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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. - PowerPoint PPT Presentation
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Design ExampleThe 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
5 ft 10 ft
10 ft 7 ft
Analysis (Masonry Frame Properties)
3 + 0.36 Kips /ft25 Kips
5 ft
10 ft
W1:Cross-Section Area = 10” x (15’x12)
= 1800 sq.in. Inertia = 10” x (15’x12)^3 / 12 =
4860000 in4.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. Inertia = 10” x (10’x12)^3 / 12 =
1440000 in4.Weight = 0.144 x (1200 / 144) x 10”
= 12 Kips Load from slab = 3 x 10’ = 30 Kips
Beam: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
W1 W2
45 kip 30 kip
Analysis ResultsAxial 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
Slenderness ratio = H/W = (10x12)/10 = 12ks = 0.84Area of element = 1800/144 = 12.5 sq. ft (>2.5 sq. ft)ka = 1Assuming height of a unit as 6 in & 10 in width: height to width
ratio = 6/10 = 0.6 (<0.75)kp = 1
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)
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 9.6.6.1 (Table 9-b)Section 9.6.6.1.1 (Table 9-b)
Section 9.6.6.1.2 (Table 9-b)
Section 9.6.6.1.3 (Table 9-b)
Section 9.6.6.2
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
Slenderness ratio = H/W = (10x12)/10 = 12ks = 0.84Area of element = 1200/144 = 8.33 sq. ft (>2.5 sq. ft)ka = 1Assuming height of a unit as 6 in & 10 in width: height to width
ratio = 6/10 = 0.6 (<0.75)kp = 1
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)
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 9.6.6.1 (Table 9-b)Section 9.6.6.1.1 (Table 9-b)
Section 9.6.6.1.2 (Table 9-b)
Section 9.6.6.1.3 (Table 9-b)
Section 9.6.6.2
Section 9.6.6.3
Beam Compressive stress:
P/A = 5.3 / 360 = 0.0147 Ksi or 14.7 psiBasic compressive stress = 0.96 MPa or 139.23 psi
Slenderness ratio = H/W = (5x12)/10 = 6 ks = 1 Area of element = 360 /144 = 2.5 sq. ft (=2.5 sq. ft) 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
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 psiAllowable compressive stress = 116.9 psi (OK)Allowable tensile stress = 20 psi (Not OK) (Provide reinforcement)
Shear Stresses:V/A = 6.7 / 360 = 0.0186 Ksi or 18.6 psiAllowable stress = fs = 0.1 + fd /6
= 0.1 + (14.7x0.0068)/6 = 0.116 MPa or 16.82 psi (Not OK) (Inc. Section)
Section 9.6.6.1 (Table 9-b)Section 9.6.6.1.1 (Table 9-b)Section 9.6.6.1.2 (Table 9-b)Section 9.6.6.1.3 (Table 9-b)
Section 9.6.6.2
Section 9.6.6.3
Bearing Stress Check (Lintel)Bearing length = 4 in or 1/10th of span
(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. inP/A = 8.4/60 = 0.14 Ksi or 140 psiAllowable 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 psiAllowable Compressive stress = 139.23 psi (OK)
Section 9.6
Section 9.6
