6.6. Design the vertical stirrups for a beam having the shear diagram shown in Figure 6.37 assumin $b_{w} = 14 in . (356 mm) d_{w} = 20 in . (508 mm)$ $V_{u 1} V_{u 2} V_{u 3} f_{c} f_{y} = 75, 000 lb (334 kN) = 60, 000 lb (267 kN) = 45, 000 lb (200 kN) = 4000 psi (27.6 MPa), normal-weight concrete = 60, 000 psi (414 MPa)$

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6.6. Design the vertical stirrups for a beam having the shear diagram shown in Figure 6.37 assumin

b_{w} = 14 in . (356 mm) d_{w} = 20 in . (508 mm)

V_{u 1} V_{u 2} V_{u 3} f_{c} f_{y} = 75, 000 lb (334 kN) = 60, 000 lb (267 kN) = 45, 000 lb (200 kN) = 4000 psi (27.6 MPa), normal-weight concrete = 60, 000 psi (414 MPa)

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