0 Wall Form Design ExampleDesign forms for 14 ft high wall to be concreted at the rate of 3 ft per hour, internally vibrated. Assume the mix is made wit h Type I cement, with no pozzolans or admixtures, and that the temperature of concrete at placing is 60°F. Slump is 4 in. The forms will be used only once, so short-term loading stresses will apply. Form grade plywood sheathing ¾ in. thick is available in 4x8-ft sheets, and 4500-lb coil ties are on hand. Framing lumber of No. 2 Douglas Fir-Larch is to be purchased as required.
Design forms for 14 ft high wall to beconcreted at the rate of 3 ft per hour, internallyvibrated. Assume the mix is made with Type Icement, with no pozzolans or admixtures, and
that the temperature of concrete at placing is60°F. Slump is 4 in. The forms will be usedonly once, so short-term loading stresses willapply.
Form grade plywood sheathing ¾ in. thick isavailable in 4x8-ft sheets, and 4500-lb coil tiesare on hand. Framing lumber of No. 2 DouglasFir-Larch is to be purchased as required.
STEP 2: SHEATHING.4x8 sheets of plywood will beused. Use plywood the "strongway" (face grain parallel toplywood span). Design for uniformly spaced supports at1-ft center-to-center.
CHECK BENDINGConsider a 12-in. wide strip of plywood.For continuous beams (more than three supports)the following equation is used:
Wall Form Design Example From Table 4-2, the bending stress for plywood is 1545 psi.
The problem states that the forms will be used only once(single-use form), the bending stress must be multiplied byan adjustment factor of 1.25 for short term loading. Hence,the allowable stress:
Again considering a 12-in. width of plywoodsheathing, check the maximum allowable deflection(( ) of the sheathing for l /360 of the span and 1 /16 in.,whichever is less.
From Table 4-2, the values for modulus of elasticityfor plywood can be found as E = 1,500,000 psi, andTable 4-3 renders the value for the moment of inertiafor plies parallel to the span as I = 0.197 in. 4.
From Table 4-2, allowable Fv (rolling shear stress) can be found to be F v = 57 psi, whichshould be multiplied by 1.25 for short-termloading. Therefore, the allowable rollingshear stress is:From Table 4-3, the value of the rolling shear constant, Ib/Q , can be found as 6.762.
Use the equation for maximum shear for acontinuous plyform and solve for L :
From the above calculations, the smallestvalue obtained for l is 12.61 in. (bendinggoverns), meaning that the studs
CANNOT be placed any further than 12.61inches apart.W e are using 8-ft.-wide plywood sheets.The sheets should have stud support atthe joints. Therefore an equal-spacing of studs at 12-inches satisfies all conditions.
4 S4S studs. Find themaximum span that can support alateral pressure of 600 psf.Equivalent uniform load, w, is the max.lateral pressure times the stud spacing.Hence:
Studs can be considered as continuous beamssubjected to uniform loading. Like the previousset of calculations, check for allowable span for
Wall Form Design Example The first adjustment factor is the short - term l oading f actor of 1.25. The secondadjustment factor is the size f actor obtainedfrom Table 4-2B, which is 1.5. Therefore:
From Table 4-2, allowable F v (rolling shear stress) can be found to be F s = 95 psi, whichshould be multiplied by 1.25 for short-termloading, as well as by a factor of 2.0 for horizontal shear adjustment. Therefore, theallowable shear stress is: psi 22525.118 0 !v!d
V F
A 2x4 S4S has an actual b = 1 ½ in. and d = 3 ½in., which is obtained from Table 4-1B. Usethe equation for maximum shear for acontinuous beam and solve for l :
ALESFrom the stud spans calculated above, theshortest span is based on bending whichis 32.1 inches.This means the wales, which are the studsupports CANNOT be spaced more than32.1 inches apart (this span can be increasednear the top, since in the top 4 ft., the pressuredecreases linearly from 600 psf to 0).
The top and bottom wales are often setabout 1 ft from top and bottom of wall
Place wales 12 inches from both topand bottom of the wall form.Then, 14' 1' 1' = 12 ft. or 144 inches
remains for spacing the other wales,which can be no more than 32.1 inchesapart.Set them at 30 in., except one span at24 in. (smaller spans at the bottom).W e place the smaller span near thebottom of the form where theoretically
ALE SIZEand TIE SPACINGFrom the pressure diagram, theequivalent uniform load per lineal footof wale is determined to be 1500 lb /lf.The problem statement indicates that4500-lb coil ties are available and will
b is the allowable stress in theextreme fiber and was calculated to be1640 psi (refer to top of page 16 of the
handout). The span, l , is 3 ft. or 36inches, and w = 1500 lb /lf.Therefor the required section modulus, S , can be calculated using the aboveequation:
Wall Form Design Example In order to avoid drilling of timbers, theycommonly use double-member wale. Sothe required section modulus of 9.6 in. 3 isfor two members.
Referring to Table 4-1B, double 2 x 4s willyield a section modulus of 2 x 3.06 or 6.12in. 3, which is less than 9.6 in. 3, andtherefore not acceptable.
Checking the next larger size, 3 x 4, willresult in: S = 2x 5.10 = 10.20 in. 3 > 9.6 in. 3,which satisfies the section modulusrequirements. Use double 3 x 4 wales .
Wall Form Design Example CHECK SHEARTo check the horizontal shear for the double 3 x 4wales, use the horizontal shear stress formulafor a uniformly loaded continuous beam.
Consider the necessary bracing for awall form 14 ft. high, above grade, in anarea where the local building code
specifies a minimum 20 psf windloading.Table 5-7 indicated that 140 lb per linealfoot should be used for design of bracing, since the wind force prescribedby local code gives a value larger thanthe 100 lb /ft minimum established byACI Committee 347.
Wall Form Design Example Bracing for Lateral Loads
Wall Form Design Example Bracing for Lateral Loads
Strut BracingIf wooden bracing is attached 1 or 2 feetbelow the top of the wall, the bracing mustcarry more than the 140 lb per ft loadapplied at the top.
Wall Form Design Example Bracing for Lateral Loads
Strut BracingH ¶ the horizontal bracing force 2 feet fromthe top of the wall would have to be
(14¶ /12¶)(140 lb /ft) = 163 lb per ft
in order to balance the 140 lb /lf designload applied at the top of the wall.If end of the brace is put 8 feet from thewall, use the relationship between sides of the right triangle to find the the length of brace and load it must carry.
Wall Form Design Example Bracing for Lateral Loads
If struts are spaced every 8 feet alongthe wall, then 8 x 294 = 2352lb must becarried by each brace.M any wood members strong enoughto carry this load in compression willalso be adequate in tension. However,
the strength of connections (nails,etc.) must be made adequate for thetension load.
