SOLUTIONS OF ASSIGNMENT NO THE RESULTANT FORCE CONCURRENT

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					                 SOLUTIONS OF ASSIGNMENT NO. (4)


                          THE RESULTANT FORCE

             OF A NON-CONCURRENT FORCE SYSTEM
Problem 4.111

Replace the force system acting on the beam by an equivalent force and
couple moment.


                                                1.5 KN
                             2.5 KN
                                                                        3 KN
                                  5                    30°
                                      3
                                  θ
             A                                                             B




                     2m               4m                           2m




                                                    1.5 cos 30°
                      2.5 sin θ
                                                                        3 KN


                 2.5 cos θ            1.5 sin 30°
             A                                                                 B

                                                             (1)
                                                                        FRy
                                          ≡


                                                                                   FRx
                 A                        (2)                            B MB

cos θ = 4/5 , sin θ = 3/5

FRx = ∑ Fx



                                                ١
   = 1.5 sin 30° - 2.5 cos θ

   = 1.5 sin 30° – 2.5 * (4/5)

   = - 1.25 KN

FRx = 1.25 KN ←

FRy = ∑ Fy

    = - 3 – 1.5 cos 30° – 2.5 sin θ

    = - 3 – 1.5 cos 30° – 2.5 * (3/5)

   = - 5.8 KN

FRy = 5.8 KN ↓

FR =   (1.252 + 5.82 )

                             5.8
FR = 5.93 KN     , tanθ =
                            1.25

θ = 77.8°

∑ MB = 1.5 cos 30 (2) + 2.5 *(3/5) (6)

∑ MB = 11.6 KN. m (counterclockwise)




                                        ٢
Problem 4.113

Replace the three forces acting on the shaft by a single resultant force. Specify
where the force acts, measured from end B.



                     2.5 m              1.5 m         1m             2m
        A                                                                           B


                             α                                  β

                                   3                       12   13
                             5
                             4                                  5
            500 KN                           200 KN
                                                                     260 KN
                                                                                    y


   A                                                                                                    x
                                                                                            B
                     500 cos α                                      260 cos β

                                             200 KN

                                 500 sin α                 260 sin β

                                                                       (1)
                                        ≡                                       y           FRy

                                                                                        x         FRx
                                                                                            y
                                                                                                   x
         A                                                                      B
                                                                       (2)

cos α = 4/5   , sin α = 3/5 , cos β = 5/13 , sin β = 12/13

FRx = ∑ Fx| 1

     = 260 cos β – 500 cos α

     = 260 * (5/13) – 500 * (4/5)

     = - 300 N

FRx = - 300 N

                                                ٣
FRy = ∑ Fy| 1

   = - 260 sin β – 200 – 500 sin α

   = -260 * (12/13) – 200 – 500 * (3/5)

FRy = - 740 N

FRx = 300 N ←          FRy = 740 N ↓
                                                  300 N
        2        2                                              θ
FR =   FRx +    FRy
                                                          FR
FR = 300 2 + 74o 2
                                                                      740 N

FR = 798.5 N

tan θ = 740 /300 =2.467

θ = 67.9°

MB| 1 = MB| 2

260 * (12/13) * 2 + 200 * 3 + 500 * (3/5) * 4.5 = FRy (x) – FRx (y)

2430 = - 740 x + 300 y (equation of the line of action of the resultant)

At y = 0

2430 = - 740 x

x =- 3.28 m

At x = 0

2460 = 300 y

y = 8.1 m




                                     ٤
             Problem 4.114

             Replace the loading on the frame by a single resultant force. Specify where its line
             of action intersects member AB, measured from A.




         y
                                                                          y       FRy

                                  200 N           400 N                       x         FRx
             ٣٠٠ N
                0.3 m                     0.4 m
                                                                                  y
                              x                       B                                  x              B

         A                                                0.2 m           A
60 N.m
                                                                  200 N
                        (1)
                                                                                        (2)



                                                          0.7 m




                                                                                                    C
                                                              C




             ∑ Fx| 1 = ∑ Fx| 2

             FRx = ∑ Fx| 1

                 = - 200 N

             FRx = - 200 N

             ∑ Fy| 1 = ∑ Fy| 2


                                                                  ٥
FRy =∑ Fy| 1

   = - 400 – 200 – 300

   = - 900 N

FRy = - 900 N
                                                         200 N
                                                                        θ
FRx = 200 N ←             FRy = 900 N ↓
                                                                   FR
      2     2
FR = FRx + FRy
                                                                            900 N
FR = 200 2 + 900 2

FR = 922 N

tan θ =900/200 = 4.5

θ = 77.47°

MA| 1 = MA| 2

60 – 200 * (0.3) – 400 * (0.7) – 200 * (0.2) = FRy (x) – FRx (y)

- 320 = -900 x + 200 y (equation of the line of action of the resultant)

At y = 0 (intersection with member AB)

-320 = - 900 x

x = 0.356 m




                                     ٦
Problem 4.115

Replace the loading acting on the beam by a single resultant force. Specify where
the force acts measured from end A.


                                                  300 N
              450 N                                                            700 N
                                                                        30°



      A            60°                        B


                                                                               1500 N.m
              2m                     4m                      3m


                                              300 N                 700 cos 30°
                   450 sin 60°

         450 cos 60°                                      700 sin 30°
     A                                        B


                                                                              1500 N.m
              2m                     4m                      3m



          y                      ≡                            (1)

                      FRy


                                 FRx                   (2)
              x
                         y
          A                                        B                                      x




∑ Fx| 1 = ∑ Fx| 2

FRx = ∑ Fx| 1


FRx = 450 cos 60° - 700 sin 30°


                                          ٧
FRx = - 125 N


∑ Fy| 1 = ∑ Fy| 2

FRy =∑ Fy| 1                                                 125 N
                                                                 θ
     = - 450 sin 60° - 300 – 700 cos 30°                     FR
                                                                          1296 N
FRy = - 1296 N

FRx = 125 N ←               FRy = 1296 N ↓




      2     2
FR = FRx + FRy
F R = 125 2 + 1296 2

FR = 1302 N

tan θ =1296/125 = 10.368
θ = 84.5°

MA| 1 = MA| 2

- 450 sin 60° *(2) – 300 *(6) – 700 cos 30° *(9) – 1500 = FRy (x) – FRx (y)

- 9535.4 = - 1296 x + 125 y

At y = 0

- 9535.4 = -1296 x

x = 7.36 m




                                     ٨
Problem 1.121

Replace the loading on the frame by a single resultant force. Specify where its line
of action intersects member CD measured from end C.


                                        300 N                                250 N
                         1m        2m              3m                5
                                                                              3
              C               B
                                                                         4

                                  2m                             D
     400 N.m



                   60°


                                  3m


               500 N


                              A


               y


                                                              250 cos α
                                        300 N
                         1m        2m              3m
          C                   B
                                                                                       x
                                                                     250 sin α
                                  2m                             D
     400 N.m


       500 cos 60°

                                  3m
          500 sin 60°
                                                                     (1)


                              A           ≡



                                          ٩
               y
                                    FRy

                      x                        FRx

                                y

             C            B                                    x

                                                           D
                                                     (2)




                          A

sin α = 4/5 , cos α = 3/5



∑ Fx| 1 = ∑ Fx| 2

FRx = ∑ Fx| 1


FRx = - 250 sin α – 500 cos 60°


FRx = - 450 N



∑ Fy| 1 = ∑ Fy| 2

FRy =∑ Fy| 1


     = - 300 – 250 cos α – 500 sin 60


FRy = - 883 N



                                          ١٠
FRx = 450 N ←              FRy = 883 N ↓
                                                            450 N
                                                                 θ
                                                            FR
        2        2
FR =   FRx +    FRy
                                                                         833 N
F R = 450 2 + 833 2

FR = 991 N

tan θ =833/450 = 1.962

θ = 63°

MC| 1 = MC| 2

400 – 300 * (3) – 250 * (3/5) * (6) – 500 cos 60° * (2) – 500 sin 60° * (1)

= FRy (x) – FRx (y)

- 2333 = - 883 x + 450 y

At y = 0

- 2333 = - 883 x

x = 2.64 m (intersection of the line of action of the resultant with member
CD measured from end C)




                                    ١١

				
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posted:4/10/2011
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