Steel Structures - Plastic Theory

1. A simply supported beam of span “L” supports a concentrated load “W” at its midspan. If the cross section of the beam is an I section, then the length of elastic-plastic zone of the plastic hinge will be
(a) L/8
(b) L/4
(c) L/2
(d) 3L/4 (IES 1995)

2. A simply supported beam of uniform cross section has span L and is loaded by a point load P at its mid span. The length of elasto-plastic zone of the plastic hinge will be
(a) L/3
(b) 2L/5
(c) L/2
(d) 3L/4 (IES 1997)

3. For propped cantilever subjected to a point load at midspan, the value of collapse load is given by
(a) 8Mp/l
(b) 12Mp/l
(c) 6Mp/l
(d) 4Mp/l  (IES 1993, 99, 2001, GATE 1998)

4. The ratio of collapse load of a propped cantilever of span I carrying a udl throughout the span to that of a simply supported beam carrying the same loads is
(a) 1.457
(b) 1.500
(c) 2.000
(d) 3.000 (IES 1998)

5. A continuous beam of constant Mp has three equal spans and carries total uniformly distributed load “w” on each span. The value of collapse load for the beam will be
(a) 12Mp/L
(b) 8.65Mp/L
(c) 11.654Mp/L
(d) 4Mp/L (IES 1995)

6. The given figure shows a continuous beam loaded with concentrated loads W at the centre of each span. The value of W at collapse will be
(a) 3.2Mp/L
(b) 4Mp/L
(c) 5.6Mp/L
(d) 6.4Mp/L (IES 1997)

7. For the beam shown in figure, the collapse load P is given by
(a) 16Mp/L
(b) 14Mp/L
(c) 12Mp/L
(d) 10Mp/L

8. A fixed beam made of steel is shown in the given figure. At collapse, the value of load P will be equal to
(a) 10Mp/L
(b) 12Mp/L
(c) 16Mp/L
(d) 20Mp/L (IES 1997)

9. A fixed beam is shown in the given figure. The plastic failure for this load for this beam is
(a) 10Mp/L
(b) 12.5Mp/L
(c) 15Mp/L
(d) 16.5Mp/L (IES 1998)

10. For a fixed beam with span L, having plastic moment capacity of Mp, the ultimate central concentrated load will be
(a) 4Mp/L
(b) Mp/8L
(c) 6Mp/L
(d) 8Mp/L (GATE 1999)


    1. (a)    2. (a)    3. (c)    4. (a)    5. (c)
    6. (c)    7. (b)    8. (a)    9. (c)    10. (d)


5. The degree of indeterminacy r = 4 - 2 = 2
Number of plastic hinges required for collapse = r + 1 = 3
In the span AB maximum moment occurs at support B and at 0.414 L away from support.
Hence Collapse load = \frac{{11.656{M_p}}}{{{L^2}}}

6. Virtual work
W\left( {\frac{{L\theta }}{2}} \right) = {M_p}(2\theta ) + 0.8{M_p}\theta \\
\therefore \,W = \frac{{5.6{M_p}}}{L}
\frac{{WL\theta }}{2} = 0.8{M_p}(\theta  + 2\theta  + \theta )\\
\therefore \,W = \frac{{6.4{M_p}}}{L}
Lower value will be taken.


{M_p}\theta  + {M_p}(2\theta ) + 2p(\theta ) = \frac{{PL\theta }}{2}

10.  {M_p} + {M_p} = \frac{{{P_u}L}}{4}


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