Determine max S.F and B.M at quarter span(1/4) from left end when UDL longer than span of intensity 20KN/m accompanied a 100KN concentrated load crosses the span of 12m. Use IDL.

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written 6.8 years ago by

teamques10 ★ 68k

• modified 2.8 years ago

The concentrated load is free to occupy any position.

Subject : Structural Analysis 1

Topic : Influence line diagram.

Difficulty : Low

structural analysis

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written 6.8 years ago by

teamques10 ★ 68k

• modified 6.6 years ago

Quarter Span $\frac{1}{4}\times 12 = 3 m$ from left end

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$1.Max \hspace{1mm} +ve \hspace{1mm} S.F_c:$

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$SF_c=100 \times \frac{9}{12}+12 \times [\frac{1}{2} \times 9 \times \frac{9}{12}]$

$SF_c=115.5KN$

$2.Max \hspace{1mm} -ve \hspace{1mm} S.F_c:$

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$SF_c=-[100 \times \frac{3}{2}+20 \times [\frac{1}{2} \times 3 \times \frac{3}{12}]]$

$SF_c:- 157.5KN$

$3. Max \hspace{1mm} BM_c:$

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$\frac{ac}{l}=\frac{3 \times 9}{12}=2.25$

$Max \hspace{1mm} BM_c:=100 \times 2.25+20 \times \frac{1}{2} \times 3 \times 2.25+20 \times \frac{1}{2} \times 9 \times 2.25$

$Max \hspace{1mm} BM_c:=495KN$

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