Example: confidence

THE STRUT-AND-TIE MODEL

reinforcement may cause cracking parallel to the strut with normal crack width: f cu = 0.8 (0.85 f c?) = 0.68 f c? As above for skew cracking or skew reinforcement: f cu = 0.6 (0.85 f c?) = 0.51 f c? For skew cracks with extraordinary crack width – such cracks must be expected if modeling of the struts departs

Tags:

  Trust, Reinforcement, Skew, Strut and tie, Skew reinforcement

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