m:=

a 0.05m:= b 0.3m:=Solution:

Segment AB: Support Reactions

+ A=0; ME By 2 a b+( ) 0=

ByME

2a b+:= By 175 N=

+At B: Bx By150200

:= Bx 131.25 N=

Segment DB: NB Bx:= VB By:=

+ Fx=0; ND NB+ 0=

ND NB:= ND 131.25 N= Ans

+ Fy=0; VD VB+ 0=

VD VB:= VD 175 N= Ans

+ D=0; MD ME By a b+( ) 0=

MD ME By a b+( ):=

MD 8.75 N m= Ans

Problem 1-6

The beam AB is pin supported at A and supported by a cable BC. Determine the resultant internalloadings acting on the cross section at point D.

Given: P 5000N:=

a 0.8m:= b 1.2m:= c 0.6m:= d 1.6m:=

e 0.6m:=

Solution:

atanbd

:= 36.87 deg=

atana b+

d

:= 14.47 deg=

Member AB:

+ A=0; FBC sin ( ) a b+( ) P b( ) 0=

FBCP b( )

sin ( ) a b+( ):=

FBC 12.01 kN=

Segment BD: + Fx=0; ND FBC cos ( ) P cos ( ) 0=

ND FBC cos ( ) P cos ( ):=

ND 15.63 kN= Ans

+ Fy=0; VD FBC sin ( )+ P sin ( ) 0=

VD FBC sin ( ) P sin ( )+:=

VD 0 kN= Ans

+ D=0; FBC sin ( ) P sin ( )( ) d csin ( ) MD 0=

MD FBC sin ( ) P sin ( )( ) d csin ( ):=

MD 0 kN m= Ans

Note: Member AB is the two-force member. Therefore the shear force and moment are zero.

Problem 1-7

Solve Prob. 1-6 for the resultant internal loadings acting at point E.

Given: P 5000N:=

a 0.8m:= b 1.2m:= c 0.6m:= d 1.6m:=

e 0.6m:=

Solution:

atanbd

:= 36.87 deg=

atana b+

d

:= 14.47 deg=

Member AB:

+ A=0; FBC sin ( ) a b+( ) P b( ) 0=

FBCP b( )

sin ( ) a b+( ):=

FBC 12.01 kN=

Segment BE:

+ Fx=0; NE FBC cos ( ) P cos ( ) 0=

NE FBC cos ( ) P cos ( ):=

NE 15.63 kN= Ans

+ Fy=0; VE FBC sin ( )+ P sin ( ) 0=

VE FBC sin ( ) P sin ( )+:=VE 0 kN= Ans

+ E=0; FBC sin ( ) P sin ( )( ) e ME 0=

ME FBC sin ( ) P sin ( )( ) e:=

ME 0 kN m= Ans

Note: Member AB is the two-force member. Therefore the shear force and moment are zero.

Problem 1-8

The boom DF of the jib crane and the column DE have a uniform weight of 750 N/m. If the hoist andload weigh 1500 N, determine the resultant internal loadings in the crane on cross sections throughpoints A, B, and C.

Given: P 1500N:= w 750Nm

:=

a 2.1m:= b 1.5m:=

c 0.6m:= d 2.4m:= e 0.9m:=

Solution:

Equations of Equilibrium: For point A

+ Fx=0; NA 0:= Ans

+VA w e P 0=Fy=0;VA w e P+:= VA 2.17 kN= Ans

+ A=0; MA w e( ) 0.5 e( ) P e( ) 0=

MA w e( ) 0.5 e( ) P e( ):= MA 1.654 kN m= Ans

Note: Negative sign indicates that MA acts in the opposite direction to that shown on FBD.

Equations of Equilibrium: For point B

+ Fx=0; NB 0:= Ans

+VB w d e+( ) P 0=Fy=0;VB w d e+( ) P+:= VB 3.98 kN= Ans

+ B=0; MB w d e+( )[ ] 0.5 d e+( )[ ] P d e+( ) 0=

MB w d e+( )[ ] 0.5 d e+( )[ ] P d e+( ):=

MB 9.034 kN m= Ans

Note: Negative sign indicates that MB acts in the opposite direction to that shown on FBD.

Equations of Equilibrium: For point C

+ Fx=0; VC 0:= Ans

+NC w b c+ d+ e+( ) P 0=Fy=0;

NC w b c+ d+ e+( ) P:=

NC 5.55 kN= Ans

+ B=0; MC w c d+ e+( )[ ] 0.5 c d+ e+( )[ ] P c d+ e+( ) 0=MC w c d+ e+( )[ ] 0.5 c d+ e+( )[ ] P c d+ e+( ):=

MC 11.554 kN m= Ans

Note: Negative sign indicates that NC and MC acts in the opposite direction to that shownon FBD.

Problem 1-9

The force F = 400 N acts on the gear tooth. Determine the resultant internal loadings on the root of thetooth, i.e., at the centroid point A of section a-a.

Given: P 400N:=

30deg:= 45deg:=

a 4mm:= b 5.75mm:=

Solution: :=Equations of equilibrium: For section a -a

+ Fx'=0; VA P cos ( ) 0=

VA P cos ( ):=

VA 386.37 N= Ans

+ Fy'=0; NA sin ( ) 0=

NA P sin ( ):=

NA 103.53 N= Ans

+ A=0; MA P sin ( ) a P cos ( ) b+ 0=

MA P sin ( ) a P cos ( ) b+:=

MA 1.808 N m= Ans

Problem 1-10

The beam supports the distributed load shown. Determine the resultant internal loadings on the crosssection through point C. Assume the reactions at the supports A and B are vertical.

Given: w1 4.5kNm

:= w2 6.0kNm

:=

a 1.8m:= b 1.8m:= c 2.4m:=

d 1.35m:= e 1.35m:=

Solution: L1 a b+ c+:=

L2 d e+:=

Support Reactions:

+ A=0; By L1 w1 L1( ) 0.5 L1( ) 0.5w2 L2( ) L1L23

+

0=

By w1 L1( ) 0.5( ) 0.5w2 L2( ) 1L2

3 L1+

+:= By 22.82 kN=

+ Fy=0; Ay By+ w1 L1 0.5w2 L2 0=

Ay By w1 L1+ 0.5w2 L2+:= Ay 12.29 kN=

Equations of Equilibrium: For point C

+ Fx=0; NC 0:= Ans

+ Fy=0; Ay w1 a b+( ) VC 0=

VC Ay w1 a b+( ) := VC 3.92 kN= Ans

+ C=0; MC w1 a b+( ) 0.5 a b+( )+ Ay a b+( ) 0=

MC w1 a b+( ) 0.5 a b+( ) Ay a b+( )+:=

MC 15.07 kN m= Ans

Note: Negative sign indicates that VC acts in the opposite direction to that shown on FBD.

Problem 1-11

The beam supports the distributed load shown. Determine the resultant internal loadings on the crosssections through points D and E. Assume the reactions at the supports A and B are vertical.

Given: w1 4.5kNm

:= w2 6.0kNm

:=

a 1.8m:= b 1.8m:= c 2.4m:=

d 1.35m:= e 1.35m:=

Solution: L1 a b+ c+:=

L2 d e+:=

Support Reactions:

+ A=0; By L1 w1 L1( ) 0.5 L1( ) 0.5w2 L2( ) L1L23

+

0=

By w1 L1( ) 0.5( ) 0.5w2 L2( ) 1L2

3 L1+

+:= By 22.82 kN=

+ Fy=0; Ay By+ w1 L1 0.5w2 L2 0=

Ay By w1 L1+ 0.5w2 L2+:= Ay 12.29 kN=

Equations of Equilibrium: For point D

+ Fx=0; ND 0:= Ans

+ Fy=0; Ay w1 a( ) VD 0=

VD Ay w1 a( ):= VD 4.18 kN= Ans

+ D=0; MD w1 a( ) 0.5 a( )+ Ay a( ) 0=

MD w1 a( ) 0.5 a( ) Ay a( )+:= MD 14.823 kN m= Ans

Equations of Equilibrium: For point E

+ Fx=0; NE 0:= Ans

+ Fy=0; VE 0.5w2 0.5 e( ) 0=

VE 0.5w2 0.5 e( ):= VE 2.03 kN= Ans

+ D=0; ME 0.5w2 0.5 e( ) e3

0=

ME 0.5 w2 0.5 e( ) e3

:= ME 0.911 kN m= Ans

Note: Negative sign indicates that ME acts in the opposite direction to that shown on FBD.

Problem 1-12

Determine the resultant internal loadings acting on (a) section a-a and (b) section b-b. Each section islocated through the centroid, point C.

Given: w 9kNm

:= 45deg:=

a 1.2m:= b 2.4m:=

Solution: L a b+:=

Support Reactions:

+ A=0; Bx L sin ( ) w L( ) 0.5 L( )+ 0=

Bxw L( ) 0.5 L( )

L sin ( ):= Bx 22.91 kN=

+ Fy=0; Ay w L sin ( ) 0=

Ay w L sin ( ):= Ay 22.91 kN=

+ Fx=0; Bx w L cos ( ) Ax+ 0=

Ax w L cos ( ) Bx:= Ax 0 kN=

(a) Equations of equilibrium: For Section a - a :

+ Fx=0; NC Ay sin ( )+ 0=

NC Ay sin ( ):=

+ Fy=0; VC Ay cos ( )+ w a 0=

VC Ay cos ( ) w a+:= VC 5.4 kN= Ans

+ A=0; MC w a( ) 0.5 a( ) Ay cos ( ) a+ 0=

MC w a( ) 0.5 a( ) Ay cos ( ) a:= MC 12.96 kN m= Ans

(b) Equations of equilibrium: For Section b - b :

+ Fx=0; NC w a cos ( )+ 0=

NC w a cos ( ):= NC 7.64 kN= Ans

+ Fy=0; VC w a sin ( ) Ay+ 0=

VC w a sin ( ) Ay:= VC 15.27 kN= Ans

+ A=0; MC w a( ) 0.5 a( ) Ay cos ( ) a+ 0=

MC w a( ) 0.5 a( ) Ay cos ( ) a:= MC 12.96 kN m= Ans

NC 16.2 kN= Ans

Problem 1-13

Determine the resultant internal normal and shear forces in the member at (a) section a-a and (b)section b-b, each of which passes through point A. Take = 60 degree. The 650-N load is appliedalong the centroidal axis of the member.

Given: P 650N:= 60deg:=

(a) Equations of equilibrium: For Section a - a :

+ Fy=0; P Na_a 0=

Na_a P:=

Na_a 650 N= Ans

+ Fx=0; Va_a 0:= Ans

(b) Equations of equilibrium: For Section b - b :

+ Fy=0; Vb_b P cos 90deg ( )+ 0=

Vb_b P cos 90deg ( ):=

Vb_b 562.92 N= Ans

+ Fx=0; Nb_b P sin 90deg ( ) 0=

Nb_b P sin 90deg ( ):=

Nb_b 325 N= Ans

Problem 1-14

Determine the resultant internal normal and shear forces in the member at section b-b, each as a

function of . Plot these results for 0o 90o . The 650-N load is applied along the centroidal axisof the member.

Given: P 650N:= 0:=

Equations of equilibrium: For Section b - b :

+ Fx0; Nb_b P cos ( ) 0=

Nb_b P cos ( ):= Ans

+ Fy=0; Vb_b P cos ( )+ 0=

Vb_b P cos ( ):= Ans

Problem 1-15

The 4000-N load is being hoisted at a constant speed using the motor M, which has a weight of 450 N.Determine the resultant internal loadings acting on the cross section through point B in the beam. Thebeam has a weight of 600 N/m and is fixed to the wall at A.

Given:

W1 4000N:= w 600Nm

:=

W2 450N:=

a 1.2m:= b 1.2m:= c 0.9m:=

d 0.9m:= e 1.2m:=

f 0.45m:= r 0.075m:=

Solution:

Tension in rope: TW12

:=

T 2.00 kN=

Equations of Equilibrium: For point B

+ Fx=0; NB T( ) 0=

NB T:= NB 2 kN= Ans

+ Fy=0; VB w e( ) W1 0=

VB w e( ) W1+:= VB 4.72 kN= Ans

+ B=0; MB w e( )[ ] 0.5 e( ) W1 e r+( ) T f( )+ 0=

MB w e( )[ ] 0.5 e( ) W1 e r+( ) T f( )+:=

MB 4.632 kN m= Ans

Problem 1-16

Determine the resultant internal loadings acting on the cross section through points C and D of thebeam in Prob. 1-15.

Given: W1 4000N:=w 600

Nm

:=W2 450N:=

a 1.2m:= b 1.2m:= c 0.9m:=

d 0.9m:= e 1.2m:=

f 0.45m:= r 0.075m:=

Solution:

Tension in rope: TW12

:= T 2.00 kN=

Equations of Equilibrium: For point C LC d e+:=+ Fx=0; NC T( ) 0=

NC T:= NC 2 kN= Ans

+ Fy=0; VC w LC( ) W1 0=VC w LC( ) W1+:= VC 5.26 kN= Ans

+ C=0; MC w LC( ) 0.5 LC( ) W1 LC r+( ) T f( )+ 0=

MC w LC( ) 0.5 LC( ) W1 LC r+( ) T f( )+:=MC 9.123 kN m= Ans

Equations of Equilibrium: For point D LD b c+ d+ e+:=+ Fx=0; ND 0:= ND 0 kN= Ans

+ Fy=0; VD w LD( ) W1 W2 0=VD w LD( ) W1+ W2+:= VD 6.97 kN= Ans

+ C=0; MD w LD( ) 0.5 LD( ) W1 LD r+( ) W2 b( ) 0=MD w LD( ) 0.5 LD( ) W1 LD r+( ) W2 b( ):=

MD 22.932 kN m= Ans

Problem 1-17

Determine the resultant internal loadings acting on the cross section at point B.

Given: w 900kNm

:=

a 1m:= b 4m:=

Solution: L a b+:=

Equations of Equilibrium: For point B

+ Fx=0; NB 0:=

NB 0 kN= Ans

+ Fy=0; VB 0.5 wbL

b( ) 0=

VB 0.5 wbL

b( ):=

VB 1440 kN= Ans

+ B=0; MB 0.5 wbL

b( )b3

0=

MB 0.5 wbL

b( )b3:=

MB 1920 kN m= Ans

Problem 1-18

The beam supports the distributed load shown. Determine the resultant internal loadings acting on thecross section through point C. Assume the reactions at the supports A and B are vertical.

Given: w1 0.5kNm

:= a 3m:=

w2 1.5kNm

:=

Solution: L 3 a:= w w2 w1:=

Support Reactions:

+ A=0; By L w1 L( ) 0.5 L( ) 0.5 w( ) L[ ]2L3

0=

By w1 L( ) 0.5( ) 0.5 w( ) L[ ]23

+:=

By 5.25 kN=

+ Fy=0; Ay By+ w1 L 0.5 w( ) L 0=

Ay By w1 L+ 0.5 w( ) L+:=

Ay 3.75 kN=

Equations of Equilibrium: For point C

+ Fx=0; NC 0:= NC 0 kN= Ans

+ Fy=0; VC w1 a+ 0.5 waL

a( )+ Ay 0=

VC w1 a 0.5 waL

a( ) Ay+:=

VC 1.75 kN= Ans

+ C=0; MC w1 a( ) 0.5 a( )+ 0.5 waL

a( )a3

+ Ay a 0=

MC w1 a( ) 0.5 a( ) 0.5 waL

a( )a3

Ay a+:=

MC 8.5 kN m= Ans

Problem 1-19

Determine the resultant internal loadings acting on the cross section through point D in Prob. 1-18.

Given: w1 0.5kNm

:= a 3m:=

w2 1.5kNm

:=

Solution: L 3 a:= w w2 w1:=

Support Reactions:

+ A=0; By L w1 L( ) 0.5 L( ) 0.5 w( ) L[ ]2L3

0=

By w1 L( ) 0.5( ) 0.5 w( ) L[ ]23

+:=

By 5.25 kN=

+ Fy=0; Ay By+ w1 L 0.5 w( ) L 0=

Ay By w1 L+ 0.5 w( ) L+:=

Ay 3.75 kN=

Equations of Equilibrium: For point D

+ Fx=0; ND 0:= ND 0 kN= Ans

+ Fy=0; VD w1 2a( )+ 0.5 w2aL

2a( )+ Ay 0=

VD w1 2a( ) 0.5 w2 aL

2a( ) Ay+:=

VD 1.25 kN= Ans

+ D=0; MD w1 2a( ) a( )+ 0.5 w2 aL

2a( )2a3

+ Ay 2a( ) 0=

MD w1 2a( ) a( ) 0.5 w2 aL

2a( )2a3

Ay 2a( )+:=

MD 9.5 kN m= Ans

Problem 1-20

The wishbone construction of the power pole supports the three lines, each exerting a force of 4 kNon the bracing struts. If the struts are pin connected at A, B, and C, determine the resultant internalloadings at cross sections through points D, E, and F.

Given: P 4kN:= a 1.2m:= b 1.8m:=Solution:

Support Reactions: FBD (a) and (b).

Given+ A=0; By a( ) Bx 0.5 b( )+ P a( ) 0= [1]

+ C=0; Bx 0.5 b( ) P a( )+ By a( ) P a( ) 0= [2]

Solving [1] and [2]: Initial guess: Bx 1kN:= By 2kN:=Bx

By

Find Bx By,( ):=

Bx

By

2.67

2

kN=

From FBD (a):+ Fx=0; Bx Ax 0=

Ax Bx:= Ax 2.67 kN=

+ Fy=0; Ay P By 0=Ay P By+:= Ay 6 kN=

From FBD (b):+ Fx=0; Cx Bx 0=

Cx Bx:= Cx 2.67 kN=

+ Fy=0; Cy By+ P P 0=Cy 2P By:= Cy 6 kN=

Equations of Equilibrium: For point D [FBD (c)].+ Fx=0; VD 0:= VD 0 kN= Ans

+ Fy=0; ND 0:= ND 0 kN= Ans

+ D=0; MD 0:= MD 0 kN m= Ans

For point E [FBD (d)].

+ Fx=0; VF Ax Cx+ 0= VF Ax Cx+:= VF 0 kN= Ans

+ Fy=0; NF Ay Cy 0= NF Ay Cy+:= NF 12 kN= Ans

+ F=0; MF Ax Cx+( ) 0.5 b( ) 0= MF Ax Cx+( ) 0.5 b( ):= MF 4.8 kN m= Ans

+ Fx=0; Ax VE 0= VE Ax:=

+ Fy=0; NE Ay 0= NE Ay:=

+ E=0; ME Ax 0.5 b( ) 0= ME Ax 0.5 b( ):= ME 2.4 kN m= Ans

For point F [FBD (e)].

VE 2.67 kN= Ans

NE 6 kN= Ans

Problem 1-21

The drum lifter suspends the 2.5-kN drum. The linkage is pin connected to the plate at A and B. Thegripping action on the drum chime is such that only horizontal and vertical forces are exerted on thedrum at G and H. Determine the resultant internal loadings on the cross section through point I.

Given: P 2.5 kN:= 60deg:=

a 200mm:= b 125mm:= c 75mm:=

d 125mm:= e 125mm:= f 50mm:=Solution:

Equations of Equilibrium: Memeber Ac and BD aretwo-force members.

Fy=0; P 2 F sin ( ) 0= [1]

FP

2 sin ( ):= [2]

F 1.443 kN=

Equations of Equilibrium: For point I.

+ Fx=0; VI F cos ( ) 0= VI F cos ( ):= VI 0.722 kN= Ans

+ Fy=0; NI F sin ( )+ 0= NI F sin ( ):= NI 1.25 kN= Ans

+ I=0; MI F cos ( ) a( )+ 0=

MI F cos ( ) a( ):= MI 0.144 kN m= Ans

Problem 1-22

Determine the resultant internal loadings on the cross sections through points K and J on the drum lifterin Prob. 1-21.

Given: P 2.5 kN:= 60deg:=

a 200mm:= b 125mm:= c 75mm:=

d 125mm:= e 125mm:= f 50mm:=

Solution:

Equations of Equilibrium: Memeber Ac and BD aretwo-force members.

Fy=0; P 2 F sin ( ) 0= [1]

FP

2 sin ( ):= [2]

F 1.443 kN=

Equations of Equilibrium: For point J.

+ Fy'=0; VI 0:= VI 0 kN= Ans

+ Fx'=0; NI F+ 0= NI F:= NI 1.443 kN= Ans

+ J=0; MJ 0:= MJ 0 kN m= Ans

Note: Negative sign indicates that NJ acts in the opposite direction to that shown on FBD.

Support Reactions: For Member DFH :

+ H=0; FEF c( ) F cos ( ) a b+ c+( ) F sin ( ) f( )+ 0=

FEF F cos ( )a b+ c+

c

F sin ( ) fc

:=

FEF 3.016 kN=

Equations of Equilibrium: For point K.

+ Fx=0; NK FEF+ 0= NK FEF:= NK 3.016 kN= Ans

+ Fy=0; VK 0:= VK 0 kN= Ans

+ K=0; MK 0:= MK 0 kN m= Ans

Problem 1-23

The pipe has a mass of 12 kg/m. If it is fixed to the wall at A, determine the resultant internal loadingsacting on the cross section at B. Neglect the weight of the wrench CD.

Given: g 9.81m

s2:= 12

kgm

:= P 60N:=

a 0.150m:= b 0.400m:=

c 0.200m:= d 0.300m:=

Solution: w g:=

Fx=0; NBx 0N:= Ans

Fy=0; VBy 0N:= Ans

Fz=0; VBz P P+ w b c+( ) 0=

VBz P P w b c+( )+:=

VBz 70.6 N= Ans

x=0; TBx P b( )+ P b( ) w b( ) 0.5 b( ) 0=

TBx P b( ) P b( )+ w b( ) 0.5 b( )+:= TBx 9.42 N m= Ans

y=0; MBy P 2a( ) w b( ) c( )+ w c( ) 0.5 c( )+ 0=

MBy P 2 a( ) w b( ) c( ) w c( ) 0.5 c( ):= MBy 6.23 N m= Ans

z=0; MBz 0N m:= Ans

Problem 1-24

The main beam AB supports the load on the wing of the airplane. The loads consist of the wheelreaction of 175 kN at C, the 6-kN weight of fuel in the tank of the wing, having a center of gravity atD, and the 2-kN weight of the wing, having a center of gravity at E. If it is fixed to the fuselage at A,determine the resultant internal loadings on the beam at this point. Assume that the wing does nottransfer any of the loads to the fuselage, except through the beam.

Given: PC 175kN:= PE 2kN:= PD 6kN:=

a 1.8m:= b 1.2m:= e 0.3m:=

c 0.6m:= d 0.45m:=

Solution:

Fx=0; VAx 0kN:= Ans

Fy=0; NAy 0kN:= Ans

Fz=0; VAz PD PE PC+ 0=

VAz PD PE PC+:=

VAz 167 kN= Ans

x=0;

MAx PD a( ) PE a b+ c+( )+ PC a b+( ):= MAx 507 kN m= Ans

y=0; TAy PD d( )+ PE e( ) 0=

TAy PD d( ) PE e( )+:= TAy 2.1 kN m= Ans

z=0; MAz 0kN m:= Ans

MAx PD a( ) PE a b+ c+( ) PC a b+( )+ 0=

Problem 1-25

Determine the resultant internal loadings acting on the cross section through point B of the signpost.The post is fixed to the ground and a uniform pressure of 50 N/m2 acts perpendicular to the face of thesign.

Given: a 4m:= d 2m:= p 50N

m2:=

b 6m:= e 3m:=

c 3m:=

Solution: P p c( ) d e+( ):=

Fx=0; VBx P 0=

VBx P:=

VBx 750 N= Ans

Fy=0; VBy 0N:= Ans

Fz=0; NBz 0N:= Ans

x=0; MBx 0N m:= Ans

y=0; MBy P b 0.5 c+( ) 0=

MBy P b 0.5 c+( ):=

MBy 5625 N m= Ans

z=0; TBz P e 0.5 d e+( )[ ] 0=

TBz P e 0.5 d e+( )[ ]:=

TBz 375 N m= Ans

Problem 1-26

The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to thepulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section throughpoint D. The 400-N forces act in the -z direction and the 200-N and 80-N forces act in the +ydirection. The journal bearings at A and B exert only y and z components of force on the shaft.

Given: P1z 400N:= P2y 200N:= P3y 80N:=

a 0.3m:= b 0.4m:= c 0.3m:= d 0.4m:=

Solution: L a b+ c+ d+:=Support Reactions:

z=0; 2P3y d( ) 2P2y c d+( )+ Ay L( ) 0=

Ay 2P3ydL 2P2y

c d+L

+:=

Ay 245.71 N=

Fy=0; Ay By 2 P2y+ 2 P3y+ 0=By Ay 2 P2y+ 2 P3y+:= By 314.29 N=

y=0; 2P1z b c+ d+( ) Az L( ) 0=

Az 2P1zb c+ d+

L:= Az 628.57 N=

Fz=0; Bz Az+ 2 P1z 0=Bz Az 2 P1z+:= Bz 171.43 N=

Equations of Equilibrium: For point D.

Fx=0; NDx 0N:= Ans

Fy=0; VDy By 2 P3y+ 0=

VDy By 2 P3y:=

VDy 154.3 N= Ans

Fz=0; VDz Bz+ 0=

VDz Bz:=

VDz 171.4 N= Ans

x=0; TDx 0N m:= Ans

y=0; MDy Bz d 0.5 c+( )+ 0=

MDy Bz d 0.5 c+( ) := MDy 94.29 N m= Ans

z=0; MDz By d 0.5 c+( )+ 2 P3y 0.5 c( ) 0=

MDz By d 0.5 c+( ) 2 P3y 0.5 c( )+:= MDz 148.86 N m= Ans

Problem 1-27

The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to thepulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section throughpoint C. The 400-N forces act in the -z direction and the 200-N and 80-N forces act in the +ydirection. The journal bearings at A and B exert only y and z components of force on the shaft.

Given: P1z 400N:= P2y 200N:= P3y 80N:=

a 0.3m:= b 0.4m:= c 0.3m:= d 0.4m:=Solution: L a b+ c+ d+:=

Support Reactions:

z=0; 2P3y d( ) 2P2y c d+( )+ Ay L( ) 0=

Ay 2P3ydL 2P2y

c d+L

+:= Ay 245.71 N=

Fy=0; Ay By 2 P2y+ 2 P3y+ 0=By Ay 2 P2y+ 2 P3y+:= By 314.29 N=

y=0; 2P1z b c+ d+( ) Az L( ) 0=

Az 2P1zb c+ d+

L:= Az 628.57 N=

Fz=0; Bz Az+ 2 P1z 0=Bz Az 2 P1z+:= Bz 171.43 N=

Equations of Equilibrium: For point C.

Fx=0; NCx 0N:= Ans

Fy=0; VCy Ay 0=

VCy Ay:=

VCy 245.7 N= Ans

Fz=0; VCz Az+ 2P1z 0=

VCz Az 2P1z+:=

VCz 171.4 N= Ans

x=0; TCx 0N m:= Ans

y=0; MCy Az a 0.5 b+( ) 2 P1z 0.5 b( )+ 0=

MCy Az a 0.5 b+( ) 2 P1z 0.5 b( ):= MCy 154.29 N m= Ans

z=0; MCz Ay a 0.5 b+( )+ 0=

MCz Ay a 0.5 b+( ):= MCz 122.86 N m= Ans

Problem 1-28

Determine the resultant internal loadings acting on the cross section of the frame at points F and G.The contact at E is smooth.

Given: a 1.2m:= b 1.5m:= c 0.9m:= P 400N:=

d 0.9m:= e 1.2m:= 30deg:=

Solution: L d2 e2+:=

Member DEF :

+ D=0; NE b( ) P a b+( ) 0=

NE Pa b+

b:= NE 720 N=

Member BCE :

+ B=0; FACeL

d( ) NE sin ( ) c d+( ) 0=

FACL

e d

NE sin ( ) c d+( ) :=

FAC 900 N=

+ Fx=0; Bx FACdL

+ NE cos ( ) 0=

Bx FACdL

NE cos ( )+:=

Bx 83.54 N=

+ Fy=0; By FACeL

+ NE sin ( ) 0=

By FACeL

NE sin ( ):=

By 360 N=

Equations of Equilibrium: For point F.

+ Fy'=0; NF 0:= NF 0 N= Ans

+ Fx'=0; VF P 0= VF P:= VF 400 N= Ans

+ F=0; MF P 0.5 a( ) 0= MF P 0.5 a( ):= MF 240 N m= Ans

Equations of Equilibrium: For point G.+

Fx=0; Bx NG 0= NG Bx:= NG 83.54 N= Ans

+ Fy=0; VG By 0= VG By:= VG 360 N= Ans

+ G=0; MG By 0.5 d( )+ 0= MG By 0.5 d( ):= MG 162 N m= Ans

Problem 1-29

The bolt shank is subjected to a tension of 400 N. Determine the resultant internal loadings acting onthe cross section at point C.

Given:

P 400N:=

r 150mm:=

90deg:=

Solution:

Equations of Equilibrium: For segment AC.

+Fx=0; NC P+ 0= NC P:= NC 400 N= Ans

+ Fy=0; VC 0:= VC 0 N= Ans

+ G=0; MC P r( )+ 0= MC P r( ):= MC 60 N m= Ans

Problem 1-30

The pipe has a mass of 12 kg/m. If it is fixed to the wall at A, determine the resultant internal loadingsacting on the cross section through B.

Given: P 750N:= MC 800N m:=

12kgm

:= g 9.81m

s2:=

a 1m:= b 2m:= c 2m:=

Solution:Py

45

P:= Pz35

P:=

Equations of Equilibrium: For point B.

Fx=0; VBx 0kip:= Ans

Fy=0; NBy Py+ 0=

NBy Py:= Ans

NBy 600 N= Ans

Fz=0; VBz Pz+ g c g b 0=

VBz Pz g c+ g b+:=

VBz 920.9 N= Ans

x=0; MBx Pz b( )+ g c b( ) g b 0.5 b( ) 0=

MBx Pz b( ) g c b( )+ g b 0.5 b( )+:=

MBx 1606.3 N m= Ans

y=0; TBy 0N m:= Ans

z=0; MBy Mc+ 0=

MBy MC:=

MBy 800 N m= Ans

Problem 1-31

The curved rod has a radius r and is fixed to the wall at B. Determine the resultant internal loadingsacting on the cross section through A which is located at an angle from the horizontal.

Solution: P kN:= deg:=

Equations of Equilibrium: For point A.

+ Fx=0; NA P cos ( )+ 0=

NA P cos ( ):= Ans

+ Fy=0; VA P sin ( ) 0=

VA P sin ( ):= Ans

+ A=0; MA P r 1 cos ( )( ) 0=

MA P r 1 cos ( )( ):= r Ans

Problem 1-32

The curved rod AD of radius r has a weight per length of w. If it lies in the horizontal plane, determinethe resultant internal loadings acting on the cross section through point B. Hint: The distance from thecentroid C of segment AB to point O is CO = 0.9745r.

Given: 22.5deg:= r m:= a 0.9745r:= wkN

m2:=

Solution:

Equations of Equilibrium: For point B.

Fz=0; VB4

r w 0= VB 0.785 w r:= Ans

Fx=0; NB 0:= Ans

x=0; TB4

r w 0.09968r( ) 0= TB 0.0783w r2:= Ans

y=0; MB4

r w 0.37293r( )+ 0= MB 0.293 w r2:= Ans

Problem 1-33

A differential element taken from a curved bar is shown in the figure. Show that dN/d = V, dV/d = -N, dM/d = -T, and dT/d = M,

Solution:

Problem 1-34

The column is subjected to an axial force of 8 kN, which is applied through the centroid of thecross-sectional area. Determine the average normal stress acting at section aa. Show this distributionof stress acting over the areas cross section.

Given: P 8kN:=

b 150mm:= d 140mm:= t 10mm:=

Solution:

A 2 b t( ) d t+:= A 4400.00 mm2=

PA

:= 1.82 MPa= Ans

Problem 1-35

The anchor shackle supports a cable force of 3.0 kN. If the pin has a diameter of 6 mm, determine theaverage shear stress in the pin.

Given: P 3.0kN:= d 6mm:=Solution:

+ Fy=0; 2 V P 0=

V 0.5P:=

V 1500 N=

A d2

4:= A 28.2743 mm2=

avgVA

:= avg 53.05 MPa= Ans

Problem 1-36

While running the foot of a 75-kg man is momentarily subjected to a force which is 5 times hisweight. Determine the average normal stress developed in the tibia T of his leg at the mid section a-a.The cross section can be assumed circular, having an outer diameter of 45 mm and an inner diameterof 25 mm. Assume the fibula F does not support a load.

Given: g 9.81m

s2=

M 75kg:=

do 45mm:= di 25mm:=

Solution:

A4

do2 di

2:= A 1099.5574 mm

2=

5M g

A:= 3.345 MPa= Ans

Problem 1-37

The thrust bearing is subjected to the loads shown. Determine the average normal stress developed oncross sections through points B, C, and D. Sketch the results on a differential volume element locatedat each section.

Units Used: kPa 103Pa:=

Given: P 500N:= Q 200N:=

dB 65mm:= dC 140mm:= dD 100mm:=

Solution:

AB dB

2

4:= AB 3318.3 mm

2=

BP

AB:= B 150.7 kPa= Ans

AC dC

2

4:= AC 15393.8 mm

2=

CP

AC:= C 32.5 kPa= Ans

AD dD

2

4:= AD 7854.0 mm

2=

DQ

AD:= D 25.5 kPa= Ans

Problem 1-38

The small block has a thickness of 5 mm. If the stress distribution at the support developed by the loadvaries as shown, determine the force F applied to the block, and the distance d to where it is applied.

Given: a 60mm:= b 120mm:= t 5mm:=

1 0MPa:= 2 40MPa:= 3 60MPa:=

Solution:

F A

d=

F 0.5 2 a t( ) 2 b t( )+ 0.5 3 2( ) b t( )+:=F 36.00 kN= Ans

Require:

F d Ax

d=

d0.5 2 a t( )

2a3

2 b t( ) a 0.5 b+( )+ 0.5 3 2( ) b t( ) a2 b3

+

+

F:=

d 110 mm= Ans

Problem 1-39

The lever is held to the fixed shaft using a tapered pin AB, which has a mean diameter of 6 mm. If acouple is applied to the lever, determine the average shear stress in the pin between the pin and lever.

Given: a 250mm:= b 12mm:=

d 6mm:= P 20N:=

Solution:

+ O=0; V b P 2a( ) 0=

V P2ab

:=

V 833.33 N=

A d2

4:= A 28.2743 mm2=

avgVA

:= avg 29.47 MPa= Ans

Problem 1-40

The cinder block has the dimensions shown. If the material fails when the average normal stressreaches 0.840 MPa, determine the largest centrally applied vertical load P it can support.

Given: allow 0.840MPa:=

ao 150mm:= ai 100mm:=

bo 2 1 2+ 3+( ) 2+[ ] mm:=

bi 2 1 3+( )[ ] mm:=

Solution:

A ao bo ai bi:= A 1300 mm2=

Pallow allow A( ):=

Pallow 1.092 kN= Ans

Problem 1-41

The cinder block has the dimensions shown. If it is subjected to a centrally applied force of P = 4 kN,determine the average normal stress in the material. Show the result acting on a differential volumeelement of the material.

Given: P 4kN:=

ao 150mm:= ai 100mm:=

bo 2 1 2+ 3+( ) 2+[ ] mm:=

bi 2 1 3+( )[ ] mm:=

Solution:

A ao bo ai bi:= A 1300 mm2=

PA

:=

3.08 MPa= Ans

Problem 1-42

The 250-N lamp is supported by three steel rods connected by a ring at A. Determine which rod issubjected to the greater average normal stress and compute its value. Take = 30. The diameter ofeach rod is given in the figure.

Given: W 250N:= 30deg:= 45deg:=

dB 9mm:= dC 6mm:= dD 7.5mm:=

Solution: Initial guess: FAC 1N:= FAD 1N:=

Given+ Fx=0; FAC cos ( ) FAD cos ( ) 0= [1]

+ Fy=0; FAC sin ( ) FAD sin ( )+ W 0= [2]

Solving [1] and [2]:FAC

FAD

Find FAC FAD,( ):=

FAC

FAD

183.01

224.14

N=

Rod AB:

AAB dB

2

4:= AAB 63.61725 mm

2=

ABW

AAB:= AB 3.93 MPa=

Rod AD :

AAD dD

2

4:= AAD 44.17865 mm

2=

ADFADAAD

:= AD 5.074 MPa=

Rod AC:

AAC dC

2

4:= AAC 28.27433 mm

2=

ACFACAAC

:= AC 6.473 MPa= Ans

Problem 1-43

Solve Prob. 1-42 for = 45.

Given: W 250N:= 45deg:= 45deg:=

dB 9mm:= dC 6mm:= dD 7.5mm:=

Solution: Initial guess: FAC 1N:= FAD 1N:=

Given+ Fx=0; FAC cos ( ) FAD cos ( ) 0= [1]

+ Fy=0; FAC sin ( ) FAD sin ( )+ W 0= [2]

Solving [1] and [2]:FAC

FAD

Find FAC FAD,( ):=

FAC

FAD

176.78

176.78

N=

Rod AB:

AAB dB

2

4:= AAB 63.61725 mm

2=

ABW

AAB:= AB 3.93 MPa=

Rod AD :

AAD dD

2

4:= AAD 44.17865 mm

2=

ADFADAAD

:= AD 4.001 MPa=

Rod AC:

AAC dC

2

4:= AAC 28.27433 mm

2=

ACFACAAC

:= AC 6.252 MPa= Ans

Problem 1-44

The 250-N lamp is supported by three steel rods connected by a ring at A. Determine the angle oforientation of AC such that the average normal stress in rod AC is twice the average normal stress inrod AD. What is the magnitude of stress in each rod? The diameter of each rod is given in the figure.Given: W 250N:= 45deg:=

dB 9mm:= dC 6mm:= dD 7.5mm:=

Solution:

Rod AB: AAB dB

2

4:= AAB 63.61725 mm

2=

Rod AD : AAD dD

2

4:= AAD 44.17865 mm

2=

Rod AC: AAC dC

2

4:= AAC 28.27433 mm

2=

Since AC 2AD= ThereforeFACAAC

2FADAAD

=

Initial guess: FAC 1N:= FAD 2N:= 30deg:=

Given FACAAC

2FADAAD

= [1]

+ Fx=0; FAC cos ( ) FAD cos ( ) 0= [2]

+ Fy=0; FAC sin ( ) FAD sin ( )+ W 0= [3]

Solving [1], [2] and [3]:

FAC

FAD

Find FAC FAD, ,( ):=FAC

FAD

180.38

140.92

N=

56.47 deg=

ABW

AAB:= AB 3.93 MPa= Ans

ADFADAAD

:= AD 3.19 MPa= Ans

ACFACAAC

:= AC 6.38 MPa= Ans

Problem 1-45

The shaft is subjected to the axial force of 30 kN. If the shaft passes through the 53-mm diameter holein the fixed support A, determine the bearing stress acting on the collar C. Also, what is the averageshear stress acting along the inside surface of the collar where it is fixed connected tothe 52-mm diameter shaft?

Given: P 30kN:=

dhole 53mm:= dshaft 52mm:=

dcollar 60mm:= hcollar 10mm:=

Solution:

Bearing Stress:

Ab4

dcollar2 dhole

2:=

bP

Ab:= b 48.3 MPa= Ans

Average Shear Stress:

As dshaft( ) hcollar( ):=

avgPAs

:= avg 18.4 MPa= Ans

Problem 1-46

The two steel members are joined together using a 60 scarf weld. Determine the average normal andaverage shear stress resisted in the plane of the weld.

Given:

P 8kN:= 60deg:=

b 25mm:= h 30mm:=

Solution:

Equations of Equilibrium:

+ Fx=0; N P sin ( ) 0=

N P sin ( ):= N 6.928 kN=

+ Fy=0; V P cos ( ) 0=

V P cos ( ):= V 4 kN=

Ah b

sin ( ):=

NA

:= 8 MPa= Ans

avgVA

:= avg 4.62 MPa= Ans

Problem 1-47

The J hanger is used to support the pipe such that the force on the vertical bolt is 775 N. Determine theaverage normal stress developed in the bolt BC if the bolt has a diameter of 8 mm. Assume A is a pin.

Given: P 775N:=

a 40mm:= b 30mm:=

d 8mm:= 20deg:=

Solution:

Support Reaction:

FA=0; P a( ) FBC cos ( ) a b+( ) 0=

FBCP a

a b+( ) cos ( ):=

FBC 471.28 N=

Average Normal Stress:

ABC d2

4:=

FBCABC

:=

9.38 MPa= Ans

Problem 1-48

The board is subjected to a tensile force of 425 N. Determine the average normal and average shearstress developed in the wood fibers that are oriented along section a-a at 15 with the axis of the board.

Given: P 425N:= 15deg:=

b 25mm:= h 75mm:=

Solution:

Equations of Equilibrium:

+ Fx=0; V P cos ( ) 0=

V P cos ( ):= V 410.518 N=

+ Fy=0; N P sin ( ) 0=

N P sin ( ):= N 1 N=

Average Normal Stress:

Ah b

sin ( ):=

NA

:= 0.0152 MPa= Ans

avgVA

:= avg 0.0567 MPa= Ans

Problem 1-49

The open square butt joint is used to transmit a force of 250 kN from one plate to the other. Determinethe average normal and average shear stress components that this loading creates on the face of theweld, section AB.

Given: P 250kN:= 30deg:=

b 150mm:= h 50mm:=

Solution:

Equations of Equilibrium:

+ Fx=0; V P sin ( )+ 0=

V P sin ( ):= V 125 kN=

+ Fy=0; N P cos ( ) 0=

N P cos ( ):= N 216.506 kN=

Average Normal and Shear Stress:

Ah b

sin 2( ):=

NA

:= 25 MPa= Ans

avgVA

:= avg 14.434 MPa= Ans

Problem 1-50

The specimen failed in a tension test at an angle of 52 when the axial load was 100 kN. If the diameterof the specimen is 12 mm, determine the average normal and average shear stress acting on the area ofthe inclined failure plane. Also, what is the average normal stress acting on the cross section whenfailure occurs?

Given: P 100kN:=

d 12mm:= 52deg:=

Solution:

Equations of Equilibrium:

+ Fx=0; V P cos ( ) 0=

V P cos ( ):= V 61.566 kN=

+ Fy=0; N P sin ( ) 0=

N P sin ( ):= N 78.801 kN=

Inclined plane:

A4

d2

sin ( )

:=

NA

:= 549.05 MPa= Ans

avgVA

:= avg 428.96 MPa= Ans

Cross section:

Ad2

4:=

PA

:= 884.19 MPa= Ans

avg 0:= avg 0 MPa= Ans

Problem 1-51

A tension specimen having a cross-sectional area A is subjected to an axial force P. Determine themaximum average shear stress in the specimen and indicate the orientation of a section on which itoccurs.

Solution:

Equations of Equilibrium:

+ Fy=0; V P cos ( ) 0=

V P cos ( )=

Inclined plane:

AinclA

sin ( )=

V

Aincl=

P cos ( ) sin ( )A

= P sin 2( )

2A=

dd

P cos 2( )A

=dd

0=

cos 2( ) 0=

2 90deg=

45deg:= Ans

maxP sin 90( )

2A= max

P2A

= Ans

Problem 1-52

The joint is subjected to the axial member force of 5 kN. Determine the average normal stress actingon sections AB and BC. Assume the member is smooth and is 50-mm thick.

Given: P 5kN:= 45deg:= 60deg:=

dAB 40mm:= dBC 50mm:= t 50mm:=

Solution: 90deg := 30.00 deg=

AAB t dAB:=

ABC t dBC:=

+ Fx=0; NAB cos ( ) P cos ( ) 0=

NABP cos ( )cos ( )

:=

NAB 4.082 kN=

+ Fy=0; NAB sin ( ) P sin ( )+ NBC 0=

NBC NAB sin ( ) P sin ( )+:=

NBC 1.494 kN=

ABNABAAB

:= AB 2.041 MPa= Ans

BCNBCABC

:= BC 0.598 MPa= Ans

Problem 1-53

The yoke is subjected to the force and couple moment. Determine the average shear stress in the boltacting on the cross sections through A and B. The bolt has a diameter of 6 mm. Hint: The couplemoment is resisted by a set of couple forces developed in the shank of the bolt.

Given: P 2.5kN:= M 120N m:=

ho 62mm:= hi 50mm:=

d 6mm:= 60deg:=

Solution:

As a force on bolt shank is zero, then

A 0:= Ans

Equations od Equilibrium:

Fz=0; P 2Fz 0=

Fz 0.5P:= Fz 1.25 kN=

Mz=0; M Fx hi( ) 0=

FxMhi

:= Fx 2.4 kN=

Average Shear Stress: A

d2

4:=

The bolt shank subjected to a shear force of VB Fx2 Fz

2+:=

BVBA

:= B 95.71 MPa= Ans

Problem 1-54

The two members used in the construction of an aircraft fuselage are joined together using a 30fish-mouth weld. Determine the average normal and average shear stress on the plane of each weld.Assume each inclined plane supports a horizontal force of 2 kN.

Given:

P 4.0kN:=

b 37.5mm:= hhalf 25m:=

30deg:=

Solution:

Equations of Equilibrium:

+ Fx=0; V 0.5P cos ( )+ 0=

V 0.5P cos ( ):= V 1.732 kN=

+ Fy=0; N 0.5P sin ( ) 0=

N 0.5P sin ( ):= N 1 kN=

Average Normal and Shear Stress:

Ahhalf( ) bsin ( )

:=

NA

:= 533.33 Pa= Ans

avgVA

:= avg 923.76 Pa= Ans

Problem 1-55

The row of staples AB contained in the stapler is glued together so that the maximum shear stress theglue can withstand is max = 84 kPa. Determine the minimum force F that must be placed on theplunger in order to shear off a staple from its row and allow it to exit undeformed through the grooveat C. The outer dimensions of the staple are shown in the figure. It has a thickness of 1.25 mmAssume all the other parts are rigid and neglect friction.

Given: max 0.084MPa:=

a 12.5mm:= b 7.5mm:=

t 1.25mm:=Solution:

Average Shear Stress:

A a b a 2t( ) b t( )[ ]:=

maxVA

= V max( ) A:=

V 2.63 N=

Fmin V:=

Fmin 2.63 N= Ans

Problem 1-56

Rods AB and BC have diameters of 4mm and 6 mm, respectively. If the load of 8 kN is applied to thering at B, determine the average normal stress in each rod if = 60.

Given: W 8kN:= 60deg:=

dA 4mm:= dC 6mm:=

Solution:

Rod AB: AAB dA

2

4:=

Rod BC : ABC dC

2

4:=

+ Fy=0; FBC sin ( ) W 0=

FBCW

sin ( ):=

FBC 9.238 kN=

+ Fx=0; FBC cos ( ) FAB 0=

FAB FBC cos ( ):=

FAB 4.619 kN=

ABFABAAB

:= AB 367.6 MPa= Ans

BCFBCABC

:= BC 326.7 MPa= Ans

Problem 1-57

Rods AB and BC have diameters of 4 mm and 6 mm, respectively. If the vertical load of 8 kN isapplied to the ring at B, determine the angle of rod BC so that the average normal stress in each rodis equivalent. What is this stress?

Given: W 8kN:=

dA 4mm:= dC 6mm:=

Solution:

Rod AB: AAB dA

2

4:=

Rod BC : ABC dC

2

4:=

+ Fy=0; FBC sin ( ) W 0=

+ Fx=0; FBC cos ( ) FAB 0=

Since FAB AAB=

FBC ABC=

Initial guess: 100MPa:= 50deg:=

Given ABC sin ( ) W 0= [1]

ABC cos ( ) AAB 0= [2]

Solving [1] and [2]:

Find ,( ):=

63.61 deg= Ans

315.85 MPa= Ans

Problem 1-58

The bars of the truss each have a cross-sectional area of 780 mm2. Determine the average normalstress in each member due to the loading P = 40 kN. State whether the stress is tensile or compressive.

Given: P 40kN:=

a 0.9m:= b 1.2m:= A 780mm2:=

Solution: c a2 b2+:= c 1.5 m=

hbc

:= vac

:=

Joint A:

+ Fy=0; v( ) FAB P 0= FABPv

:= FAB 66.667 kN=

+ Fx=0; h( ) FAB FAE 0= FAE h( ) FAB:= FAE 53.333 kN=

ABFAB

A:= AB 85.47 MPa= (T) Ans

AEFAE

A:= AE 68.376 MPa= (C) Ans

Joint E:

+ Fy=0; FEB 0.75P 0= FEB 0.75P:= FEB 30 kN=

+ Fx=0; FED FAE 0= FED FAE:= FED 53.333 kN=

EBFEB

A:= EB 38.462 MPa= (T) Ans

EDFED

A:= ED 68.376 MPa= (C) Ans

Joint B:

+ Fy=0; v( ) FBD v( ) FAB FEB 0= FBD FABFEB

v

+:= FBD 116.667 kN=

+ Fx=0; FBC h( )FAB h( )FBD 0= FBC h( )FAB h( )FBD+:= FBC 146.667 kN=

BCFBC

A:= BC 188.034 MPa= (T) Ans

BDFBD

A:= BD 149.573 MPa= (C) Ans

Problem 1-59

The bars of the truss each have a cross-sectional area of 780 mm2. If the maximum average normalstress in any bar is not to exceed 140 MPa, determine the maximum magnitude P of the loads that canbe applied to the truss.

allow 140MPa:=Given:

a 0.9m:= b 1.2m:= A 780mm2:=

Solution: c a2 b2+:= c 1.5 m=

hbc

:= vac

:=

For comparison purpose, set P 1kN:=

Joint A:

+ Fy=0; v( ) FAB P 0= FABPv

:= FAB 1.667 kN=

+ Fx=0; h( ) FAB FAE 0= FAE h( ) FAB:= FAE 1.333 kN=

ABFAB

A:= AB 2.137 MPa= (T)

AEFAE

A:= AE 1.709 MPa= (C)

Joint E:

+ Fy=0; FEB 0.75P 0= FEB 0.75P:= FEB 0.75 kN=

+ Fx=0; FED FAE 0= FED FAE:= FED 1.333 kN=

EBFEB

A:= EB 0.962 MPa= (T)

EDFED

A:= ED 1.709 MPa= (C)

Joint B:

+ Fy=0; v( ) FBD v( ) FAB FEB 0= FBD FABFEB

v

+:= FBD 2.917 kN=+ Fx=0; FBC h( )FAB h( )FBD 0= FBC h( )FAB h( )FBD+:= FBC 3.667 kN=

BCFBC

A:= BC 4.701 MPa= (T)

BDFBD

A:= BD 3.739 MPa= (C)

Since the cross-sectional areas are the same, the highest stress occurs in the member BC,which has the greatest force

Fmax max FAB FAE, FEB, FED, FBD, FBC,( ):= Fmax 3.667 kN=

PallowP

Fmax

allow A( ):= Pallow 29.78 kN= Ans

Problem 1-60

The plug is used to close the end of the cylindrical tube that is subjected to an internal pressure of p =650 Pa. Determine the average shear stress which the glue exerts on the sides of the tube needed tohold the cap in place.

Given: p 650Pa:= a 25mm:=

di 35mm:= do 40mm:=

Solution:

Ap di

2

4:= As do( ) a( ):=

P a( ) FBC cos ( ) a b+( ) 0=

P p Ap( ):= P 0.625 N=

Average Shear Stress:

avgPAs

:=

avg 199.1 Pa= Ans

Problem 1-61

The crimping tool is used to crimp the end of the wire E. If a force of 100 N is applied to the handles,determine the average shear stress in the pin at A. The pin is subjected to double shear and has adiameter of 5 mm. Only a vertical force is exerted on the wire.

Given: P 100N:=

a 37.5mm:= b 50mm:= c 25mm:=

d 125mm:= dpin 5mm:=

Solution:

From FBD (a):+ Fx=0; Bx 0:=

Bx 0 N=

+ D=0; P d( ) By c( ) 0=

By Pdc:=

By 500 N=

From FBD (b):+ Fx=0; Ax 0:=

Ax 0 N=

+ E=0; Ay a( ) By a b+( ) 0=

Ay Bya b+

a:=

Ay 1166.67 N=

Average Shear Stress:

Apin dpin

2

4:=

VA 0.5 Ay( ):= VA 583.333 N=

avgVA

Apin:= avg 29.709 MPa= Ans

Problem 1-62

Solve Prob. 1-61 for pin B. The pin is subjected to double shear and has a diameter of 5 mm.

Given: a 37.5mm:= b 50mm:= c 25mm:=

d 125mm:= dpin 5mm:= P 100N:=Solution:

From FBD (a):+ Fx=0; Bx 0:=

Bx 0 N=

+ D=0; P d( ) By c( ) 0=

By Pdc:=

By 500 N=

Average Shear Stress: Pin B is subjected to doule shear

Apin dpin

2

4:=

VB 0.5 By( ):= VB 250 N=

avgVB

Apin:= avg 12.732 MPa= Ans

Problem 1-63

The railcar docklight is supported by the 3-mm-diameter pin at A. If the lamp weighs 20 N, and theextension arm AB has a weight of 8 N/m, determine the average shear stress in the pin needed tosupport the lamp. Hint: The shear force in the pin is caused by the couple moment required forequilibrium at A.

Given: w 8Nm

:= P 20N:=

a 900mm:= h 32mm:=

dpin 3mm:=

Solution:

From FBD (a):+ Fx=0; Bx 0:=

+ A=0; V h( ) w a( ) 0.5a( ) P a( ) 0=

V w a( ) 0.5ah

Pah+:= V 663.75 N=

Average Shear Stress:

Apin dpin

2

4:=

avgV

Apin:= avg 93.901 MPa= Ans

Problem 1-64

The two-member frame is subjected to the distributed loading shown. Determine the average normalstress and average shear stress acting at sections a-a and b-b. Member CB has a square cross sectionof 35 mm on each side. Take w = 8 kN/m.

Given: w 8kNm

:=

a 3m:= b 4m:= A 0.0352( )m2:=Solution: c a2 b2+:= c 5 m=

hac

:= vbc

:=

Member AB:

MA=0; By a( ) w a( ) 0.5a( ) 0=

By 0.5w a:= By 12 kN=

+ Fy=0; v( ) FAB By 0=

FABByv

:= FAB 15 kN=

Section a-a:

a_aFAB

A:= a_a 12.24 MPa= Ans

a_a 0:= a_a 0 MPa= Ans

Section b-b:

+ Fx=0; N FAB h( ) 0= N FAB h( ):= N 9 kN=

+ Fy=0; V FAB v( ) 0= V FAB v( ):= V 12 kN=

Ab_bAh

:=

b_bN

Ab_b:= b_b 4.41 MPa= Ans

b_bV

Ab_b:= b_b 5.88 MPa= Ans

Problem 1-65

Member A of the timber step joint for a truss is subjected to a compressive force of 5 kN. Determinethe average normal stress acting in the hanger rod C which has a diameter of 10 mm and in member Bwhich has a thickness of 30 mm.

Given: P 5kN:= 60deg:= 30deg:=

drod 10mm:= h 40mm:= t 30mm:=

Solution:AB t h:=

Arod4

drod2:=

+ Fx=0; P cos ( ) FB 0=FB P cos ( ):= FB 2.5 kN=

+ Fy=0; Fc P sin ( ) 0=FC P sin ( ):= FC 4.33 kN=

Average Normal Stress:

BFBAB

:= B 2.083 MPa= Ans

CFC

Arod:= C 55.133 MPa= Ans

Problem 1-66

Consider the general problem of a bar made from m segments, each having a constant cross-sectionalarea Am and length Lm. If there are n loads on the bar as shown, write a computer program that can beused to determine the average normal stress at any specified location x. Show an application of theprogram using the values L1 = 1.2 m, d1 = 0.6 m, P1 = 2 kN, A1 = 1875 mm2, L2 = 0.6 m, d2 = 1.8 m,

P2 = -1.5 kN, A2 = 625 mm2.

Problem 1-67

The beam is supported by a pin at A and a short link BC. If P = 15 kN, determine the average shearstress developed in the pins at A, B, and C. All pins are in double shear as shown, and each has adiameter of 18 mm.

Given: P 15kN:=

a 0.5m:= b 1m:= c 1.5m:=

d 1.5m:= e 0.5m:=

30deg:= dpin 18mm:=

Solution: L a b+ c+ d+ e+:=Support Reactions: A=0; By L( ) P L a( )+ 4P c d+ e+( )+ 4P d e+( )+ 2P e( )+ 0=

By PL a

L 4 P

c d+ e+L

+ 4 Pd e+

L+ 2 P

eL+:=

By 82.5 kN=

+ Fy=0; By P+ 4 P+ 4 P+ 2 P+ Ay 0=

Ay By P+ 4 P+ 4P+ 2 P+:=

Ay 82.5 kN=

FBCBy

sin ( ):= FBC 165 kN=

Ax FBC cos ( ):= Ax 142.89 kN=

Average Shear Stress:

Apin dpin

2

4:=

For Pins B and C:

B_and_C0.5FBC

Apin:= B_and_C 324.2 MPa= Ans

For Pin A:

FA Ax2 Ay

2+:= FA 165 kN=

A0.5FAApin

:= A 324.2 MPa= Ans

Problem 1-68

The beam is supported by a pin at A and a short link BC. Determine the maximum magnitude P of theloads the beam will support if the average shear stress in each pin is not to exceed 80 MPa. All pins arein double shear as shown, and each has a diameter of 18 mm.

Given: allow 80MPa:=

a 0.5m:= b 1m:= c 1.5m:=

d 1.5m:= e 0.5m:=

30deg:= dpin 18mm:=

Solution: L a b+ c+ d+ e+:=

For comparison purpose, set P 1kN:=Support Reactions:

A=0; By L( ) P L a( )+ 4P c d+ e+( )+ 4P d e+( )+ 2P e( )+ 0=

By PL a

L 4 P

c d+ e+L

+ 4 Pd e+

L+ 2 P

eL+:= By 5.5 kN=

+ Fy=0; By P+ 4 P+ 4 P+ 2 P+ Ay 0=

Ay By P+ 4 P+ 4P+ 2 P+:= Ay 5.5 kN=

FBCBy

sin ( ):= FBC 11 kN=

Ax FBC cos ( ):= Ax 9.53 kN=

FA Ax2 Ay

2+:= FA 11 kN=

Require:

Fmax max FBC FA,( ):= Fmax 11 kN=

Apin dpin

2

4:=

PallowP

Fmax

allow 2Apin( ) := Pallow 3.70 kN= Ans

Problem 1-69

The frame is subjected to the load of 1 kN. Determine the average shear stress in the bolt at A as a

function of the bar angle . Plot this function, 0 90o , and indicate the values of for which thisstress is a minimum. The bolt has a diameter of 6 mm and is subjected to single shear.

Given: P 1kN:= dbolt 6mm:=

a 0.6m:= b 0.45m:= c 0.15m:=

Solution:

Support Reactions:

C=0; FAB cos ( ) c( ) FAB sin ( ) a( )+ P a b+( ) 0=

FABP a b+( )

cos ( ) c( ) sin ( ) a( )+=

Average Shear Stress: Pin B is subjected to doule shear

FABAbolt

= Abolt dbolt

2

4:=

4P a b+( )

cos ( ) c( ) sin ( ) a( )+ dbolt2

=

dd

4P a b+( )

dbolt2

sin ( ) c( ) cos ( ) a( )cos ( ) c( ) sin ( ) a( )+ 2

=

dd

0= sin ( ) c( ) cos ( ) a( ) 0= tan ( ) ac

=

atanac

:=

75.96 deg= Ans

Problem 1-70

The jib crane is pinned at A and supports a chain hoist that can travel along the bottom flange of thebeam, 1ft x 12ft . If the hoist is rated to support a maximum of 7.5 kN, determine the maximumaverage normal stress in the 18-mm-diameter tie rod BC and the maximum average shear stress in the16-mm-diameter pin at B.

Given: P 7.5kN:= xmax 3.6m:=

a 3m:= 30deg:=

drod 18mm:= dpin 16mm:=

Solution:

Support Reactions:

C=0; FBC sin ( ) a( ) P x( ) 0=

FBCP x( )

sin ( ) a( )=

Maximum FBC occurs when x= xmax . Therefore,

FBCP xmax( )sin ( ) a( )

:= FBC 18.00 kN=

Arod drod

2

4:= Apin

dpin2

4:=

pin0.5 FBC

Apin:= pin 44.762 MPa= Ans

rodFBCArod

:= rod 70.736 MPa= Ans

Problem 1-71

The bar has a cross-sectional area A and is subjected to the axial load P. Determine the average normaland average shear stresses acting over the shaded section, which is oriented at from the horizontal.

Plot the variation of these stresses as a function of (0o 90o ).

Solution:

Equations of Equilibrium:

+ Fx=0; V P cos ( ) 0=

V P cos ( )=

+ Fy=0; N P sin ( ) 0=

N P sin ( )=

Inclined plane:

AA

sin ( )=

NA

= PA

sin ( )2= Ans

avgVA

= avgP

2Asin 2( )= Ans

Problem 1-72

The boom has a uniform weight of 3 kN and is hoisted into position using the cable BC. If the cable hasa diameter of 15 mm, plot the average normal stress in the cable as a function of the boom position for

0o 90o .

Given: W 3kN:=

a 1m:=

do 15mm:=

Solution: Angle B: B 0.5 90deg +( )=

B 45deg 0.5+=Support Reactions:

A=0; FBC sin B( ) a( ) W 0.5a( ) cos ( ) 0=

FBC0.5W cos ( )

sin 45deg 0.5+( )=

Average Normal Stress:

BCFABABC

= ABC do

2

4:=

BC2W

do2

cos ( )sin 45deg 0.5+( )= Ans

Problem 1-73

The bar has a cross-sectional area of 400 (10-6) m2. If it is subjected to a uniform axial distributedloading along its length and to two concentrated loads as shown, determine the average normal stress inthe bar as a function of for 0 x< 0.5m .

Given: P1 3kN:= P2 6kN:=

w 8kNm

:= A 400 10 6( ) m2:=a 0.5m:= b 0.75m:=

Solution: L a b+:=

+ Fx=0; N P1+ P2+ w L x( )+ 0=

N P1 P2+ w L x( )+=

Average Normal Stress:

NA

=

P1 P2+ w L x( )+

A= Ans

Problem 1-74

The bar has a cross-sectional area of 400 (10-6) m2. If it is subjected to a uniform axial distributedloading along its length and to two concentrated loads as shown, determine the average normal stress inthe bar as a function of for 0.5m x< 1.25m .

Given: P1 3kN:= P2 6kN:=

w 8kNm

:= A 400 10 6( ) m2:=a 0.5m:= b 0.75m:=

Solution: L a b+:=

+ Fx=0; N P1+ w L x( )+ 0=

N P1 w L x( )+=

Average Normal Stress:

NA

=

P1 w L x( )+

A= Ans

Problem 1-75

The column is made of concrete having a density of 2.30 Mg/m3. At its top B it is subjected to an axialcompressive force of 15 kN. Determine the average normal stress in the column as a function of thedistance z measured from its base. Note: The result will be useful only for finding the average normalstress at a section removed from the ends of the column, because of localized deformation at the ends.

Given: P 3kN:= 2.3 103( ) kgm3

:= g 9.81m

s2:=

r 180mm:= h 0.75m:=

Solution:A r2:= w g A:=

+ Fz=0; N P w h z( ) 0=

N P w h z( )+=

Average Normal Stress:

NA

=

P w h z( )+

A= Ans

Problem 1-76

The two-member frame is subjected to the distributed loading shown. Determine the largestintensity of the uniform loading that can be applied to the frame without causing either the averagenormal stress or the average shear stress at section b-b to exceed = 15 MPa, and = 16 MParespectively. Member CB has a square cross-section of 30 mm on each side.Given: allow 15MPa:= allow 16MPa:=

a 4m:= b 3m:= A 0.0302( )m2:=Solution: c a2 b2+:= c 5 m=

vbc

:=hac

:=

Set wo 1kNm

:=

Member AB:

MA=0; By b( ) wo b( ) 0.5b( ) 0=By 0.5wo b:=

By 1.5 kN=Section b-b:

By FBC h( )= FBCByh

:=

FBC 1.88 kN=

+ Fx=0; FBC h( ) Vb_b 0= Vb_b FBC h( ):=

Vb_b 1.5 kN=

+ Fy=0; Nb_b FBC v( )+ 0= Nb_b FBC v( ):=

Nb_b 1.125 kN=Ab_b

Av

:=

b_bNb_bAb_b

:= b_b 0.75 MPa=

b_bVb_bAb_b

:= b_b 1 MPa=

Assume failure due to normal stress: wallow woallowb_b

:= wallow 20.00

kNm

=

Assume failure due to shear stress: wallow woallowb_b

:= wallow 16.00

kNm

= Ans

Controls !

Problem 1-77

The pedestal supports a load P at its center. If the material has a mass density , determine the radialdimension r as a function of z so that the average normal stress in the pedestal remains constant. Thecross section is circular.

Solution:

Require: P w1+

A=

P w1+ dw+

A dA+=

P dA w1 dA+ A dw=

dwdA

P w1+

A=

dwdA

= [1]

dA r dr+( )2 r2=

dA 2r dr=

dw r2 g( ) dz=

From Eq.[1],r2 g( ) dz

2r dr=

r g( ) dz2dr

=

g2 0

zz1( )

d

r1

r

r1r

d=

g z2

lnrr1

= r r1 e

g2

z=

However,

P

r12

=

Ansr r1 e

r12

g

2P

z

=

Problem 1-78

The radius of the pedestal is defined by r = (0.5e-0.08y2) m, where y is given in meters. If the materialhas a density of 2.5 Mg/m3, determine the average normal stress at the support.

Given:ro 0.5m:= h 3m:= g 9.81

m

s2=

r ro e0.08 y2m= 2.5 103( ) kg

m3:=

yunit 1m:=

Solution:

dr e 0.08 y2

dy=

Ao ro2:= Ao 0.7854 m

2=

dV r2( ) dy=

dV ro2 e 0.08 y

2

2

dy=

V0

3

yro2 e 0.08 y

2

2

yunit( )

d:=

V 1.584 m3=

W g V:=

W 38.835 kN=

WAo

:=

0.04945 MPa= Ans

Problem 1-79

The uniform bar, having a cross-sectional area of A and mass per unit length of m, is pinned at itscenter. If it is rotating in the horizontal plane at a constant angular rate of , determine the averagenormal stress in the bar as a function of x.

Solution:

Equation of Motion :

+ Fx=MaN=M r2 ;

N mL2

x

x12

L2

x

+

=

Nm

8L2 4 x2( )=

Average Normal Stress:

NA

=

m 8A

L2 4 x2( )= Ans

Problem 1-80

Member B is subjected to a compressive force of 4 kN. If A and B are both made of wood and are10mm. thick, determine to the nearest multiples of 5mm the smallest dimension h of the support so thatthe average shear stress does not exceed allow = 2.1 MPa.Given: P 4kN:=

t 10mm:= allow 2.1MPa:=

a 300mm:= b 125mm:=

Solution: c a2 b2+:= c 325 mm=

hac

:= vbc

:=

V P v( ):= V 1.54 kN=

allowVt h

=

hV

t allow:= h 73.26 mm=

Use h 75mm:= h 75 mm= Ans

Problem 1-81

The joint is fastened together using two bolts. Determine the required diameter of the bolts if the failureshear stress for the bolts is fail = 350 MPa. Use a factor of safety for shear of F.S. = 2.5.Given: P 80kN:= fail 350MPa:= 2.5:=

Solution:

allowfail

:= allow 140 MPa=

Vbolt 0.5P2

:= Vbolt 20 kN=

AboltVboltallow

=4

d2Vboltallow

=

d4

Vboltallow

:=

d 13.49 mm= Ans

Problem 1-82

The rods AB and CD are made of steel having a failure tensile stress of fail = 510 MPa. Using a factorof safety of F.S. = 1.75 for tension, determine their smallest diameter so that they can support the loadshown. The beam is assumed to be pin connected at A and C.

Given: P1 4kN:= P2 6kN:= P3 5kN:=

a 2m:= b 2m:= c 3m:= d 3m:=

1.75:= fail 510MPa:=

Solution: L a b+ c+ d+:=

Support Reactions:

A=0; FCD L( ) P1 a( ) P2 a b+( ) P3 a b+ c+( ) 0=

FCD P1aL

P2a b+

L+ P3

a b+ c+L

+:= FCD 6.70 kN=

C=0; FAB L( ) P1 b c+ d+( )+ P2 c d+( )+ P3 d( )+ 0=

FAB P1b c+ d+

L P2

c d+L

+ P3dL

+:= FAB 8.30 kN=

Average Normal Stress: Design of rod sizes

allowfail

:= allow 291.43 MPa=

For Rod AB

AboltFABallow

=4

dAB2

FABallow

=

dAB4

FABallow

:= dAB 6.02 mm= Ans

For Rod CD

AboltFCDallow

=4

dCD2

FCDallow

=

dCD4

FCDallow

:= dCD 5.41 mm= Ans

Problem 1-83

The lever is attached to the shaft A using a key that has a width d and length of 25 mm. If the shaft isfixed and a vertical force of 200 N is applied perpendicular to the handle, determine the dimension d ifthe allowable shear stress for the key is allow = 35 MPa.

Given: P 200N:= allow 35MPa:=

L 500mm:= a 20mm:=

b 25mm:=

Solution:

A=0; Fa_a a( ) P L( ) 0=

Fa_a PLa

:=

Fa_a 5000 N=

For the key

Aa_aFa_aallow

= b dFa_aallow

=

d1b

Fa_aallow

:=

d 5.71 mm= Ans

Problem 1-84

The fillet weld size a is determined by computing the average shear stress along the shaded plane,which has the smallest cross section. Determine the smallest size a of the two welds if the forceapplied to the plate is P = 100 kN. The allowable shear stress for the weld material is allow = 100 MPa.

Given: P 100kN:= allow 100MPa:=

L 100mm:= 45deg:=

Solution:

Shear Plane in the Weld: Aweld L a sin ( )=

Aweld0.5Pallow

=

L a sin ( ) 0.5Pallow

=

a1

L sin ( )0.5Pallow

:=

a 7.071 mm= Ans

Problem 1-85

The fillet weld size a = 8 mm. If the joint is assumed to fail by shear on both sides of the block alongthe shaded plane, which is the smallest cross section, determine the largest force P that can be appliedto the plate. The allowable shear stress for the weld material is allow = 100 MPa.

Given: a 8mm:= L 100mm:=

45deg:= allow 100MPa:=

Solution:

Shear Plane in the Weld: Aweld L a sin ( )=

P allow 2Aweld( )=P allow 2 L a sin ( )( ) :=

P 113.14 kN= Ans

Problem 1-86

The eye bolt is used to support the load of 25 kN. Determine its diameter d to the nearest multiples of5mm and the required thickness h to the nearest multiples of 5mm of the support so that the washerwill not penetrate or shear through it. The allowable normal stress for the bolt is allow = 150 MPa andthe allowable shear stress for the supporting material is allow = 35 MPa.

Given: P 25kN:= dwasher 25mm:=

allow 150MPa:= allow 35MPa:=

Solution:

Allowable Normal Stress: Design of bolt size

AboltP

allow=

4

d2P

allow=

d4

Pallow

:=

d 14.567 mm=

Use d 15mm:= d 15 mm= Ans

Allowable Shear Stress: Design of support thickness

AsupportP

allow= dwasher( ) h

Pallow

=

h1

dwasher( )P

allow

:=

h 9.095 mm=

Use d 10mm:= d 10 mm= Ans

Problem 1-87

The frame is subjected to the load of 8 kN. Determine the required diameter of the pins at A and B ifthe allowable shear stress for the material is allow = 42 MPa. Pin A is subjected to double shear,whereas pin B is subjected to single shear.

Given: P 8kN:= allow 42MPa:=

a 1.5m:= b 1.5m:= c 1.5m:= d 0.6m:=

Solution: BC 45deg:=

Support Reactions: From FBD (a),

D=0; FBC sin BC( ) c( ) P c d+( ) 0=

FBC Pc d+

sin BC( ) c( ):=

FBC 15.839 kN=

From FBD (b),

A=0; Dy a b+( ) P c d+( ) 0=

Dy Pc d+a b+:= Dy 5.6 kN=

+ Fx=0; Ax P 0= Ax P:= Ax 8 kN=

+ Fy=0; Dy Ay 0= Ay Dy:=

FA Ax2 Ay

2+:= FA 9.77 kN=

Apin0.5FAallow

=4

d20.5FAallow

=

d4

0.5FAallow

:= d 12.166 mm= Ans

For pin B: Pin A is subjected to single shear, and FB FBC:=

ApinFB

allow=

4

d2FB

allow=

d4

FBallow

:= d 21.913 mm= Ans

Problem 1-88

The two steel wires AB and AC are used to support the load. If both wires have an allowable tensilestress of allow = 200 MPa, determine the required diameter of each wire if the applied load is P = 5kN.

Given: P 5kN:= allow 200MPa:=

a 4m:= b 3m:= 60deg:=

Solution:c a2 b2+:= h

ac

:= vbc

:=

At joint A:

Initial guess: FAB 1kN:= FAC 2kN:=

Given

+ Fx=0; FAC h( ) FAB sin ( ) 0= [1]

+ Fy=0; FAC v( ) FAB cos ( )+ P 0= [2]

Solving [1] and [2]:FAB

FAC

Find FAB FAC,( ):=

FAB

FAC

4.3496

4.7086

kN=

For wire AB

AABFABallow

=4

dAB2

FABallow

=

dAB4

FABallow

:= dAB 5.26 mm= Ans

For wire AC

AACFACallow

=4

dAC2

FACallow

=

dAC4

FACallow

:= dAC 5.48 mm= Ans

Problem 1-89

The two steel wires AB and AC are used to support the load. If both wires have an allowable tensilestress of allow = 180 MPa, and wire AB has a diameter of 6 mm and AC has a diameter of 4 mm,determine the greatest force P that can be applied to the chain before one of the wires fails.

Given: allow 180MPa:=

a 4m:= b 3m:= 60deg:=

dAB 6mm:= dAC 4mm:=

Solution:c a2 b2+:= h

ac

:= vbc

:=

Assume failure of AB: FAB AAB( ) allow=

FAB4

dAB2 allow:= FAB 5.09 kN=

At joint A:

Initial guess: P1 1kN:= FAC 2kN:=

Given

+ Fx=0; FAC h( ) FAB sin ( ) 0= [1]

+ Fy=0; FAC v( ) FAB cos ( )+ P1 0= [2]

Solving [1] and [2]:P1

FAC

Find P1 FAC,( ):=

P1

FAC

5.8503

5.5094

kN=

Assume failure of AC: FAC AAC( ) allow=

FAC4

dAC2 allow:= FAC 2.26 kN=

At joint A:

Initial guess: P2 1kN:= FAB 2kN:=

Given

+ Fx=0; FAC h( ) FAB sin ( ) 0= [1]

+ Fy=0; FAC v( ) FAB cos ( )+ P2 0= [2]

Solving [1] and [2]:FAB

P2

Find FAB P2,( ):=

FAB

P2

2.0895

2.4019

kN=

Chosoe the smallest value: P min P1 P2,( ):= P 2.40 kN= Ans

Problem 1-90

The boom is supported by the winch cable that has a diameter of 6 mm and an allowable normal stressof allow = 168 MPa. Determine the greatest load that can be supported without causing the cable to failwhen = 30 and = 45. Neglect the size of the winch.

Given: allow 168MPa:= do 6mm:=

30deg:= 45deg:=

Solution:

For the cable:Tcable Acable( ) allow=

Tcable4

do2 allow:=

Tcable 4.7501 kN=At joint B:

Initial guess: FAB 1 kN:= W 2 kN:=

Given

+ Fx=0; Tcable cos ( ) FAB cos ( )+ 0= [1]

+ Fy=0; W FAB sin ( )+ Tcable sin ( ) 0= [2]

Solving [1] and [2]:FAB

W

Find FAB W,( ):=

FAB

W

5.818

1.739

kN= Ans

Problem 1-91

The boom is supported by the winch cable that has an allowable normal stress of allow = 168 MPa. Ifit is required that it be able to slowly lift 25 kN, from = 20 to = 50, determine the smallestdiameter of the cable to the nearest multiples of 5mm. The boom AB has a length of 6 m. Neglect thesize of the winch. Set d = 3.6 m.

Given: allow 168MPa:= W 25 kN:=

d 3.6m:= a 6m:=Solution:

Maximum tension in canle occurs when 20deg:=

sin ( )a

sin ( )d

= asinda

sin ( )

:= 11.842 deg=

At joint B:

Initial guess: FAB 1 kN:= Tcable 2 kN:=

Given +:=

+ Fx=0; Tcable cos ( ) FAB cos ( )+ 0= [1]

+ Fy=0; W FAB sin ( )+ Tcable sin ( ) 0= [2]

Solving [1] and [2]:FAB

Tcable

Find FAB Tcable,( ):=

FAB

Tcable

114.478

103.491

kN=

For the cable:

AcableP

allow=

4

do2

Tcableallow

=

do4

Tcableallow

:=

do 28.006 mm=

Use do 30mm:= do 30 mm= Ans

Problem 1-92

The frame is subjected to the distributed loading of 2 kN/m. Determine the required diameter of thepins at A and B if the allowable shear stress for the material is allow = 100 MPa. Both pins aresubjected to double shear.

Given: w 2kNm

:= allow 100MPa:=

r 3m:=

Solution: Member AB is atwo-force member

45deg:=

Support Reactions:

A=0; FBC sin ( ) r( ) w r( ) 0.5r( ) 0=

FBC0.5w rsin ( )

:= FBC 4.243 kN=

+ Fy=0; Ay FBC sin ( )+ w r 0=

Ay FBC sin ( ) w r+:= Ay 3 kN=

+ Fx=0; Ax FBC cos ( ) 0=

Ax FBC cos ( ):= Ax 3 kN=

Average Shear Stress: Pin A and pin B are subjected to double shear

FA Ax2 Ay

2+:= FA 4.243 kN=

FB FBC:= FB 4.243 kN=

Since both subjected to the same shear force V = 0.5 FA and V 0.5FB:=

ApinV

allow=

4

dpin2

Vallow

=

dpin4

Vallow

:=

dpin 5.20 mm= Ans

Problem 1-93

Determine the smallest dimensions of the circular shaft and circular end cap if the load it is required tosupport is P = 150 kN. The allowable tensile stress, bearing stress, and shear stress is (t)allow = 175MPa, (b)allow = 275 MPa, and allow = 115 MPa.

Given: P 150kN:= t_allow 175MPa:=

allow 115MPa:= b_allow 275MPa:=

d2 30mm:=

Solution:

Allowable Normal Stress: Design of end cap outer diameter

AP

t_allow=

4

d12 d2

2

Pt_allow

=

d14

Pt_allow

d22+:= d1 44.62 mm= Ans

Allowable Bearing Stress: Design of circular shaft diameter

AP

b_allow=

4

d32

Pb_allow

=

d34

Pb_allow

:= d3 26.35 mm= Ans

Allowable Shear Stress: Design of end cap thickness

AP

allow= d3( ) t Pallow

=

t1

d3P

allow

:= t 15.75 mm= Ans

Problem 1-94

If the allowable bearing stress for the material under the supports at A and B is (b)allow = 2.8 MPa,determine the size of square bearing plates A' and B' required to support the loading. Take P = 7.5 kN.Dimension the plates to the nearest multiples of 10mm. The reactions at the supports are vertical.

Given: b_allow 2.8MPa:= P 7.5 kN:=

P1 10 kN:= P2 10 kN:=

P3 15 kN:= P4 10 kN:=

a 1.5m:= b 2.5m:=

Solution: L 3 a b+:=

Support Reactions:

A=0; By 3a( ) P2 a( ) P3 2a( ) P4 3a( ) P L( ) 0=

By P2a3a

P32a3a

+ P43a3a

+ PL3a

+:= By 35 kN=

B=0; Ay 3 a( ) P1 3 a( )+ P2 2 a( )+ P3 a( )+ P b( ) 0=

Ay P13a3a

P22a3a

+ P3a3a

+ Pb3a

:= Ay 17.5 kN=

For Plate A:

Aplate_AAy

b_allow= aA

2 Ayb_allow

=

aAAy

b_allow:=

aA 79.057 mm=

Use aA x aA plate: aA 80mm= Ans

For Plate B

Aplate_BBy

b_allow= aB

2 Byb_allow

=

aBBy

b_allow:=

aB 111.803 mm=

Use aB x aB plate: aB 120mm= Ans

ULoHSText BoxRb = 35kN

(b is in subscript)

Problem 1-95

If the allowable bearing stress for the material under the supports at A and B is (b)allow = 2.8 MPa,determine the maximum load P that can be applied to the beam. The bearing plates A' and B' havesquare cross sections of 50mm x 50mm and 100mm x 100mm, respectively.

Given: b_allow 2.8MPa:=

P1 10 kN:= P2 10 kN:=

P3 15 kN:= P4 10 kN:=

a 1.5m:= b 2.5m:=

aA 50mm:= aB 100mm:=

Solution: L 3 a b+:=

Support Reactions:

A=0; By 3a( ) P2 a( ) P3 2a( ) P4 3 a( ) P L( ) 0=

By P2a3a

P3 2a3a

+ P43 a3a

+ PL3a

+=

B=0; Ay 3 a( ) P1 3 a( )+ P2 2 a( )+ P3 a( )+ P b( ) 0=

Ay P13a3a

P22a3a

+ P3a3a

+ Pb3a

=

For Plate A: Ay aA2

b_allow:=

aA2

b_allow P1

3a3a

P22a3a

+ P3a3a

+ Pb3a

=

P P13ab

P22ab

+ P3ab

+ aA2

b_allow

3ab

:= P 26.400 kN=

Pcase_1 P:=For Plate B: By aB

2

b_allow:=

aB2

b_allow P2

a3a

P3 2a3a

+ P43 a3a

+ PL3a

+=

P P2aL

P3 2aL

P43 aL

aB2

b_allow

3aL

+:= P 3.000 kN=

Pcase_2 P:=

Pallow min Pcase_1 Pcase_2,( ):= Pallow 3 kN= Ans

Problem 1-96

Determine the required cross-sectional area of member BC and the diameter of the pins at A and B ifthe allowable normal stress is allow = 21 MPa and the allowable shear stress is allow = 28 MPa.Given: allow 21MPa:= allow 28MPa:=

P 7.5kip:= 60deg:=

a 0.6m:= b 1.2m:= c 0.6m:=

Solution: L a b+ c+:=

Support Reactions:

A=0; By L( ) P a( ) P a b+( ) 0=

By PaL P

a b+L

+:=

FBCBy

sin ( ):= FBC 38.523 kN=

By 33.362 kN=

Bx FBC cos ( ):= Bx 19.261 kN=

+ Fy=0; By P+ P+ Ay 0= Ay By P+ P+:= Ay 33.362 kN=

+ Fx=0; Bx Ax 0= Ax Bx:= Ax 19.261 kN=

FA Ax2 Ay

2+:= FA 38.523 kN=

Member BC: ABC

FBCallow

:= ABC 1834.416 mm2= Ans

Pin A:AA

FAallow

=4

dA2

FAallow

=

dA4

FAallow

:= dA 41.854 mm= Ans

Pin B:AB

0.5FBCallow

=4

dB2

0.5FBCallow

=

dB4

0.5FBCallow

:= dB 29.595 mm= Ans

Problem 1-97

The assembly consists of three disks A, B, and C that are used to support the load of 140 kN.Determine the smallest diameter d1 of the top disk, the diameter d2 within the support space, and thediameter d3 of the hole in the bottom disk. The allowable bearing stress for the material is (allow)b =350 MPa and allowable shear stress is allow = 125 MPa.

Given: P 140kN:=

allow 125MPa:= b_allow 350MPa:=

hB 20mm:= hC 10mm:=

Solution:

Allowable Shear Stress: Assume shear failure dor disk C

AP

allow= d2( ) hC Pallow

=

d21

hCP

allow

:= d2 35.65 mm= Ans

Allowable Bearing Stress: Assume bearing failure dor disk C

AP

b_allow=

4

d22 d3

2

Pb_allow

=

d3 d22 4

Pb_allow

:= d3 27.60 mm= Ans

Allowable Bearing Stress: Assume bearing failure dor disk B

AP

b_allow=

4

d12

Pb_allow

=

d14

Pb_allow

:= d1 22.57 mm=

Since d3 > d1, disk B might fail due to shear.

PA

= P

d1 hB:= 98.73 MPa= < allow (O.K.!)

Therefore d1 22.57 mm= Ans

Problem 1-98

Strips A and B are to be glued together using the two strips C and D. Determine the required thicknesst of C and D so that all strips will fail simultaneously. The width of strips A and B is 1.5 times that ofstrips C and D.

Given: P 40N:= t 30mm:=

bA 1.5m:= bB 1.5m:=

bC 1m:= bD 1m:=

Solution:

Average Normal Stress: Requires,

A B= B C= C D=

NbA( ) t

0.5NbC( ) tC( )

=

tC0.5 bA( ) t

bC:= tC 22.5 mm= Ans

Problem 1-99

If the allowable bearing stress for the material under the supports at A and B is (b)allow = 2.8 MPa,determine the size of square bearing plates A' and B' required to support the loading. Dimension theplates to the nearest multiples of 10mm. The reactions at the supports are vertical. Take P = 7.5 kN.

Given: b_allow 2.8MPa:= P 7.5kN:=

w 10kNm

:= a 4.5m:= b 2.25m:=

Solution: L a b+:=

Support Reactions:

A=0; By a( ) w a( ) 0.5 a( ) P L( ) 0=

By w 0.5 a( ) PLa

+:= By 33.75 kN=

B=0; Ay a( ) w a( ) 0.5 a( )+ P L( ) 0=

Ay w 0.5 a( ) Pba

:= Ay 18.75 kN=

Allowable Bearing Stress: Design of bearing plates

For Plate A:

AreaAy

b_allow= aA

2 Ayb_allow

=

aAAy

b_allow:=

aA 81.832 mm=

Use aA x aA plate: aA 90mm:= aA 90 mm= Ans

For Plate B

AreaBy

b_allow= aB

2 Byb_allow

=

aBBy

b_allow:=

aB 109.789 mm=

Use aB x aB plate: aB 110mm:= aB 110.00 mm= Ans

Problem 1-100

If the allowable bearing stress for the material under the supports at A and B is (b)allow = 2.8 MPa,determine the maximum load P that can be applied to the beam. The bearing plates A' and B' havesquare cross sections of 50mm x 50mm and 100mm x 100mm, respectively.

Given: b_allow 2.8MPa:=

w 10kNm

:= a 4.5m:= b 2.25m:=

aA 50mm:= aB 100mm:=

Solution: L a b+:=

Support Reactions:

A=0;By a( ) w a( ) 0.5 a( ) P L( ) 0=

P ByaL

waL

0.5 a( )=

B=0; Ay a( ) w a( ) 0.5 a( )+ P b( ) 0=

P Ayab

wab

0.5 a( )+=

Allowable Bearing Stress:

Assume failure of material occurs under plate A. Ay aA2

b_allow:=

P aA2

b_allow

ab

w a( )0.5a

b+:= P 31 kN=

Pcase_1 P:=

Assume failure of material occurs under plate B. By aB2

b_allow:=

P ByaL

waL

0.5 a( ):= P 3.67 kN=

Pcase_2 P:=

Pallow min Pcase_1 Pcase_2,( ):= Pallow 3.67 kN= Ans

Problem 1-101

The hanger assembly is used to support a distributed loading of w = 12 kN/m. Determine the averageshear stress in the 10-mm-diameter bolt at A and the average tensile stress in rod AB, which has adiameter of 12 mm. If the yield shear stress for the bolt is y = 175 MPa, and the yield tensile stress forthe rod is y = 266 MPa, determine the factor of safety with respect to yielding in each case.

Given: y 175MPa:= w 12kNm

:=

y 266MPa:=

a 1.2m:= b 0.6m:= e 0.9m:=

do 10mm:= drod 12mm:=

Solution:

c a2 e2+:= hac

:= vec

:=

Support Reactions: L a b+:=

C=0; FAB v( ) a( ) w L( ) 0.5 L( ) 0=

FAB wL

a v

0.5 L( ):=

FAB 27 kN=

For bolt A: Bolt A is subjected to double shear, and V 0.5FAB:= V 13.5 kN=

A4

do2:=

VA

:= 171.89 MPa= Ans

FSy

:= FS 1.02= Ans

For rod AB: N FAB:= N 27 kN=

A4

drod2:=

NA

:= 238.73 MPa= Ans

FSy

:= FS 1.11= Ans

Problem 1-102

Determine the intensity w of the maximum distributed load that can be supported by the hangerassembly so that an allowable shear stress of allow = 95 MPa is not exceeded in the 10-mm-diameterbolts at A and B, and an allowable tensile stress of allow = 155 MPa is not exceeded in the12-mm-diameter rod AB.

Given: allow 95MPa:= allow 155MPa:=

a 1.2m:= b 0.6m:= e 0.9m:=

do 10mm:= drod 12mm:=

Solution: c a2 e2+:= hac

:= vec

:=

Support Reactions: L a b+:=

C=0; FAB v( ) a( ) w L( ) 0.5 L( ) 0=

FAB wL

a v

0.5 L( )=

Assume failure of pin A or B:

V 0.5FAB= V allow A= A4

do2:=

0.5 wL

a v

0.5 L( ) allow4

do2

=

wa v

0.5L( )2allow

4

do2

:=

w 6.632kNm

= (controls!) Ans

Assuming failure of rod AB:

N FAB= N allow A= A4

drod2:=

wL

a v

0.5 L( ) allow4

drod2

=

wa v

0.5L2allow

4

drod2

:=

w 7.791kNm

=

Problem 1-103

The bar is supported by the pin. If the allowable tensile stress for the bar is (t)allow = 150 MPa, andthe allowable shear stress for the pin is allow = 85 MPa, determine the diameter of the pin for whichthe load P will be a maximum. What is this maximum load? Assume the hole in the bar has the samediameter d as the pin. Take t =6 mm and w = 50 mm.

Given: allow 85MPa:=

t_allow 150MPa:=

t 6mm:= w 50mm:=

Solution: GivenAllowable Normal Stress: The effective cross-sectionalarea Ae for the bar must be considered here by takinginto account the reduction in cross-sectional areaintroduced by the hole. Here, effective area Ae is equalto (w - d) t, and allow equals to P /Ae .

t_allowP

w d( ) t= [1]

Allowable Shear Stress: The pin is subjected to doubleshear and therefore the allowable equals to 0.5P /Apin,

and the area Apin is equal to ( /4) d2.

allow2

P

d2

= [2]

Solving [1] and [2]: Initial guess: d 20mm:= P 10kN:=

P

d

Find P d,( ):= P 31.23 kN= Ans

d 15.29 mm= Ans

Problem 1-104

The bar is connected to the support using a pin having a diameter of d = 25 mm. If the allowabletensile stress for the bar is (t)allow = 140 MPa, and the allowable bearing stress between the pin andthe bar is (b)allow =210 MPA, determine the dimensions w and t such that the gross area of the cross

section is wt = 1250 mm2 and the load P is a maximum. What is this maximum load? Assume the holein the bar has the same diameter as the pin.

Given: t_allow 140 MPa:= b_allow 210 MPa:=

A 1250mm2:= d 25mm:=

Solution: A w t= Given

Allowable Normal Stress: The effective cross-sectionalarea Ae for the bar must be considered here by takinginto account the reduction in cross-sectional areaintroduced by the hole. Here, effective area Ae is equalto (w - d) t, that is (A- d t) and allow equals to P /Ae .

t_allowP

A d t= [1]

Allowable Bearing Stress: The projected area Ab is equalto (d t), and allow equals to P /Ab .

b_allowP

d t

= [2]

Solving [1] and [2]: Initial guess: t 0.5in:= P 1kip:=

P

t

Find P t,( ):= P 105.00 kN= Ans

t 20.00 mm= Ans

And : wAt

:= w 62.50 mm= Ans

Problem 1-105

The compound wooden beam is connected together by a bolt at B. Assuming that the connections atA, B, C, and D exert only vertical forces on the beam, determine the required diameter of the bolt at Band the required outer diameter of its washers if the allowable tensile stress for the bolt is (t)allow =150 MPa. and the allowable bearing stress for the wood is (b)allow = 28 MPa. Assume that the hole inthe washers has the same diameter as the bolt.Given: P1 3kN:= P2 1.5kN:= P3 2kN:=

t_allow 150MPa:= a 2m:=

b_allow 28MPa:= b 1.5m:=

Solution:

From FBD (a): Given

D=0; Cy 4 b( ) By 3b( )+ P2 2 b( )+ P3 b( )+ 0= [1]

From FBD (b):

A=0; By 2 a b+( ) Cy 2 a( ) P1 a( ) 0= [2]

Solving [1] and [2]: Initial guess: By 1kN:= Cy 2kN:=

By

Cy

Find By Cy,( ):=

By

Cy

4.4

4.55

kN=

For bolt:

AraeBy

t_allow=

4

dB2

Byt_allow

=

dB4

Byt_allow

:=

dB 6.11 mm= Ans

For washer:

AreaBy

b_allow=

4

dw2 dB

2

Byb_allow

=

dw dB2 4

Byb_allow

+:=

dw 15.41 mm= Ans

Problem 1-106

The bar is held in equilibrium by the pin supports at A and B. Note that the support at A has a singleleaf and therefore it involves single shear in the pin, and the support at B has a double leaf and thereforeit involves double shear. The allowable shear stress for both pins is allow = 150 MPa. If a uniformdistributed load of w = 8 kN/m is placed on the bar, determine its minimum allowable position x fromB. Pins A and B each have a diameter of 8 mm. Neglect any axial force in the bar.Given: allow 150MPa:= do 8mm:=

a 2m:=w 8

kNm

:=b 2m:=Solution:

A=0; By a( ) w b x( ) a x+ 0.5 b x( )+[ ] 0= [1]

B=0; Ay a( ) w b x( ) x 0.5 b x( )+[ ] 0= [2]

Assume failure of pin A:

AraeAy

allow=

4

do2

Ayallow

=

Ay4

do2

allow( ):= Ay 7.5398 kN=

Substitute value of force A into Eq [2],

Given Ay a( ) w b x1( ) x1 0.5 b x1( )+ 0= [2]Initial guess: x1 0.3m:= x1 Find x1( ):= x1 0.480 m=

xcase_1 x1:=Assume failure of pin B:

Arae0.5Byallow

=4

do2

0.5Byallow

=

By 24

do2

allow( ):= By 15.0796 kN=

Substitute value of force A into Eq [1],

Given By a( ) w b x2( ) a x2+ 0.5 b x2( )+ 0= [1]Initial guess: x2 0.3m:= x2 Find x2( ):= x2 0.909 m=

xcase_2 x2:=

Choose the larger x value: x max xcase_1 xcase_2,( ):=x 0.909 m= Ans

Problem 1-107

The bar is held in equilibrium by the pin supports at A and B. Note that the support at A has a singleleaf and therefore it involves single shear in the pin, and the support at B has a double leaf and thereforeit involves double shear. The allowable shear stress for both pins is allow = 125 MPa. If x = 1 m,determine the maximum distributed load w the bar will support. Pins A and B each have a diameter of 8mm. Neglect any axial force in the bar.

Given: allow 125MPa:= x 1m:= do 8mm:=

a 2m:= b 2m:=Solution:

wokNm

:=

Given

A=0; Bw a( ) wo b x( ) a x+ 0.5 b x( )+[ ] 0= [1]

B=0; Aw a( ) wo b x( ) x 0.5 b x( )+[ ] 0= [2]

Initial guess: Bw 1kN:= Aw 1kN:=

Solving [1] and [2]:Bw

Aw

Find Bw Aw,( ):=

Bw

Aw

1.75

0.75

kN=

For pin A: Ay w1Awwo

=

AraeAy

allow=

4

do2

w1allow

Awwo

=

w14

do2

allow( )

woAw

:= w1 8.378

kNm

=

For pin B By wBwwo

=

Arae0.5Byallow

=2

do2

w2allow

Bwwo

=

w22

do2

allow( )

woBw

:= w2 7.181

kNm

=

The smalleer w controls ! w min w1 w2,( ):=

w 7.181kNm

= Ans

Problem 1-108

The bar is held in equilibrium by the pin supports at A and B. Note that the support at A has a singleleaf and therefore it involves single shear in the pin, and the support at B has a double leaf and thereforeit involves double shear. The allowable shear stress for both pins is allow = 125 MPa. If x = 1 m and w= 12 kN/m, determine the smallest required diameter of pins A and B. Neglect any axial force in thebar.Given: allow 125MPa:= x 1m:= do 8mm:=

a 2m:= b 2m:=

Solution:w 12

kNm

:=

Given

A=0; By a( ) w b x( ) a x+ 0.5 b x( )+[ ] 0= [1]

B=0; Ay a( ) w b x( ) x 0.5 b x( )+[ ] 0= [2]

Initial guess: By 1kN:= Ay 1kN:=

Solving [1] and [2]:By

Ay

Find By Ay,( ):=

By

Ay

21

9

kN=

For pin A:

AraeAy

allow=

4

dA2

Ayallow

=

dA4

Ayallow

:=

dA 9.57 mm= Ans

For pin B

Arae0.5Byallow

=2

dB2

Byallow

=

dB2

Byallow

:=

dB 10.34 mm= Ans

Problem 1-109

The pin is subjected to double shear since it is used to connect the three links together. Due to wear,the load is distributed over the top and bottom of the pin as shown on the free-body diagram.Determine the diameter d of the pin if the allowable shear stress is allow = 70 MPa and the load P = 40kN. Also, determine the load intensities w1 and w2 .

Given: allow 70 MPa:= P 40kN:=

a 37.5mm:= b 25mm:=Solution:

+Pin: Fy=0; P w1 a( ) 0=

w1Pa

:=

w1 1066.67kNm

= Ans

+Link: Fy=0; P 2 0.5w2( ) b( ) 0=

w2Pb

:=

w2 1600.00kNm

= Ans

Shear Stress

Area0.5Pallow

=2

d2P

allow=

d2

Pallow

:=

d 19.073 mm= Ans

Problem 1-110

The pin is subjected to double shear since it is used to connect the three links together. Due to wear,the load is distributed over the top and bottom of the pin as shown on the free-body diagram.Determine the maximum load P the connection can support if the allowable shear stress for thematerial is allow = 56 MPa and the diameter of the pin is 12.5 mm. Also, determine the load intensitiesw1 and w2 .

Given: allow 56 MPa:= d 12.5mm:=

a 37.5mm:= b 25mm:=Solution:

Shear Stress

Area0.5Pallow

=2

d2P

allow=

P2

d2 allow( ):=

P 13.7445 kN= Ans

Pin:

+ Fy=0; P w1 a( ) 0=

w1Pa

:=

w1 366.52kNm

= Ans

Link:

+ Fy=0; P 2 0.5w2( ) b( ) 0=

w2Pb

:=

w2 549.78kNm

= Ans

Problem 1-111

The cotter is used to hold the two rods together. Determine the smallest thickness t of the cotter andthe smallest diameter d of the rods. All parts are made of steel for which the failure tensile stress is fail= 500 MPa and the failure shear stress is fail = 375 MPa. Use a factor of safety of (F.S.)t = 2.50 intension and (F.S.)s = 1.75 in shear.

Given: fail 500MPa:= fail 375MPa:= P 30kN:=

d2 40mm:= h 10mm:=

FSt 2.50:= FSs 1.75:=

Solution:

Allowable Normal Stress : Design of rod size

allowfailFSt

:= allow 200 MPa=

AreaP

allow=

4

d2P

allow

=

d4

Pallow:=

d 13.82 mm= Ans

Allowable Shear Stress : Design of cotter size

allowfailFSs

:= allow 214.29 MPa=

Area0.5Pallow

= h t0.5Pallow

=

t1h

0.5Pallow:=

t 7 mm= Ans

Problem 1-112

The long bolt passes through the 30-mm-thick plate. If the force in the bolt shank is 8 kN, determinethe average normal stress in the shank, the average shear stress along the cylindrical area of the platedefined by the section lines a-a, and the average shear stress in the bolt head along the cylindrical areadefined by the section lines b-b.

Given: