Which of the following equations is correct for both, subjected to both combined shear and tension?

V = Applied shear at service load

V_{sdf }= Design shear strength

T_{e} = Externally applied tension at service load

T_{ndf} = Design tension strength

This question was previously asked in

MPSC AE CE Mains 2018 Official (Paper 1)

Option 1 : \({\left( {\frac{V}{{{V_{sdf}}}}} \right)^2} + {\left( {\frac{{{T_e}}}{{{T_{ndf}}}}} \right)^2} \le 1\)

ST 1: Building Construction and Materials

1095

20 Questions
40 Marks
24 Mins

**Explanation:**

Clause No. 10.3.6 of IS 800: 2007 states that a bolt required to resist both design shear force and design tensile force at the same time shall satisfy:

\({\left( {\frac{V}{{{V_{sdf}}}}} \right)^2} + {\left( {\frac{{{T_e}}}{{{T_{ndf}}}}} \right)^2} \le 1\)

V = Applied shear at service load

Vsdf = Design shear strength

Te = Externally applied tension at service load

Tndf = Design tension strength

__Important Points__

Type of Stress |
Formula |

Shear Strength |
V V f γ |

Bearing Strength |
V V k smaller of (e/3d f e = end distance p = pitch distance d = nominal diameter of the bolt d t = thickness of the connected plates experiencing bearing stress f f γ |

Tensile Strength |
T T f f A A γ γ |