# Stringers and Shores

The joist span selected will be based on the spacing of the string­ers. We can follow the same procedure that is used for analyzing joists. Again, an integer or modular value is selected for stringer spacing.

After calculating the stringer spacing, the span of the stringer is checked against the capacity of the shore. The load on each shore is equal to the shore spacing multiplied by the load per unit foot of the stringer. The maximum shore spacing will be the lower of these two values (based on joists loading or shore spacing).

In calculating the design load of stringers, we do not consider the effect of joists on stringers. Instead, as a good approximation, we calculate load transmitted directly from sheathing to stringers. Hence, it is necessary to check for crushing at the point where the joists rest on stringers. Finally, shores are designed as columns (compression member). In checking the capacity of the different elements of the form, we use the same equations that will be used in the analysis of the wall form design. These equations are shown in Tables 3.15 and 3.16.

 Type One span Two spans Three spans Bending moment (in.-lb) M = wl 96 m = wl 96 m = wl 120 Shear (lb) V = wl = 24 1 1 ‘O V = wl = 20 Deflection (in.) 5wl4 Д = 4608EI . wl4 Д = 2220EI wl4 Д = 1740EI
 Table 3.15 Design Equations for Different Support Conditions

 Notation: l = length of span (in.) w = uniform load per foot of span (lb/ft) E = modules of elasticity (psi) I = moment of inertia (in.4) Source: Reproduced from the 1998 edition of Construction Methods and Management by S. W. Nunnally, with the permission of the publisher, Prentice-Hall. Table 12-3, pp. 340­341.

EXAMPLE

Design formwork to support flat slab floor of 8-in. thickness and conventional density concrete. Sheathing will be plywood that has the following characteristics: [16]

 If д If д

 l = 1.57 l = 1.37

 1/3

 l = 2.10 l = 1.83

 1/3

 l 240 l 360

 1/3

 1/3

Notation:

l = length of span, center to center of supports (in.)

Fb = allowable unit stress in bending (psi)

FbKS = plywood section capacity in bending (lb x in./ft)

Fc = allowable unit stress in compression parallel to grain (psi)

Fcl = allowable unit stress in compression perpendicular to grain (psi)

FJb/Q = plywood section capacity in rolling shear (lb/ft)

Fv = allowable unit stress in horizontal shear (psi)

= actual unit stress in compression parallel to grain (psi)

= actual unit stress in compression perpendicular to grain (psi)

= actual unit stress in tension (psi)

= area of section (in.2)[17]

= modulus of elasticity (psi)

= moment of inertia (in.4)*

= applied force (compression to tension) (lb)

= section modulus (in.3)*

= deflection (in.)

= width of member (in.)

= depth of member (in.) w = uniform load per foot of span (lb/ft)

*For a rectangular member: A = bd, S = bd2/6,1 = bd3/12.

Reprinted by permission of Prentice Hall

SOLUTION

• Dead load = weight of concrete + weight of the form­work.

= 8/12 in. of concrete x 150 + 5.0 (assumed) = 105 lb/ft2

• Live load = 75 lb/ft2 (according to ACI = 347).

• Total vertical load = 75 + 105 = 180 lb/ft2

Sheathing:

• Thickness = 11/8-in. sanded panels.

• From Tables 3.12 to 3.14, we get the following properties:

Grade Stress Level S-2.

A = 3.854 in.2 (KS) = 0.820 in.3/ft.

I = 0.548 in.4

lb/Q = 9.883 in.2/ft.

Fb = 1200 psf Fv = 140 psf.

E = 1.2 x 106 psf •

 From Table 3.15, using 3 or more spans: Bending:

 get

 l1 = 25.612 in.

 Shear: Ws

 153.736 in.

 Deflection:

 І3 _ 180 x 13 , 360 1743 x 1.2 x 106 x 0.548 3

 26.055 in.

 Bending governs. Allowable span — 25.612 in., which is rounded down to 24 in. (2 ft). Hence, joist spacing — 2 ft. Joists: From Tables 3.6 to 3.8, we can get the following values for Red­wood (select-structural):

 • Fb — 1350 psi • Fv — 80 psi • Fcl — 650 psi • E — 1.4 x 106 psi • Temperature factor Ct — 1.0. • Load duration factor CD — 1.25 • Shear stress factor CH — 2.0 (no split case) Choose 2 x 6 Redwood which has the following characteristics: • A — 8.25 in.2 • S — 7.563 in.3 • I — 20.8 in.4 • d — 5.5 in. • Cf — 1.20

Load/ft of joist = 180 x 2 = 360 lb/ft Bending:

F’b = Fb( Cf)(Ct)(CD)

= 1350 x 1.3 x 1.0 x 1.25 = 2193.75 lb/ft From Table 3.16

Shear:

FV = F„ (Ch)(C,)(Cd) = 80 x 2 x 1.0 x 1.25 = 200 psi

= 13.5 x 200 x 8’25 + 2 x 5.5 = 72.875 in.

Deflection:

E’ = 1.4 x 106 psi

 Shear governs. Span of joists = 72.875 in. = 6 ft (OK)

Stringers:

Load on a stringer is = 180 x 6 = 1,080 lb/ft Use 2 x 8 (two stringers)

• b = 1.5 in.

• d = 7.25 in.

• I = 47.63 in.4

• A = 10.87 in.2

• S = 13.14 in.3

• E = 1.4 x 106 psi

• Cf = 1.2

Bending:

F’b = 1350 x 1.2 x 1.0 x 1.25 = 2025 psi

Shear:

I = 13.3 x |Fv^Aj + 2 x d

_ 13.3 x 200 x (10.87 x 2 ^ two stringers) 1080

+ 2 x 7.5 in. = 68.045 in.

 Note that we did not multiply the second term by 2 because, ac­cording to NDS, shear is checked at d/ 2 from the face of the sup­port. Deflection:

 1/3

 1.4 x 106 x 47.63 x 2

 I = 1.69 x

 84.157 in.

Shear governs. Maximum span <68.045 in. (5.67 ft). Take stringer spacing = 5.5 ft.

Check for Crushing (joist on stringer):

Force transmitted from joist to stringer is equal to load of joist/ ft multiplied by the span of the joist, force = 6 x 360 = 2160 lb.

Area through which this force is transmitted = 1.5 x 3 = 4.5 in.2

Crushing stress = 2160/4.5 = 480 lb/in.2 From Table 6.a, we get the following properties:

• Fc± = 650 lb/in.2

• Bearing area factor = 1.25 (b = 1.5 in.)

• Temperature factor = 1.0

F’c± = Fcl (Cb)(Ct) = 650 x 1.25 = 812.5 psi since 480 < 812.5 ^ safe in crushing.

Shore strength:

Required shore spacing = (stringer span)

Shore strength = span of stringer x load of stringer = 5.5 x 1080 = 5940 lb

So we use shores whose strength is larger than 6000 lb.