Design the simply supported beam that carries two concentrated factored loads of 214 kips each on a clear span of 12 ft as shown in Figure 1. The beam has a width of 14 in. and a 48 in. overall depth. The width of the bearing plate at each concentrated load location is 16 in. Neglect the selfweight.
Use and
Figure 1 (Click here to view a larger image)
Check Bearing Stress at Points of Loading and Supports:
The area of bearing plate is The bearing stresses at points of loading and at supports are
Since these bearing stresses are less than their corresponding limits, i.e. at points of loading and at supports, the area of bearing plates provided is adequate.
Select and Establish the StrutandTie Model:
Assume that the loads are carried by a strutandtie model consisting of two trusses. One truss uses a direct strut running from the load to the support. The other truss uses stirrups as vertical ties. The geometry of the assumed truss is shown in Figure 2.
The center of the horizontal ties is assumed to be located 4 in. from the bottom of the beam. Thus, d = 48 – 4 = 44 in. The centroid of the vertical tie (stirrups) BC is assumed to be at the middle of the shear span, i.e. 56/2 = 28 in. from point A. The location of strut DD’ centerline can be found by determining the strut stress limit, calculating the required compressive force in. strut DD’, N_{DD’}, and imposing moment equilibrium about point A to obtain the strut width a as follows:
Figure 2 (Click here to view a larger image)
Thus, node strut DD’ is located 9.12/2 = 4.56 in. from the top of the beam and N_{DD’} = 33.32(9.12) = 304 kips. This fixes the geometry of the truss.
Determine the Required Truss Forces:
Since the truss shown in Figure 2 is statically indeterminate, it is necessary first to select the amount and position of the vertical tie BC (stirrups) and assume that the stirrups yield. The truss then becomes statically determinate and all the member forces can be found easily by statics.
Assume that 50 % of the loads, i.e. 214/2 = 107 kips, is transmitted by the stirrups at yield and the other 50 % of the loads is transmitted by the direct strut.
The required forces in all the members of the truss are given in the following table. Note that positive indicates tension, negative compression.
Member 
AB 
AC 
AD 
CC' 
CD 
BC 
BD 
DD' 
Force (kips) 
131 
+228 
186 
+304 
131 
+107 
76.0 
304 
Slope (deg) 
54.6 
0 
35.2 
0 
54.6 
90 
0 
0 

Select the Steel Reinforcement for the Ties:
Try to use 5 #4 twolegged stirrups at 6 in. o.c. for the vertical tie BC. This corresponds to a capacity of
and is very close to the assumed load. Hence provide 5 #4 twolegged stirrups at 6 in., According the AASHTO LRFD, the minimum reinforcement for horizontal tie CC’ and AC is
The required area of steel reinforcement for tie CC’ is and the required area of reinforcement for tie AC is Thus, choose 2 layers of 4 #8 bars for tie CC’, and choose 2 layers of 3 #8 bars for tie AC
Check the Struts:
The struts will be checked by computing the strut widths and checked whether they will fit in the space available.
By neglecting the tensioning effects, the average tensile strain in tie BC can be estimated as Similarly, the average tensile strain in tie AC can be taken as
The bottom part of strut AB is crossed by tie AC. The tensile strain perpendicular to strut AB due to tensile strain in this tie AC is = = 0.00350. Thus, the stress limit at the bottom of strut AB is The top part of strut AB is crossed by tie BC and the tensile strain perpendicular to strut AB due to tie BC is Thus, the stress limit at the top of strut AB becomes By taking the smaller stress limit, the required width for strut AB is Choose a width of 8 in. for strut AB width.
The bottom of strut AD is crossed by tie AC and the tensile strain perpendicular to strut AD due to tensile strain in tie AC is Thus, the stress limit at the bottom of strut AD is The middle part of strut AD is crossed by tie BC and the tensile strain perpendicular to strut AD due to tie BC is Thus, the stress limit at the middle of strut AD is The required width for strut AD is Choose a width of 11 in. for strut AD width.
The bottom part of strut CD is mostly influenced by tie BC and can be assumed to be the same as the top part of strut AB. Thus, the stress limit and the required width for strut CD are 1176 psi and 7.96 in. respectively. Choose also a width of 8 in. for strut CD.
Strut BD is mostly crossed by tie BC and the tensile strain perpendicular to strut BD due to tensile strain in tie BC is Thus, the stress limit for strut BD is Take The required width for strut BD is Choose a width of 3 in. for strut BD width.
The required width for short strut transmitting the applied load to node D is Choose a width equal to the bearing plate width for this strut, i.e. 16 in. The required width for short strut transmitting the force meeting at node A to the support is Choose a width of 13 in. for this short strut.
The stress demands, stress limits, and the widths of the struts are summarized in Figure 3. As shown in Figure 3, most of the strut widths fit into the outline of the beam region except that struts AB and AD near node A overlap and struts BD, AD, and CD near node D overlap. To ensure that the overlapping struts in those regions do not exceed the stress limit, the stresses due to the force resultants are checked against the corresponding stress limit. The force resultant of struts AB and AD is with a slope of and the available width is (Figure 4(a)). The stress due to this force resultant is then This force resultant zone crosses both ties AC and BC. The tensile strain perpendicular to this force resultant due to tensile strain in tie AC is while the tensile strain perpendicular to the force resultant due to tie BC is By taking the larger tensile strain, it gives the lower stress limit of which is greater than the stress demand, i.e. 1518 psi.
Similarly, the force resultant of struts BD, AD, and CD is with a slope of and the available width is (Figure 4(b)). The stress due to this force resultant is Part of force resultant zone crosses tie BC. The tensile strain perpendicular to this force resultant due to tensile strain in tie BC is This gives a stress limit of which is greater than the stress demand, i.e. 1594 psi.
Figure 3 (Click here to view a larger image)
Figure 4 (Click here to view a larger image)
Design the Nodal Zones and Check the Anchorages:
The width a of nodal zone D was chosen to satisfy the stress limit on that nodal zone. The stresses of the nodal zone A and C are limited to and respectively. To satisfy the stress limit of nodal zone C, the tie reinforcement must engage an effective depth of concrete at least equal to and to satisfy the stress limit of nodal zone A, the tie reinforcement must engage an effective depth of concrete at least equal to: These limits are easily satisfied since the nodal zone available is 8 in. The required anchorage length for tie AC is Since this is less than the available length, i.e. 16 – 2.5 = 13.5 in, then anchorage length is adequate.
Calculate the Minimum Reinforcement Required for Crack Control:
According AASHTO LRFD, a uniformly distributed reinforcement in vertical and horizontal directions near each face must be provided with minimum of volumetric ratio of 0.003 in each direction and the minimum bar spacing for each direction is 12 in. Try pairs of #4 bars with spacing of 9 in. for both vertical and horizontal reinforcement. The reinforcement ratio is The ratio is greater than 0.003. Hence use pairs of #4 bars @ 9 in.
o.c. in each direction.
Summary of the Design:
The reinforcement details for the deep beam designed using the strutandtie model according to AASHTO LRFD are shown in Figure 5.
Figure 5 (Click here to view a larger image)
