kontribusi kekakuan balok sekunder

Tulisan ini merupakan kelanjutan dari sebelumnya, mengenai distribusi beban dua arah pada struktur beton bertulang dimana saat ini ditinjau pengaruh balok sekunder/pembagi yang biasa ditemui pada bangunan lokal.

The actual distribution of loads on these elements is quite complex. ETABS uses the concept of tributary loads as a simplifying assumption for transforming the loads (CSi,1999)

I took simple frame models to check, it contains secondary beams with different stiffness near to 4 times lower compared to main beams.

Figures above shows extruded 3d views, left models with automatic area loads to frame done by SAP another using finite element models with considering beam to slab eccentricity.

Bending moment diagrams, right models has low values.

Zero axial force in beams for left models, contrary high axial force (tension) exist in right models due to eccentricity modeling with rigid links.

Slab act in compression by couple tension force in beams, it may be figuring as tee-beam with flanges effective width.

(source: Chiewanichakorn, etal. 2001)

Using tributary areas load concept will ignore variable stiffness of beams  in a panel being considered. FE models do betters in prediction but required  additional nodes and computational effort. In case slab panel  with secondary beams, bending moment will varies depending geometry and beams sectional properties. Another approach is remove secondary beams with equivalent slab stiffness with modified flexural rigidities of plates may required for this cases to avoid unacceptable results and it’s behavior.

Calculation pads,

‘ Tributary area concept
‘ Beam-Y
‘ B30/60
M_1 = 69.7;
‘ B30/40
M_2 = 69.56;
‘ Beam-X
‘ B30/60
M =116.49;
‘ FE approximation
‘ Beam-Y
‘ B30/60
M_a=177.06*0.24= 42.49
M=M_e+M_a= 83.08 kN.m
‘ B30/40
M_a=168.63*0.14= 23.61
M=M_e+M_a= 55.93 kN.m
‘ Beam-X
‘ B30/60
M_a=214.46*0.24 = 51.47
M=M_e+M_a = 119.40 kN.m
‘ Flexural Reinforcement (hand calculations)
‘ Tributary approximation
‘ Beam-Y
‘ A_s1 < A_smin = 590.10 mm2 (B30/60)
‘ A_s2 = 634.89 mm2 (B30/40)
‘ Beam-X
‘ A_s = 672.62 mm2
‘ FE approximation
‘ Beam-Y
‘ A_s1 < A_smin = 590.10 mm2 (B30/60)
‘ A_s2 = 502.10 mm2 (B30/40)
‘ Beam-X
‘ A_s = 690.11 mm2 (B30/60)

‘ Beam Stiffness Ratio (middle & edge)
E_c=(4700*Sqrt(25))*10.19716= 239633.26 kgf/cm2
L = 600 cm
b1 = 30 cm
h1 = 60 cm
Ix1=1/12*b1*h1^3= 540000.00 cm4
b2 = 30 cm
h2 = 40 cm
Ix2=1/12*b2*h2^3= 160000.00 cm4
K_db1 = (48*E_c*Ix1)/L= 10352156832.00
K_db2 = (48*E_c*Ix2)/L= 3067305728.00
R_db=K_db1/K_db2= 3.38 (~4 times stiffness)
‘ Beam internal moment ratio (secondary beams 30/40)
M_x1 = 69.56;
M_x2 = 55.93;
R_mx = M_x1/M_x2= 1.24 (~ 25%higher)

‘ Beam internal moment ratio (main beams 30/60)

M_x3 = 69.70;
M_x4 = 83.08;
R_mx = M_x3/M_x4= 0.84 (~15%lower)

Leave a Reply

Please log in using one of these methods to post your comment:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s