cpe 461 assignment uitm

8/9/2019 CPE 461 Assignment Uitm

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Free Bo dy Diagram

∑ F X +→

=0 ;

A cos !0 " # cos !0 = 0

A=#$$$$$$$$$..%&'uation1(

∑ F y+↑

=0 ;

A sin !0 ) # sin !0 " * =0 $$$$$$$$$..%&'uation+(

,i-en *=800 lb then subtitled &'uation 1 into &'uation +

+ A sin !0 =800 lb

A = /!1.880 lb

# = A = /!1.880lb.

800lb

!02!02

# A



Question 2

A cur-ed slender bar is loaded and supported as shown in Figure +.Determine the

reactions at support A.

Free Body Diagram

∑ F X +→

=0 ;

Ax = 0

∑ F y+↑

=0 ;

Ay ) +30 lb = 0

Ay = +30 lb

∑ M A$% & 0 ; Moment =Force x distance

M A ) +30 lb % 4ft( = 0

M A = 5630 lb.ft

y

+30lb

4ft

x

M A A



Question 3

Free Body Diagram

7he lawn mo-er shown in Figure 4 weight 43 lbf .Determine the forces re'uired to mo-e

the mo-er at a constant -elocity and the forces exerted on the front and rear wheels by the

inclined surfaces.

+↗ ∑F=0 ; %9nclined( tan40=x:4/inch

,i-en *= 43 lbf

cos 40 " 43 sin 13 = 0

=35sin15

cos30

= 10./! lbf $$$$$$..%&'uation 1(

+↖ ∑Fy=0

5 sin 40 " 43 cos 13 ) A ) # = 0 ................%&'uation +(

ubstitute = 10./! lb into &'uation +

5 10./! lbf sin 40 " 43 cos 13 ) A ) # = 0

A ) # = 4<.0/ lbf

∑ M A+↻=0 ; Moment =Force x distance

/inch y

x

132, #

#

14 inch

1/inch 402

A

A

4/inch

x=1<.!4 i

x

40

4/inch



0 = 5#%14)1/()*cos13%14( "*sin13%/() cos40%1<.!4(" sin 40%4/(

,i-en *= 43 lbf = 10./! lbf

0 = 5#%14)1/()43cos13%14("43sin13%/()10./!cos40%1<.!4("10./!sin 40%4/(

# = 1/.</ lbf

A = 4<.0/ lbf 51/.</ lb =+/.1 lbf

1 lbf = /./3

A= +/.1 lbf =106.+/3

# = 1/.</ lbf = !!./84

= 10./! lbf = /!.3/6



Question 4

Determine the forces in members D> >F D, of the bridge truss as shown in Figure /.

Free Body Diagram

+→ ∑F=0 ;

+↑ ∑Fy=0 ;

A ) & 510 ?ip 5+0 ?ip = 0

A ) & = 40?ip =144.3?

∑ M A$% & 0 ;

)10 ?ip %13ft( )4ft %+0?ip( 5/3ft %&(=0

& = 1!.!6 ?ip

A = 40 ?ip " 1!.!6 ?ip = 14.44 ?ip = 3<.4183?

>

D#

A!02 !02402 402

, F

#

#

+0?ip10?ip A



At point A

+↑ ∑Fy=0 ;

7 A# sin 40 )14.44 ?ip =0

7 A# = 5 +!.!! ?ip = 118.!46 ?

+→ ∑F=0 ;

7 A# cos 40 ) 7 A, =0

5 +!.!! ?ip cos 40 ) 7 A, =0

7 A, = 5+4.0< ?ip=5 10+.63?

At point D

+↑ ∑Fy=0 ; +→ ∑F=0 ;

7D& sin 40 = 1!.!6 ?ip

7D&=44.4/ ?ip = 1/8.4!4?

7D& cos 40 "7&F= 0

7&F= +8.86 ?ip= 1+8./6 ?

7 A#

7 A,

7 A# sin 40

402

A= 14.44 ?ip

7D&

&

402

& = 1!.!6 ?ip7&F



7>D5 7D& =0

7>D = 7D&

7>D= 544.4/ ?ip = 51/8.4!4 ?

At point F

+→ ∑F=0 ; +↑ ∑Fy=0 ;

5 7>F cos !057F, =0

7F, = 16.4+ ?ip

=66.06/?

7>F sin !0 5+0 ?ip=0

7>F= +4.0< ?ip

=10+.6303?

7D&

D

7>D

10 ?ip

7F,

7>F

+0?ip

!02



Question 5

A drum of mass +00 @g is supported by a pair of frames. 7he second frame is behind the

one shown. 7he system is as shown in Figure 3. Determine the forces acting on member

A>&.

Free #ody Diagram 5 7he drum and itBs support

RC =W =mg

¿( 200×9.81

1000 )kN

¿1.962kN

∑ F y+↖=0

R D sin 45°+ R E sin 45°−W =0

R D sin 45°− R E sin 45°−1.962 kN =0 $$..%&'uation 1(

∑ F X +↗=0

R Dcos45°− R Ecos 45°=0

R D= R E %&'uation +(

/32

xy&

D

*=mg=c A #



ubstituted e'uation + in to e'uation 1

2⌊ R E (sin 45° ) ⌋=1.962

R E=1.3873 kN

R D=1.3873 kN

C A#

7an /32=1m:%1:+A#(A#=+m

∑ F Y +↑=0

R A+ RB−W =0

R A+ R B=1.962 kN %&'uation 1(

∑ M A+↻=0

− RB (2 )+W (1 )=0

RB=−1.962

−2

¿0.981kN %&'uation +(

ub e'uation + to e'uation 1

R A=1.962−0.981

R A=0.981 kN

1m

/32

1m

/32 /32*=1.<!+?

A #

+m



E A$)= 0

5+# = 51.<!+ %1(

# = 0.<81 ?

A ) 0.<81 = 1.<!+

A = 0.<81 ?

Fx) = 0

7 A# cos /3 ) A cos /3 " >x = 0

Fy)G = 0

>y ) A sin /3 " 7 A# sin /3 ) & = 0

E A$)= 0

5>y % √ 2 ( " & %0.8

tan 45 ( ) √ 2 = 0

√ 2 >y " 1.486 %+.+1/+( = 0

>y = 5+.16+ ?

5+.16+ ) 0.<81 sin /3 " 7 A# sin /3 ) 1.486 = 0



*1 *+

W 1=( 50×9.81

1000 )kN

¿0.4905=0.491kN

W 1=(50×9.81

1000 )kN

¿0.4905=0.491kN

∑ F Y +↖=0

−W 1−W

2+ R A+ R Bcos27°=0

−0.491−0.491+ R A+ R B=0

R A+ RB=−0.982 $..&'uation +

Centred of gravityW =580

2

¿290mm

∑ M A+↻

=0

27°

120sin¿=0

− R B (580)+W 2 (290cos27° )−W

1¿

27°

120sin¿=0

− RB (580)+0.4891 (290cos27° )−0.491¿

RB (580)=100.1213

RB

=0.1726 kN … …… E!ation2

ubstituted &'uation + to &'uation 1

R A=0.982−0.1726cos27°

R A=0.82821 kN



Question 7

7he reaction between the crutch and the ground is 43 9b as shown in the diagram

below. Determine the internal forces acting on section a5a

Free Body Diagram:

∑ F X +↗=0

A X −35sin25°=0

A X =14.79 lbf

∑ F y+↖=0

A 43 sin +3 2F=43 lbf

F=43 lbf 43 cos +32

+3 2 +32

/32

#

yx



A y+35cos25°=0

A y=−31.72 lbf

∑ M A

+↻=0

M a "a+(35sin25 ° ) (2 ft )=0

M a "a=29.58 lbf.ft

Question 8

Determine the force re'uired to push the 1135 @g cylinder o-er the small bloc? shown in

the diagram below

Free Body Diagram:

'

14

"220

RB

RBѲ

220 20*P

sin =1/3mm:++0mmѲ

= /1.+4Ѳ °

RA

+&1#1,N20*

+&1#1,N ./

+&1#1,N

20*

+&1#1,N S



Determine force

W =115×9.81/1000

¿1.13 kN

∑ F X +↗=0

#cos20°− RBcos41.23 °−1130sin 20°=0 $$$$$.&'uation 1

∑ F Y +↖=0

− # sin 20°+ RB sin41.23 °−1130 cos20°+ R A=0 $$..&'uation +

∑ M

A

+↻=0

# cos20° (220)− #$in 20° (0)+1130sin 20° (220)− RB sin41.23 ° (165.45 )− RB cos41.23° (75)=0

$$$.&HIA79J 4

ol-e the e'uation 1 and e'uation 4 to get and #.

0.9397 #−0.7521 RB−386.4828=0

206.7324 #−165.4835 R B−85026.2076=0

= 488./383 #= 5+8.3168

−388.4585sin20°−28.5178sin 41.23 °−1130 cos20°+ R A =0

A=1+14.308<



Question 9

An axial load is applied to a timber bloc? as shown. Determine the normal and shear

stress on the plane of the grain if = 1300 9b

Assume &ffecti-e Area= 1ft+

∑ F Y +↘=0

F N − # sin 14°=0

F N = # sin 14°

sin1/2

1/2

x cos 1/ 2

y

1/2



¿362.883 lbf

∑ F X +↗=0

F % − #cos14°=0

F % = #cos14°

¿1455.44 lbf

& =( F N

Eff " Area)

¿362.883

1

¿362.883 '(

ft 2

) =( F t

Eff " Area )

¿1455.444

1

¿1455.444 '(

ft 2

Question 1

7he steel bar shown will be used to carry an axial tensile load of /00 @. 9f the thic?ness

of the bar is /3 mm determine the normal and shearing stresses on the plane a5a



Free #ody Diagram

sin 53°=75

*

*=( 75

sin 53° )m

¿93.91m

Effe+tive " Area, A=45mm X (93.91mm )

¿4225.96mm2

∑ F -+↗=0

− F N −400cos37°=0

F7F

x

/00 ?

462462

63mmy



F N =−319454.204 N

∑ F y+↖=0

+ F % −400sin37°=0

F t =240726 N

!orma" #tress:$

& = F N

Eff " Area

¿319454.204

4225.96

¿75.59 M#a

#%ear #tress:$

) = F t

Eff " Area

¿ 240726

4225.96

¿56.96 M#a

Question 11



Free #ody Diagram

tan °=1.52

¿36.87 °

AD= √ 22+√ 1.52

AD= +.3 m

D

y 1.3m

> A

+m #x +m18 kN

A



∑ F y+↑=0

A) # 518 kN =0

A)#=18kN

∑ M A+↻=0 ; Moment =Force x distance

%+m( 18 kN 5#%/m( =0

#=%185<( kN =< kN

At point A

∑ F y+↑=0

A5F AD sin 4!.862=0

A=F ADsin 4!.862

F AD=<? : sin4!.862 = 13 kN

∑ F -+→=0

F A#=F AD cos 4!.86 2

F AD=13 kN cos 4!.86 2= 1+ kN

2.25 M#a=12kN

100a

a=53.07mm

D

A#

4!.862

A



Quest&on 12

Area , A=5∈×6∈¿

¿30¿2

3 &22

F )

Affe+tive Area=225 .$i

4in.

H

4 in.



F ) =225 .$i - Affe+tive Area

F ) =225 .$i - 30¿2

F ) =6750

.$i

¿2

∑ F -+→=0

/−¿ F ) =0

/= F )

H ¿6750 '(f

sin❑= 5

13

¿ sin−1 5

13

¿22.62°

∑ F y+↑=0

/− #cos22.62°=0

#= 6750

cos22.62°=7312.51'(f

1+

3

H



Question 13

*ith reference to the diagram as shown determine

a(7he resultant single load

b(7he center of gra-ity of the system

d( 7he moment at point KJL

e( 7he internal forces and moment at the center of the bridge

a( esultant single load

∑ A = ∫0

3

1

3 -

2

dx

=M1

3 N M -

3

3 N3

0

∑ A=3

kN

b( b(center of gra-ity of the system

> A= ∫ -y dx formula

A

J

4?:m

w=1

3 % -2

(

4m



> A=1

3 ∫0

3

-2

" - d -

> =

1

3

∫0

3

-2 - d-

1

3 ∫

0

3

-2

d-

> =

∫0

3

X 3 d-

∫0

3

-2

d-

>=

[ -4

4 ]30[ -

3

3 ]30

>=

20.25

9=2.25¿ .oint A

c( moment at point KJL

∑ M A+↻

=0 ; Moment =Force x distance−¿ M A +( +.+3m x 4? (= 0

M A =6.75kN.m

d( 9nternal forces and moment of the center if bridge

> A= ∫ -y dx formula

> A=1

3 ∫0

1.5

-2

" - d -

> =

1

3 ∫

0

1.5

-2 - d-

1

3 ∫

0

1.5

-2

d-

4?:m

J

+.+3m0.63m

4m A



> =

∫0

1.5

X 3 d-

∫0

1.5

-2

d-

> =

[ -4

4 ]1.50[ -

3

3 ]1.50

>=1.125 ¿ .oint >

Ooad at centre

∑ A = ∫0

1.5

1

3 -

2

dx

=M1

3 N M -

3

3 N1.5

0

∑ A=0.375 kN

∑ M A+

+↻

=0

; Moment =Force x distance

−¿ Mc +( 1.1+3m x 0.463? (= 0

M c =0.422kN.m

Question 14

eplace the loading by an e'ui-alent resultant force and specify its location

on the beam measured from point #.

Free Body Diagram

0.463kN

1.1+3m

c

800lb:ft

12

3

# A 1+ft < ft

300lb:ft



1.

'oad due to triang"e1 (entroid o) triang"e 1

1

2 1+ ft 800 lb:ft =/800 lb1

3 x 1+ ft = /f from #

+.

'oad due to triang"e 2 (entroid o) triang"e21

2 < ft 800 lb:ft =1430 lb1

3 x < ft = 4f from #

4.

A

#

/ ft from point #

# A

/800 lb

4 ft

1430 lb

9ft

# < ft

4 ft

/300lb



'oad due to *+(,A!-.'A* (entroid o) re/tangu"ar

< ft 300 lb:ft =/300 lb 1

2 x < ft = 4f from #

Free Body Diagram

#

A

1+ft < ft

1430lb

/300lb

/800lb:ft 3

12

4ft/ft A

#

cpe 461 assignment uitm

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