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          p 44
        
        
          
            Chapter 2
          
        
        
          Helical Foundation Systems
        
        
          
            CHAPTER 2
          
        
        
          HELICAL FOUNDATION SYSTEMS
        
        
          
            2.11.2 Helical Tiebacks
          
        
        
          Example 3
        
        
          Helical tiebacks are being considered to stabilize
        
        
          an existing reinforced concrete retaining wall.
        
        
          The tiebacks can extend no further than 20 feet
        
        
          from the front face of the wall due to property
        
        
          line issues. A geotechnical investigation found
        
        
          the retained soils to consist of silty sand. The
        
        
          design engineer proposed an HA150 shaft
        
        
          (1.5-inch solid square) with a 10/12/14 helix
        
        
          plate configuration. The soil parameters and
        
        
          preliminary tieback design are shown on
        
        
          Figure
        
        
          2.11.2.a
        
        
          . The engineer must determine the
        
        
          allowable tieback capacity so tieback spacing
        
        
          can be established.
        
        
          
            Q
          
        
        
          
            u
          
        
        
          
            = ∑A
          
        
        
          
            h
          
        
        
          
            (q’N
          
        
        
          
            q
          
        
        
          
            )
          
        
        
          
            A
          
        
        
          
            14”
          
        
        
          = 1.05 ft
        
        
          2
        
        
          
            A
          
        
        
          
            12”
          
        
        
          = 0.77 ft
        
        
          2
        
        
          
            A
          
        
        
          
            10”
          
        
        
          = 0.53 ft
        
        
          2
        
        
          
            q’
          
        
        
          
            14”
          
        
        
          = (120)(7.17) = 860 lb/ft
        
        
          2
        
        
          
            q’
          
        
        
          
            12”
          
        
        
          = (120)(7.67) = 920 lb/ft
        
        
          2
        
        
          
            q’
          
        
        
          
            10”
          
        
        
          = (120)(8.08) = 969 lb/ft
        
        
          2
        
        
          
            N
          
        
        
          
            q
          
        
        
          = 1+0.56(12
        
        
          Φ
        
        
          )
        
        
          Φ
        
        
          /54
        
        
          = 15.7
        
        
          
            Q
          
        
        
          
            u
          
        
        
          = (1.05)(860)(15.7) + (0.77)(920)(15.7) +
        
        
          (0.53)(969)(15.7) = 33,300 lb = 33.3
        
        
          kips
        
        
          
            Q
          
        
        
          
            a
          
        
        
          = 33,300 / 2 (FOS) = 16,650 lb = 16.6 kips
        
        
          The horizontal and vertical components of the
        
        
          tieback force can be calculated in accordance
        
        
          with Section 2.8.1.
        
        
          Figure 2.11.2.a
        
        
          Example 3. Helical Tieback Capacity
        
        
          Retained Soils:
        
        
          Silty Sand
        
        
          
            c= 0
          
        
        
          Φ = 30°
        
        
          γ
        
        
          moist
        
        
          = 120 Ib/ft
        
        
          3