Test: Plastic Analysis - 2 - Civil Engineering (CE) MCQ

Number of plastic hinges required for complete collapse - r + 1
= (3 - 2) + 1 = 2
The plastic hinges will be formed under the load and at the fixed end of the beam respectively.

But a = L and b = L/4

The collapse load for a prismatic beam fixed at both ends and caries UDL throughout the span is given by,

where M_p is its plastic moment.

Substituting value of S from (ii) in (iii), we get,

Substituting value of M_p from (i), we get,

Total area, A = 400 x 100 + 100 x 500
= 90000 mm²

Thus the plastic neutral axis lies outside the flange because the flange area is only 40000 mm².
Let the plastic neutral axis lies y mm below the junction of flange and web.
∴ y x 100 = 45000 - 40000
⇒ y = 50 mm
∴ Plastic neutral axis as measured from top of flange = 100 + y
= 100 + 50 = 150 mm

The static method represents the lower limit to the true ultimate load and has a maximum factor of safety.

Because maximum bending moment develops at end A and between mid span and end B.

If the number of plastic hinges in the collapse mechanism are less than (r + 1) the collapse is called partial collapse. In such a case, part of the structure may fail making it useless as a whole. If the number of plastic hinges in the collapse mechanism are (r + 1) the collapse is called complete collapse. Such a mechanism has only one degree of freedom. If the number of plastic hinges developed are more than (r + 1), the collapse is called over complete collapse. In such a case there are two or more mechanisms for which the corresponding value of the load is the same, this load value being the actual coljapse load.
If the degree of indeterminacy is r, and the number of plastic hinges developed is N then,
N < (r + 1) Partial collapse
N = r + 1 Complete collapse .
N > r + 1 Overcomplete collapse

For collapse in IJ

For collapse in HI

⇒ P = 8M_P/L
∴ Minimum load for collapse = 6M_P/L

With increase in allowable stress, the factor of safety will reduce to 1.5/1.2 = 1.25.
∴ Load Factor = FS x shape factor
= 1.25 x 1.12
= 1.40