Wall Form Design Part I Lecture 3 Bearing or rushing Bearing Stresses (ompression Perpendicular to the Grain) Allowable stresses for compression perpendicular to the grain are available from tables providing wood properties for various species and grades of lumber. These allowable stresses may be modified (increased) if both of the following criteria are satisfied: Bearing is applied 3 inches or more from the end of the member being stressed. Bearing length is less than 6 inches. When criteria are met, the allowable stresses are modified by the following factor: l 0. 375 l Where l is the bearing length in inches measured along the grain of the wood. For round washers, assume l is equal to the diameter of washer. 2 Spring Quarter 2018 1
Bearing or rushing To check for a bearing failure (crushing of wood fibers), divide the imposed load by the area of contact and compare this determined actual bearing stress to the allowable bearing stress. If the actual bearing stress exceeds the allowable bearing stress, a failure results. The multiplying factors for indicated lengths of bearing on such small areas plates and washers are shown below. Bearing is applied 3 inches or more from the end of the member being stressed. Length of bearing, in 1/2 1 1 1/2 2 3 4 6 0r more Factor. 1.79 1.37 1.25 1.19 1.13 1.09 1.00 3 Design of olumn / Braces Wood members subjected to axial compression (compression parallel to the grain). The capacity of a wood column is dependent on the following properties: ross-sectional area. Slenderness Ratio. Allowable compressive stress parallel to the grain (the basic allowable stress depends on the wood species and grade. This allowable stress may be modified depending on the slenderness ratio). 4 Spring Quarter 2018 2
The slenderness Ratio The slenderness ratio is the ratio of the unsupported length (l) of a member to the width (d) of the face of the member under consideration. Two values of the slenderness ratio (l/d) must be calculated for wood members used in construction because buckling can occur about either axis of the cross-section. If a column is unbraced, the controlling slenderness ratio (the larger one) will be the one determined by using the dimension of the narrower face. For wood members, l/d cannot exceed 50. 5 The slenderness Ratio The following examples will illustrate the calculation of the slenderness ratios: 1. Unbraced olumn: a. Slenderness ratio parallel to narrow face: l d 72 1.5 48.0 6 ft or 72" 2 4 b. Slenderness ratio parallel to wide face: l d 72 3.5 20.57 The larger, thus controlling, slenderness ratio is 48 6 Spring Quarter 2018 3
The slenderness Ratio Since the column is unbraced in both dimensions, it is intuitive that the slenderness ratio on the narrow face would control. Note that if the column were unbraced and 7 foot long, the controlling slenderness ratio would be 56 (over 50) and column would not be permitted without modification: larger member section additional lateral bracing l d 84 1.5 56 7 The slenderness Ratio 2. Braced olumn: a. Slenderness ratio parallel to narrow face: l d 108 3.5 30.86 9' or 108" 4 6 Note: Use longest unbraced length - 9 feet 16' or 192".. Brace b. Slenderness ratio parallel to wide face: l d 192 5.5 34.91 The controlling slenderness ratio is 34.91 7' or 84" 5.5" 3.5" 8 Spring Quarter 2018 4
Sheathing 9 Wall Forms Lateral Pressure (psf) P = wh w: Unit weight of fresh concrete (pcf) h: Depth of plastic concrete (ft) Lateral Pressure Values hemistry oefficient, Unit Weight oefficient, W Use Table 5-4 for and W Maximum Lateral Pressure: P Max W P 10 Spring Quarter 2018 5
Wall Forms Maximum Lateral Pressure (psf) For wall height less than 14 feet and placement rate of less than 7 ft/hr: P Max W 9000R 150 T For all walls with placement rate of 7 to 15 ft/hr; and for walls placed at less than 7 ft/hr, but having a placement height greater than 14 feet: P Max W 43,400 2,800R 150 T T R = Rate of Placement, feet per hour T = Temperature of concrete during placing, deg F 11 Wall Forms Table 5-5B gives the base values of lateral pressure 12 Spring Quarter 2018 6
Design forms for 14-foot high wall to be concreted at the rate of 3 feet per hour, internally vibrated. Assume the mix is made with 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 inch thick is available in 4x8-foot sheets, and 4500-lb coil ties are on hand. Framing lumber of No. 2 Douglas Fir-Larch is to be purchased as required. 13 Solution Step 1: Finding oncrete Pressure; Step 2: Sheathing; Step 3: Stud Size and Spacing of Wales; Step 4: Wale Size and Tie Spacing; Step 5: Tie Design; Step 6: Bearing heck; and Step 7: Lateral Bracing 14 Spring Quarter 2018 7
Refer to Table 5-4: Type I cement with no pozzolans or admixtures 15 STEP 1: FIND PRESSURE (using Table 5-5B) The concrete used for this project satisfied the conditions of Table 5-4. Using Table 5-5A, for R=3 ft/hr, and T=60F, the minimum pressure for design is: P = 600 psf Then the depth of the hydrostatic load zone, for a concrete with a unit weight of 150 pcf is: 600 psf H 4 feet 150 pcf 16 Spring Quarter 2018 8
STEP 1: FIND PRESSURE (Using Pressure Equation) P Max W 9000R 150 T From Table 5-4 W = 1 = 1 90003 P 150 150 450 600 psf 60 H 600 psf 150 pcf 4 feet 17 The diagram of lateral pressure on wall form is shown here: 4 14 10 600 psf 18 Spring Quarter 2018 9
19 STEP 2: SHEATHING 4x8 sheets of plywood will be used. Use plywood the "strong way" (face grain parallel to plywood span). Design for uniformly spaced supports at 1-ft center-to-center. HEK BENDING onsider a 12-in. wide strip of plywood. For continuous beams (more than three supports) the following equation is used: l 10. 95 fs w 20 Spring Quarter 2018 10
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 by an adjustment factor of 1.25 for short term loading. Hence, the allowable stress: f 1.251545 1930 psi 21 From Table 4-3, the section modulus, S, for ¾- in. plywood is: 0.412 in. 3 w, loading of the beam for a 1-ft wide strip of plywood is p 600 psf w 600 lb/lf 12 in. 12 in. 22 Spring Quarter 2018 11
Substituting in the equation: l 10. 95 fs w l 1930 0.412 10.95 10.95 1.33 12.61 600 in. 23 HEK DEFLETION: Again considering a 12-in. width of plywood sheathing, check the maximum allowable deflection (D) 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 elasticity for plywood can be found as E = 1,500,000 psi, and Table 4-3 renders the value for the moment of inertia for plies parallel to the span as I = 0.197 in. 4. For D = l/360: l EI 1500000 0.197.69 3 1.69 1.69 7.897 13.35 w 600 1 3 For D = 1/16 : l EI 1500000 0.197.234 3.23 3.23 4.41 15.22 w 600 3 4 in. in. 24 Spring Quarter 2018 12
HEK ROLLING SHEAR: From Table 4-2, allowable Fs (rolling shear stress) can be found to be F s = 57 psi, which should be multiplied by 1.25 for short-term loading. Therefore, the allowable rolling shear stress is: F S 57 1.25 71 psi 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 a continuous plyform and solve for L: FS L 0.6w Ib Q 71 6.762 1.33 ft. 16.0 in. 12" OK 0.6600 25 SPAING OF THE STUDS: From the above calculations, the smallest value obtained for l is 12.61 in. (bending governs), meaning that the studs ANNOT be placed any further than 12.61 inches apart. We are using 8-ft.-wide plywood sheets. The sheets should have stud support at the joints. Therefore an equal-spacing of studs at 12- inches satisfies all conditions. USE STUDS WITH SPAING OF 1-FT. 26 Spring Quarter 2018 13
Wall Form Example 27 Questions? Kamran M. Nemati nemati@uw.edu Architecture Hall Room 130J Spring Quarter 2018 14