Q21GATE 2007MCQ

The plastic collapse load $W_{p}$ for the propped cantilever supporting two point loads as shown in fig. in terms of plastic moment capacity, $M_{p}$ is given by

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📖 Explanation

$D_{s}=\left ( 3+1 \right )-3 =1$ No. of plastic hinges formed $=D_{s}+1=2$ $1^{st}$ case: Plastic hinge formed at A and B

\begin{aligned} \frac{L}{3}\theta&=\frac{2L}{3}\alpha \\ \theta&=2\alpha \\ \text{By virtual work method}&\\ \text{External work}&=\text{Internal work}\\ W\left ( \frac{L}{3}\theta \right )+W\left ( \frac{L}{3}\alpha \right ) &=M_{P}\theta +M_{P}\left ( \theta +\alpha \right )\\ \text{Put } \; \theta &=2\alpha \\ W_{u}&=\frac{5M_{P}}{L} \end{aligned}

$2^{nd}$ case: Plastic hinge at (A) and (C)

\begin{aligned} \frac{2L}{3}\theta &=\frac{L}{3}\alpha \\ \alpha &=2\theta\\ \text{External work} &=\text{Internal work} \\ \frac{WL}{3}\theta +\frac{2WL}{3}\theta &=M_{P}\theta+M_{P}\left ( \theta +\alpha \right ) \\ \text{Put }\; \alpha &=2\theta \\ W_{u} &=\frac{4M_{P}}{L} \end{aligned}

Hence, collapse load $\; \; \; W_{u} =\frac{4M_{P}}{L}$

Q22GATE 2006MCQ

When the triangular section of a beam as shown below becomes a plastic hinge, the compressive force acting on the section (with $\sigma _{y}$ denoting the yield stress) becomes

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📖 Explanation

We kno that the nuetral axis of the plastified section is the equal area axis. $\therefore$ Compressive force,

\begin{aligned} &=\sigma _{y}\times \frac{\text{Area of triangular section}}{2}\\ &= \sigma _{y}\times \frac{bh}{2\times 2} = \frac{bh\sigma _{y}}{4} \end{aligned}

Q23GATE 2005MCQ

A cantilever beam of length 1, width b and depth d is loaded with a concentrated vertical load at the tip. If yielding starts at a load P, the collapse load shall be

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📖 Explanation

Yielding starts at P, therefore yielding moment at the fixed end is given by, $M_{y}=PL$ Now, if $W_{u}$ is the collapse load, then collapse moment at the fixed end is given by, $M_{p}=W_{u}L$ But, for rectangular beam,

\begin{aligned} \frac{M_{p}}{M_{y}}& =1.5 \\ \Rightarrow \; \; \frac{W_{u}L}{PL}&=1.5 \\ \Rightarrow \;\;W_{u}&=1.5P \end{aligned}

Q24GATE 2004MCQ

A propped cantilever of span L is carrying a vertical concentrated load acting at midspan. The plastic moment of the section is $M_{p}$ . The magnitude of the collapse load is

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📖 Explanation

\begin{aligned} &\text{External work done} \\ &=\text{load} \times \text{deflection} \\ &=W_{u}\times \Delta =W_{u}\times \frac{L}{2}\times \theta\\ &=\frac{W_{u}L\theta }{2} \end{aligned}

\begin{aligned} &\text{Internal work done}\\ & =\text{moment} \times \text{rotation} \\ &=M_{P}\theta +M_{P}\left ( \theta +\theta \right )\\ &=3M_{P}\theta \end{aligned}

By principle of virtual work, External work done = Internal work done $\; \frac{W_{u}L\theta }{2}=3M_{P}\theta$ $\; W_{u}= \frac{6M_{P}}{L}$

Q25GATE 2003MCQ

A steel portal frame has dimensions, plastic moment capacities and applied loads as shown in the figure. The vertical load is always twice of the horizontal load. The collapse load P required for the development of a beam mechanism is

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📖 Explanation

The possible locations of the plastic hinges are at A, B, C, D and E. Number of plastic hinges, $N= 5$ Degree of redundancy, $r=6-3=3$ Number of possible independent mechanisms, $n=N-r=5-3=2$ Thus the two independent mechanisms are 1. Beam mechanisms 2. Sway mechanism Beam mechanism: The ends B and D of the frame ABCD are fixed joints. Therefore, the beam BD acts as a fixed end beam.

\begin{aligned} &\text{External work done}\\ &=\text{load} \times \text{deflection} \\ &=2P\times \Delta =2P\times L\times \theta \\ &=2PL\theta\\ & \text{Internal work done} \\ &=\text{moment} \times \text{rotation} \\ &=M_{P}\theta +2M_{P}\left ( \theta +\theta \right )+M_{P}\theta \\ &=6M_{P}\theta \end{aligned}

By principal of virtual work, External work done = Internal work done $\; \; 2PL\theta =6M_{P}\theta$ $\Rightarrow \; P=\, \frac{3M_{P}}{L}$

Q26GATE 2002MCQ

A steel beam (with a constant El , and span L) is fixed at both ends and carries a uniformly distributed load (w kN/m), which is gradually increased till the beam reaches the stage of plastic collapse (refer to the following figure). Assuming 'B' to be at mid-span, which of he following is true,

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📖 Explanation

Initially when load is increased maximum -ve BM occurs at end $\frac{Wl^{2}}{12}$ and plastic hinges are formed at ends and when load is further increased it behaves like simply supported bean and third hinge is formed at mid span when load increases to $W_{u}=\frac{16M_{p}}{l^{2}}$ 1st two hinges are formed at (A) and (c) togather at load, $W_{u}=\frac{12M_{p}}{l^{2}}$ and then third hinge is formed at load, $W_{u}=\frac{16M_{p}}{l^{2}}$