Transcript SEISMIC BUILDING CODE OF PAKISTAN

Design Example
 The 10” TH. wall system shown in the figure below is to be checked for
a service gravity load of 3 Kips/ft and a lateral load of 25 Kips, acting in
plane to the wall.
 The lintel bearing is as per SBC 07.
 The masonry unit consist of 1500 psi solid concrete blocks with type M1
cement - sand mortar.
15 ft
10 ft
5 ft
7 ft
10 ft
Analysis (Masonry Frame Properties)
 W1:
45 kip
30 kip
Cross-Section Area = 10” x (15’x12) = 1800 sq.in.
3 + 0.36 Kips /ft
Inertia = 10” x (15’x12)^3 / 12 = 4860000 in4.
25 Kips
Weight = 0.144 x (1800 / 144) x 10” = 18 Kips
Load from slab = 3 x 15’ = 45 Kips
 W2:
 Cross-Section Area = 10” x (10’x12) = 1200 sq.in.
10 ft
 Inertia = 10” x (10’x12)^3 / 12 = 1440000 in4.
W1
W2
 Weight = 0.144 x (1200 / 144) x 10” = 12 Kips
 Load from slab = 3 x 10’ = 30 Kips
 Beam:
5 ft
 Cross-Section Area = 10” x (3’x12) = 360 sq.in.
 Inertia = 10” x (3’x12)^3 / 12 = 38880 in4.
 Weight = 0.144 x (360 / 144) = 0.36 Kips / ft




Analysis Results
Axial Force (K)
Shear Force (K)
Bending Moment (K.ft)
Wall 1
 Compressive stress:
 P/A = (45+24.7) / 1800 = 0.0387 Ksi or 38.7 psi
 Basic compressive stress = 0.96 MPa or 139.23 psi






Section 9.6.6.1 (Table 9-b)
Slenderness ratio = H/W = (10x12)/10 = 12
Section 9.6.6.1.1 (Table 9-b)
ks = 0.84
Area of element = 1800/144 = 12.5 sq. ft (>2.5 sq. ft)
Section 9.6.6.1.2 (Table 9-b)
ka = 1
Assuming height of a unit as 6 in & 10 in width: height to width ratio = 6/10 = 0.6 (<0.75)
kp = 1
Section 9.6.6.1.3 (Table 9-b)
 Allowable compressive stress = ks ka kp x Basic compressive Stress
0.84 x 1 x 1x 139.23 = 116.9 psi (OK)
 Flexural Stresses:




P/A + My/I = 0.0387 + (199x12)(7.5x12)/4860000 = 0.0809 Ksi or 80.9 psi
P/A - My/I = 0.0387 - (199x12)(7.5x12)/4860000 = -0.0035 Ksi or 3.5 psi
Allowable compressive stress = 116.9 psi (OK)
Allowable tensile stress = 20 psi (OK)
Section 9.6.6.2
 Shear Stresses:
 V/A = 19.7 / 1800 = 0.0109 Ksi or 10.9 psi
 Allowable stress = fs = 0.1 + fd /6
= 0.1 + (38.7x0.0068)/6 = 0.1155 MPa or 16.75 psi (OK)
Section 9.6.6.3
Wall 2
 Compressive stress:
 P/A = (30+22.1) / 1200 = 0.0434 Ksi or 43.4 psi
 Basic compressive stress = 0.96 MPa or 139.23 psi






Section 9.6.6.1 (Table 9-b)
Slenderness ratio = H/W = (10x12)/10 = 12
Section 9.6.6.1.1 (Table 9-b)
ks = 0.84
Area of element = 1200/144 = 8.33 sq. ft (>2.5 sq. ft)
Section 9.6.6.1.2 (Table 9-b)
ka = 1
Assuming height of a unit as 6 in & 10 in width: height to width ratio = 6/10 = 0.6 (<0.75)
kp = 1
Section 9.6.6.1.3 (Table 9-b)
 Allowable compressive stress = ks ka kp x Basic compressive Stress
= 0.84 x 1 x 1x 139.23 = 116.9 psi (OK)
 Flexural Stresses:




P/A + My/I = 0.0434 + (42.4x12)(5x12)/1440000 = 0.0646 Ksi or 64.6 psi
P/A - My/I = 0.0434 - (42.4x12)(5x12)/1440000 = -0.0222 Ksi or 22.8 psi
Allowable compressive stress = 116.9 psi (OK)
Allowable tensile stress = 20 psi (OK) (No Tensile Stresses)
Section 9.6.6.2
 Shear Stresses:
 V/A = 5.3 / 1200 = 0.0044 Ksi or 4.4 psi
 Allowable stress = fs = 0.1 + fd /6
= 0.1 + (18.4x0.0068)/6 = 0.120 MPa or 17.4 psi (OK)
Section 9.6.6.3
Beam
 Compressive stress:
 P/A = 5.3 / 360 = 0.0147 Ksi or 14.7 psi
 Basic compressive stress = 0.96 MPa or 139.23 psi






Section 9.6.6.1 (Table 9-b)
Slenderness ratio = H/W = (5x12)/10 = 6
Section 9.6.6.1.1 (Table 9-b)
ks = 1
Area of element = 360 /144 = 2.5 sq. ft (=2.5 sq. ft)
Section 9.6.6.1.2 (Table 9-b)
ka = 1
Assuming height of a unit as 6 in & 10 in width: height to width ratio = 6/10 = 0.6 (<0.75)
Section 9.6.6.1.3 (Table 9-b)
kp = 1
 Allowable compressive stress = ks ka kp x Basic compressive Stress
=1 x 1 x 1x 139.23 = 139.23 psi (OK)
 Flexural Stresses:
 My/I = (10.6x12)(1.5x12)/38880 = 0.0588 Ksi or 58.2 psi
 Allowable compressive stress = 116.9 psi (OK)
 Allowable tensile stress = 20 psi (Not OK) (Provide reinforcement) Section 9.6.6.2
 Shear Stresses:
 V/A = 6.7 / 360 = 0.0186 Ksi or 18.6 psi
 Allowable stress = fs = 0.1 + fd /6
Section 9.6.6.3
= 0.1 + (14.7x0.0068)/6 = 0.116 MPa or 16.82 psi (Not OK) (Inc. Section)
Bearing Stress Check (Lintel)
 Bearing length = 4 in or 1/10th of span
Section 9.6
 (5x12)/10 = 6 in (Take 6 in)
 Bearing load = (3 + 0.36) x 5 / 2 = 8.4 Kips
 Bearing area = 10 x 6 = 60 sq. in
 P/A = 8.4/60 = 0.14 Ksi or 140 psi
 Allowable Compressive stress = 139.23 psi (Not OK)
 Bearing area with 30° inclination. = 10 x (6+6sin30)= 90 sq.in
 P/A = 8.4/90 = 0.0933 Ksi or 93.3 psi
 Allowable Compressive stress = 139.23 psi (OK)
Section 9.6