Question

In a 4-bar linkage, if the lengths of shortest, longest and other two links are denoted by s, l, p and q, then it would result in Grashof's linkage provided

• A. l+p $<$ s+q
• B. l+s $<$ p+q
• C. l+p = s+q
• D. l+s = p+q

Hint:

A simple way to remember Grashof's Law is "Shortest + Longest \(\le\) Sum of the Others". This single rule is key to analyzing the motion of any four-bar linkage. If the condition is not met (\(s+l > p+q\)), it's a non-Grashof linkage, and no link can make a full revolution.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question asks for the condition that defines a Grashof's linkage. Grashof's law (or Grashof's criterion) is a fundamental principle in the theory of machines that predicts the rotational capabilities of a four-bar linkage based on the lengths of its links. It determines whether at least one link can make a full 360-degree revolution relative to another link.

Step 2: Key Formula or Approach:
Grashof's Law states that for a planar four-bar linkage, the sum of the lengths of the shortest and longest links must be less than or equal to the sum of the lengths of the other two links for there to be continuous relative rotation between at least two links.
Let the lengths of the four links be:

• s = length of the shortest link
• l = length of the longest link
• p = length of one of the other links
• q = length of the other link According to Grashof's Law, the condition for a Grashof linkage is:
\[ s + l \le p + q \]

Step 3: Detailed Explanation:
Let's analyze the given options based on Grashof's Law.
(A) l+p $<$ s+q: This is not the correct form of the law. It compares the longest link with another link against the shortest link with the fourth link.
(B) l+s $<$ p+q: This is the strict inequality form of Grashof's Law. If \(s+l < p+q\), the linkage is a Class I Grashof linkage, and depending on which link is fixed, it can be a crank-rocker, double-crank (drag-link), or double-rocker mechanism. The question asks for the condition for a "Grashof's linkage", and this inequality is the primary condition.
(C) l+p = s+q: This is not the correct form.
(D) l+s = p+q: This is a special case of Grashof's Law known as a "change-point" or "Class II Grashof" or "neutral" linkage. At this point, the linkage can exhibit uncertain behavior and may lock or jam. While technically satisfying \(s+l \le p+q\), the strict inequality \(s+l < p+q\) is the more general condition for predictable continuous rotation. Among the given options, (B) is the most standard and widely accepted criterion.

Step 4: Final Answer:
The condition for a Grashof's linkage is that the sum of the longest (l) and shortest (s) link lengths must be less than (or equal to) the sum of the other two link lengths (p and q). The option l+s $<$ p+q represents this condition.