Kamis, 03 Juni 2010

balok sederhana


Konstruksi Balok Sederhana


Contoh soal dan penyelesaian MK Statika pelengkung tiga sendi
                                                q= 1 t/m


                                                            S                                                 
                                                                                                                                c=4m

  A                                                                                           B            
 

                                a=4m                                     b=4m
                                                L= 8m

Ditanyakan ;       a. reaksi perletakan
                                b. bidang M,D,N

Penyelesaian:
                Analisis
q= 1 t/m


                                                            S                                                 
                                                                                                                                c=4m
   RAH                                                                                                    RBH
  A                                                                                           B
                   RAV                                                                    RBV       
 

                                a=4m                                     b=4m
                                                L= 8m

a.       Reaksi perletakan
Yang perlu selalu diingat dalam penyelasaian kasus-kasus pada statika tertentu adalah :
∑M=0
∑V=0
∑H=0
Dalam kasus ini yang diselesaikan adalah :
∑MA=0
∑MB=0

∑MS=0 (bagian A-S atau B-S, dan salah satunya digunakan untuk control atau mengecek apakah sudah ok apa belum)

Q=q.L = 1 t/m . 8m = 8t

∑MA=0
       0 = -RBV.L + Q.a
       0 = - RBV.8 + 8.4
 RBV = 32/8 = 4t

∑MB=0
       0 = -RAV.L + Q.b
       0 = - RAV.8 + 8.4
 RAV = 32/8 = 4t

∑MS=0                  (bentang A-S)
       0 = - q.a. ½ a + RAV.a – RAH.c
       0 = - 1.4. ½ 4 + 4.4 – RAH.4
       0 = - 8 + 16 – RAH.4
 RAH = 8/4 = 2t (         )

∑H=0    
       0 =  RAH – RBH
       0 = 2  – RBH
 RBH =  2t (         )

Control
∑MS=0                  (bentang B-S)
       0 = - q.b. ½ b + RBV.b – RBH.c
       0 = - 1.4. ½ 4 + 4.4 – RBH.4
       0 = - 8 + 16 – 2.4
       0 = 0  ( ok )
                q= 1 t/m


                                                            S                                                 
                                                                                                                                c=4m
   RAH = 2t                                                                                            RBH=2t
  A                                                                                           B
                   RAV= 4t                                                                             RBV= 4t               
 

                                a=4m                                     b=4m
                                                L= 8m


b.      Bidang M,D,N
Untuk menyelesaikan bidang M,D,N selalu mengadakan Potongan…dan peninjauan
Adakan potongan sejauh x dari A (bentang A-S) dan tinjau kiri
Tg α= 4/4 = 1, α= 450
                qx
                                                         y= x tg α
       RAH= 2t         450                                      


                                X
                RAV= 4t
                qx= q.x = x
                Mx = +RAV.x – qx. ½ x – RAH.y
                       = + 4x – ½ x2  – 2y      0≤ x ≤ 4
                X=0        Mx = 0
                X=4        Mx = 0
                Mx max atau titik puncak momen
                dMx/dx = 0 = 4 – x – 2
                 puncak saat x = 2 dari A
                Mx max = 4.2 – ½ 22 – 2.2 =  2 tm
               
Untuk bidang D
                Tg α= 4/4 = 1, α= 450
                qx
            RAH sinα                                          
 RAH= 2t               450                                      


                                X
      RAV= 4t          RAV cos α
                Dx= RAVcosα – RAHsinα – qxcos2α
                     = 4. ½ √2 – 2. ½ √2 – x. ½
                     = 2√2 - √2 – ½ x            0≤x≤4
                X=0        Dx= √2= 1,41
                X=4        Dx = √2 – 2= - 0,59

                Untuk bidang N


                                qx
                                                        
       RAH= 2t         450                                      
RAHcosα
+ RAVsinα                           X
                RAV= 4t
                Nx = RAHcosα+ RAVsinα – qxsin2α
                      = 2. ½ √2 + 4. ½ √2 – ½ x
                     = √2 + 2√2 – ½ x
                     = 3√2 – ½ x                     0 ≤x ≤4
                x=0         Nx = 3√2=4,23
                x=4         Nx = 2,23
               
                karena kedua bentang sama (A-S dan B-S).. maka tidak perlu susah-susah menghitung lagi karena hasilnya akan sama. Hanya saja tandanya akan berbeda dalam proses penyelesaian.

                                                                                                                q= 1 t/m


                                                                                                                S


                                                                                                                                                                                                                                                                C=4m


RAH = 2t                                                                                                                                                                                                               RBH=2t
                                                A                                                                                                                                     B

                                                RAV= 4t                                                                                                                                RBV=4t


                                                                a=4m                                                                     b=4m





                                                                                +                                                             +                                                             Bidang Momen (M)





-                        -




       +                                                                                                                      +                             Bidang Gaya Lintang (D)









                                                                                +                                                                             +                                             Bidang Gaya Normal (N)