Question

3. Figure 3 shows a rigid frame and the factored loads acting on the frame. Design the stirrups in the beam BC only. fe-5000 psi and fy 60000 psi. Hint: Solve for support reactions, draw dd n 50 kip Wu 3.1 kip/ft 50 kip 12 in by A Figure 3 Columns are 12 in. by 14 11 ft 20 ft 14 in. 14 in 1,0 kipーーグ A

Answer 1

Draw the rigid below frame with factored loads acting on the frame as shown 50 kips Wu=3.1 kips/ft 50 kips 7 kips 12 in. by 20 in ||B Columns are 12 in. by 14 in 11 ft k14 in 14 in. 1.0 kips 8.0 kips 20 ft a) Calculate shear force at mid-span using relation Pxh Shear force at mid span /, Here, P is horizontal load, h is height of frame, and /, is clear span 7 x11 Shearforce at mid-span = 20 3.85kips

Calculate shear force at the end span wind from right using the relation +Shear force at the mid span 2 Shear force at the end span wind from right 3.10 x 20 Shear force at theend span wind from right = +3.85 2 -34.85kips Calculate shear force at the end span wind from left using the relation Shear force at the end span wind from left- Wuxl Shear force at themid span 2 n 3.10 x 20 Shear force at the endspan wind from left - 2 -3.85 -27.15 kips Calculate ratio between factored design end shear and strength reduction V factor

Ratio between factored design end shear V and strength reduction factor Ratio between factored design end shear] 34.85 and strength reduction factor 0.75 46.5kips Determine the shear strength reduction factor from face of the column using the relation -+Shear forceat the mid span 2 Shear strength reduction factor from face of the column 3.10x 18.84 3.85 Shear strength reduction factor 2 from face of the column 0.75 44.10kips

Determine the shear strength reduction factor at d from face of the column using the relation Shear strength reduction factor 2 at d from face of the column 3.10x 18.54 Shear strength reduction factor 2 at d from face of the column 0.75 =38kips Draw the shear force envelope and shear-force with strength reduction factor as in Figure (2)

34.9 Wind from right 27.2 V kips 3.85 -3.85 Wind from left -27.2 |-34.9 46.5 44.1 38.0 Vu kips 5.1 -5.1 face of column d from face of column

Calculate nominal shear force carried by concrete V using the formula V 2xbxd Here, is specified compressive strength of concrete, b is width of web, and d is effective depth C Assume effective depth, d is 17.5 in 2/5,000 x 12 x 17.5 C 29,698.48(lb) kip 1,000 = 29.69kip V. Hence, V is less than Therefore, stirrups are required C Check stirrup anchorage. Use No.3 Grade 40 stirrups ACI code Sec.12.13.2.1. Allow these to be anchored by a hook around a top bar

d Check the maximum spacing as follows for 'max 4 V, <6xb,xd 46.565,000 x 12 x 17.5 < 89,095.45lb kips 46.5 1,000 46.5 89.1kips would not be suffice 4 Therefore, the maximum spacing of Calculate maximum spacing as follows: d Smax 2 17.5 max 2 = 8.75in Spacing of stirrups required to resist shear forces.

Calculate spacing s at d from face of column А, x f, хd V C Here, A is area of shear reinforcement with in a distance s and the value is 0.11in.2 which is obtained from U.S. rebar size chart No. 3 stirrups and f yt is specified yield tensile strength of non pre-stressed reinforcement 0.22 x 40 x 17.5 38 29.69 0.75 18.4in Hence, s, exceeds maximum space of 8.75 in. Therefore, the value of column face is 8.75in.. V. и. 29.69 14.8 kips 2 Terminate stirrups when the shear force is equal to C 2 Calculate the distance x where nominal shear strength occurs V. Consider " value of 46.5 and mid-span value of 5.1 and 2 C

46.5-14.8 x(10ft x 12in. ft) 46.5 5.1 =91.7in Therefore, the nominal shear stress of 14.8 kips occurs at a distance of 91.7 in. at a stirrup spacing of 8in Therefore, provide No.3, Grade 40 stirrups, starting at face of column from each end, use 1@4 in., 11@8 in.