---
title: "Design strength and design load in LSM"
subject: "Design of RCC Structure"
chapter: 3
key: "semester_7/design_of_rcc_structure:design-strength-and-design-load-in"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/design-strength-and-design-load-in
site: BCT Engineering
---

# Design strength and design load in LSM

Design of RCC Structure · Chapter 3: Limit State Method of design

- Asked in: 2081 Baisakh, 4 marks
- Video: [Characteristics Strength, Design Strength, Characteristics Load and Design Load| Lecture 16](https://www.youtube.com/watch?v=8Z7jbtTBgH8) — Tech Master Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/design-strength-and-design-load-in
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-3

## Question

Define design strength and design load for limit state design method.

## Short answer

Design strength = characteristic strength / partial safety factor (γm=1.5 for concrete giving 0.446fck, γm=1.15 for steel giving 0.87fy). Design load = characteristic load × partial safety factor (γf=1.5 for DL+LL, 1.2 for DL+LL+WL/EL, 0.9DL for favorable). Member safe when design resistance ≥ design load effect.

## Answer

#### Design strength

Design strength of a material is its characteristic strength divided by the partial safety factor for the material:

$$
f_d=\frac{f_k}{\gamma_m}
$$

For concrete $\gamma_m=1.5$ (so $f_{cd}=f_{ck}/1.5$, and in the stress block $0.446f_{ck}$), and for steel $\gamma_m=1.15$ (so $f_{yd}=f_y/1.15=0.87f_y$). The factor allows for variation in material strength, workmanship and member behaviour.

#### Design load

Design load is the characteristic load multiplied by the partial safety factor for load:

$$
F_d=\gamma_f\,F_k
$$

IS 456:2000, Table 18, gives $\gamma_f$: 1.5 for DL+LL, 1.2 for DL+LL+WL/EL, 1.5 for DL±WL/EL, and 0.9 for DL when it is favourable (0.9DL+1.5WL). The factor covers possible overloading and inaccuracies in analysis.

The member is safe when the design resistance (from design strengths) is at least the effect of the design loads: $R_d\ge S_d$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is design strength defined?
   A. Characteristic strength multiplied by gamma_m
   B. Characteristic strength divided by the material partial safety factor
   C. Mean strength
   D. Characteristic load divided by gamma_f
   Answer: B. f_d = f_k / gamma_m, with gamma_m = 1.5 for concrete and 1.15 for steel. (Key point: Design strength)

2. How is design load obtained?
   A. F_d = F_k / gamma_f
   B. F_d = gamma_f x F_k
   C. F_d = F_k + gamma_f
   D. F_d = F_k x gamma_m
   Answer: B. Characteristic loads are multiplied by the load factor gamma_f, for example 1.5 for DL+LL. (Key point: Design load)

3. What factor applies to dead load when it is favourable (combined with wind)?
   A. 1.5
   B. 1.2
   C. 0.9
   D. 1.15
   Answer: C. In 0.9DL + 1.5WL the dead load helps resist overturning, so it is reduced to 0.9. (Key point: Favourable dead load)

4. When is a member safe in limit state design?
   A. Design resistance is at least the effect of the design loads
   B. Service load equals strength
   C. Mean strength exceeds characteristic load
   D. Deflection is zero
   Answer: A. The requirement is R_d >= S_d. (Key point: Safety condition)

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